do

w
nl
oa
de
d
fro
m
te
ch
oe
du
.c
o m
MATHEMATICS - XII (Volume 2)

19. DEFINITE INTEGRALS

20. AREAS OF BOUNDED REGIONS 20.1-

21. DIFFERENTIALEQUATIONS 21 .1-

om
.c
22. ALGEBRA OF VECTORS 22.1-

du
oe
23. SCALAR OR DOT PRODUCT ch 23.1-

24. VECTOR OR CROSS PRODUCT

te
m
fro
d

26. DIRECTION COSINES AND DIRECTIG>N RATIOS

de
oa

27. STRAIGHT LINE IN SPACE

nl
w

28. THE PLANE

do

29. LINEAR PROGRAMMING

30. PROBABILITY

31. MEAN AND VARIANCE ()JI A RANl)(lM VARIABLE

APPENDIX
 ,___.__2
DEFINITE INTEGRALS
~ - -----~
FUNDAMFNTAL THE=ORrM or INTrr.RAI CALCUl us
EMENT Let t1> ( x) be the primitive or antiderivative of a continuous function f (x) defined on [a, b]
b
x fct,(x)\ =/(x). Then the de.finite integral of f(x) over[a, b] is denoted by J f(x) dx and is equalto

m
/l

o
-tj>(a)].

.c
b

du
...(i)
Jf(x) dx = q, (b) - tj> {a)
a

oe
num.bers a and b are called the limits o f integration, 'a' is called the lower limit and 'b' the
r limit. The interval [a, b] is called the interval of integration.
ch
e use the notation ( q> ( x)] : to denote ct, (b) - q, (a). Then,
te
m

b b
Jf(x) dx = [ 4> (x)] (From {i)]
fro

a a
b
f f(x)dx = (va1ueof q, (x) at x=a)
d

x= b) - ( value of tj>(x)at
de

/l

Jb f(x) dx
oa

= (Value of antid erivativ e at x = b) - (Value of antiderivative at x =a)

nl

a
~RK , In the above statement it does not matter which anti-derivative is used to evalu,1t,' th<'
w

J
ite integral, because if f(x) dx = ct,(x) + C, then
do

j f(x) dx = [ 4> (x) +CJ: ={4>(b) + C} -{ cp(a) + C} = cl> (b) - $(tr)

/l
f,.er words, to evaluate the definite integral there is no need to keep Iii<' r:lll1st,111t cf mtt'~"Tation.

RK 2 Jb J(x) dx is read as "the integral of f(x.)from n Ioli" or i11te~ml of/(,) oi•rr 111, b].
a

EVALUATION OF DEFINITE INTEGRALS

b
J
aluate the definite integral /(x) dx of a con tim10us function /(x) defined on[a, b], we m ay
a
he following algorithm.
 19.2

ALGORITHM
~TEP I Find f
the indefinite integral f(x) dx. Let this be~ (x). There is no need lo keep the co

integration.
$TEP JI Evaluate <I> (b) and cji (a).
sn:r m Calculate 4>(!1) - <J>(a).
. b
The number o/itained i11 Step Ill is the value of the definite inteKral f f(x) dx.
II

ILLUSTRATIVE EXAMPLES

LEVEL-1 I
E.XA!\1PlE. Ei1aluate:

m
-1
2
fx 2 (ii) f }_ dx

o
(i) dx
-4x

.c
1

du
. 1 1
(iii)
I
f'1I+x+./x
l dx (1v) f- - dx
2x - 3

oe
0 0

i: l:
SOLUTION We have,
ch
1x' X: 2
!_~ ~
te

(i) dx a [ a : - a a
m
fro

(ii)
d
de

(ill)
oa
nl
w

=[! (1 + x)3/2 _ ~ x3/2 J:

do

=[2(l+l)3/2
_3
_ 2(l)3/2J
3 .
-
~ (L I 0) J/ :! -,l
-1~ (()) ' -

!(23/2 1) 2 (t 0) ~ (? J" ") I l\.: 1)

" J

(iv)
1
l
J 2x- 3 dx • ½
0
l o~ (2x a)l1
l)

1
=½l log I LI log! Jll ~ ( log 1 log 3 j ~ (0 )l)g 3) ~ -2
1
EXAMPLl!2 if J( 3x2 +2x+k) dx = 0,findk.
0
 19.3
EINTEGRALS

N Wehave,

!
I
(3:r2 + 2x + k) d:r •O ⇒ [.t3 + x1 + h 1: = ⇒
o (l + 1 r k) - 0 =0 => k = - 2

II
U l I( J (3x 2 + 2:r + 1) dx .. 11,,tind rt;2/ mluts ofa.
l
~ \\'ehave,
a
J (3.x2 .... 2x .. 1)dx=ll
I

r. .

m
x3 + %2 +:rr "'11
J1

o
.c
( a:1 + a2 ... a) -(1 + 1 ... l) = 11

du
a 3 -+ a2 + a - 3 = 11
+ a 2 + a -14 = 0

oe
,1 3

(a-2)(a2 T 3,1T7)=0=a=2
ch [-: i + 3a +7 c1, 0 for anv a e R 1
t, ,,
te

If J :r 3 dx = 0 and if J :r2 dx = ~ , find a and b.

m

n ll

m01': \\'c ha\e,

fro

b
j 3
.x dx=O
d

a
de
oa
nl
w

i,4 -a4 =0 =
do

. .,
t, ')

j x~ dx =
3
II

l~l ·~
1 (b3 _ 0 3) = 2
3 3 (·: b =- a\
3 3
!, 3 -a3 : 2= (-a) -a s 2

-2a3 =2 => a 3 : -1 => a == -1

b = -a ⇒ b = 1
19.4
rt/2
EXAMPLE 5 Evaluate: f \ft - cos zx dx.
0
rr./2
SOLUTION Let I f
= .Jl - cos 2x dx. Then,
0
n/2 ~ - -
I = J✓2 sin 2 x dx
0

,
I = "2 :; f
n/2
"'f
,c/2
I sin xi dx = ✓ 2 sin x d.x
[
·: sinx~OforO ".,,
0 0

cosxJ:12 =-/2 [(-cos;J-(-cos0)]=J2(0+l)=J2°

m
⇒

o
l=fi [-

.c
du
E'\AMPLE 6 Evaluate:
rr/2 n/4
rr/4
J tan f sin 2 f sin 3x sjn 2x =

oe
2 (ii) x dx (iii)
(i) x dx
0 0 0
ch
n/4
SOLUTION (i) Let J tan 2 x dx. Then,
te

!
rr./4 2 n/ 4
:-iJ-(tan0-0) =1-; =
m

1= (sec x-l)dx = [tanx - x ] =(tan

0
fro

n/2 =
(ti) Let I= J sin2 x dx. Then,
d
de

0
1 tr./2
J
(1 - cos 2x) dx
oa

I= -
2 0
=
nl

1 1r
= 2 ~- 2sin2x
, 2
]1r.1 = 1 {(~- ~sin 1t) - (o - ½sin o)} 7t
w

O 2 4
do

n/4
(iii) Let 1 = J sin 3x sin 2x dx. Then,
0
I
n/4
1
I=
2 uJ (2 sin 3x sin 2x) dx
I
1 :t/4
I =
2 J0 (cos x cos5x) dx
,..

5l smO
· )} a I { -I- -1 ---I } 6 3
2 -[2 ( '2) S
'\/"
= - -- - = -
2 (5,fi.) 1
 EINTEGRALS
19.5
I E7 Evalunte:
. 3 x dx n/2 x/2
sm
(ii) f cos 3
X dx (iii) J sin 4 x dx
0 0
1t

(i) l ct I = Jsin 3 x dx. Then,

0

[": sin 3x = 3sin x - 4sin 3x]

= nf 3 sin x - sin 3x
I - dx
0 4
1 It
I= f (3 sin x - si.11 3x) dx

m
4
0

o
]n

.c
I = 1 [ - 3 cos x + 1 cos 3x
4 3

du
0

3cos 1 O+ ~ cos OJ} = i {( ~ ) - ( - 3 + ~)} = !

oe
I = : {( - 1t + cos 3n)-( 3 cos ch 3-

n/2
Let I = f cos 3 x dx. Then,
te

0
n/2
J ~s3x+3cosxdx
m

I= 3
(·: cos 3x = 4 cos x - 3 cos x]
fro

0 4
x/2
I = .!. f
d

(cos 3x + 3 cos x) dx
4 0
de

1 ft ]n/2
oa

I = - - sin 3x + 3 sin x
4 L3 0
.,
nl

= . 3n- + 3 sm
1 {( 1 sin
1
. n) - ( s ·m + 3 sin
· o o)} = =-
w

l 3
4 3 2 2 3
do

1'/2
) Let I = f sin 4 x dx. Then,
0
n/2 1 tr./2 2
I = .!. J (2 sin 2 x,2 dx =- J (1 - cos 2x) th
4 0 4 0
tr./2 I n/ 2 l I cos 4.Y
I= .!.4 J (1 - 2 cos 2x i
2
cos 2x) rlx .-
4
fI 2 cos 2x + --
2
ti.\
0 0

x/2 1 [ 4 1 ]n/2
r= ~ J (3 - 4 cos 2x + cos 4x) dx = 3x - sm 2x ~ sin 4x
80 8 2 4 o
3
~_,_=_½
_f{ f-2smn+¼sm2n}-{o -o+o}] 3
= ½[ 2n-o+o] = ~;
 19.6

EXA\fl'LE 8 Evaluate: n/4

(i)
;r /4
f J1 + sin 2x dx (ii) f ✓1 - sin 2x dx
0
0
n/4
SOLUTlOK (i) Let l - f Jf ~ sin 2.\ dx. Then,
0
nl~ -- - - - - - n/4 ~ - . 2
r= f \/sin 2 x + cos 2 x + 2 sin x cos x dx-:= f
,y(sin x + cosx) d:t
0 0
n/4 n/4 [ ]n/4
1= f lcosx+sinxldx= f (cosx +s.inx)dx= sinx -cosx

m
0 0 0

⇒ 1-1)- (0-1) =

o
r = (sin : - cos :J - (sin 0-coso) = ( 1

.c
du
11./4
(ii) Let I = f ✓~1---sm-·-2-x dx. Then,
oe
0
r--::-- - --=---- - -- n/4 , - - - -- - -
ch
n/4
I= J 2 2
, /sin x + cos x - 2 sin x cos x dx = ✓(cos x - s.in x)2 f
te
0 0
n/4
= J Icos x - sin xi dx
m

I=
fro

0
1t/4

[
I= J (cos x - sin x) dx ⇒·: 0 <X <n/4 :. cos x >sin x ⇒ cosx -sin :r
d

0 I cosx-sinxl =cosx-sin x
de

it/4
r
I= sin X +
oa

COS X
]O

1- lsm 4 +cos 4lt)- (s.inO+cos o)= ( ~-+ 2- )-(O+l) - 2

nl

::;,
- ( . 1t
w

-;2 .J2_ - ✓ - t=,2 - 1 2

do

n/2
EXA!¼PLE ~ Evaluat-e; J .Ji :=sin 2x dx
rr./4
n/2 •
SOLUTION Lr:t / J vft. bl ri 2x d.1 J lwn,
,r/ 4
n/2
r= I Jcn~2 X + ~JI i 2A •
n/4
1t/2
J = f ✓(CO!;,; - !!ill X)2 1/.\
n/4
 EINTEGRALS
19.7
1t/2
I = J I cos x - sin x I dx
n/4
rt/ 2
f = J - (cosx~sin x) dx ·: cosX <sinxfor ~< x <~:. cosx-sinx<Ol
n/4 [ ⇒ I cos X - sin x I = - (cos x - sin x)
n/ 2 n/ 2
1 = J (sin x - cos x) dx = [ - cos x - sin x
]
rc/ 4 rt/ 4

I = { -cos~ - sin~}-{- cos~-sm~} = (0- 1)-(-1) = Ji-1

o m
a n/ 2 a+ 1
J

.c
PLE! 10 If J ..fx dx = 2a sin 3 x dx, find the value of integral J x dx.

du
0 0 a

UTION We have,

1-Jx dx =3_ ( x3/ 2 ]a =3_a3/2

oe
... (i)
ch
o -3 3 0
rr/2
te

T= J sin 3 x dx. Then,

0
m

n/ 2 . . . n/2 1[ 1 ] n/ 2
I= J 3smx-sm 3x dx=¼ J (3 sinx-sin3x)dx= 4 3 cos3x
fro

- 3cosx +
0
4 0

:r1(- 3 cos 2 + 13 cos 231t) - (- 3+!1]=!:4. [o-(-3+.3!.)J=.!.[3

4 -.3!.]=~
3 ...
d

I = ~
n (ii)
3)
de

4
oa

n/ 2
= 2a J si.n 3 xdx
nl

0
w
do

[Using (i) and (ii)J

= 0, we obtain
OT

a+ l
1
Jn
1
xdx = J xdx = lx:
0
9 1
I = -12 - 0 ~
I
-
2

ce, J xdx = 2 or, 2

a
19.8

EXAMPLE ll Evaluate:
4 1 (ii) j l dx
J
(i) -;====dx
1J .Jx + 2x + 3
2
o [ax -x2
SOLUTION (i) Let T = J4 -:-====
1
dx. Then,
o .J.i+2x + 3
4 1
l= f r 2 dx
o ✓ x + 2x + 3

4 1
⇒ I = J .=====~dx
2 2

m
o .J<x + 1) + (-./2)

o
✓(x ~ 1)2 + (.,J2~]
- . -- - 4

.c
1 = llog (x + 1) +

du
0

✓x2 + 2x + 31]:
oe
I = [log Ix + 1 + ch
⇒ I = log (5 - ,/16 + 8 + 3) - log (1 + ./3)
te

I = log (5 + 3 ./3) - log (1 + ./3) = log (S1++./3

3
./3)
m

a
p
fro

(ii) Let J =J dx. Then,

O ax-x
d

a
J - r ~ ~ l ~ ~ = dx
de

I =
o -{ x2 _ax + a: _a:}
oa
nl

a
I = f 1 dx

~r;f-(x-;J
w
do

f It

I = sin-1 q_
2
0
EXAMPlF 12 Evaluate:

(i)
1/2
J
1/4
p x-x2
dx (ii)
I
J
2
~

I- ~ I
t/1
l
5
2

1 2
(""Ill") } 2x
2 dx (iv) J 5\;
.,
l
o 5x + 1
n ;1~ ~ •I
'"
DIIIINJTE INTEGRALS
19.9
1 2
9C)U.'l'ION (i) L~t I ., _f l d,. 1 hl'n
14\Xl.2

I
\+
4

o m
.c
1f2 [sin-I (X 1/2):11/2

du
= d, =
1· 4 1 )2 )2 ~
,l2 1 1/2

oe
-\x - 2 1/4

12 [.sm -JO - sm. -1(- l)] = [0 + sm.

ch
1 = - sm
. -1(2 x -1 )1 Jv'4 =
2
-11]
2 =6
lt
te

;;
f
m

Let l == x dx. Then,

2 X 2 +1
fro

' 4 2x
.=.:.J
2 2
2
+ 1
dx
d

X
de

1 r 2 -4
I = - Jog (x -t l)j
2 2
1[log(42+ 1)-log(22+ 1)] "' 1(log 17- log5) 2l log (17)
oa

I = =
2 5 2
nl

2
l
-f-
w

Let I =J dx. Then,

Sx -1
do

/ = 1- Jt
5 0 5x + l
10 X
dx = l log (5~ 2 + 1)
'> 2
I 1' 1
S (lug 6
I
log l) = e; l,1g tt
0
2
~ Let l :: f 5
~ dx. Then,
0 l' +4
2 S 2 J
r = J /--dx ~ J 2 dA
0 x+4 0 x~4
2 2
5 2x I
I = - 2
2 0 x<-4
J dx + 2 f 2
x+2
1/1
0
19.10
2
2 1
I =%[1og(x +4)]: +~[tan- (~)]
⇒
0
⇒
I == %(log 8 - log 4) + t{ tan - 1 (1) - tan - l O}

⇒ l= % log (:) + : (: - 0) = ~ log 2+ ;

EXAMPLE 13 Evaluate:

(i)
1
Jx ex d..,: [NCERT] (ii) f2 log x dx
x2
l
0

m
rr./2
Jx sin x dx f
(iv) 1 { xex + sjn 1tX} dx.

o
(iii) 4

.c
0 0

du
1
(v) J x log (1 + 2x) dx [NCERTEXE

oe
0
1
= I0 x e X
ch
SOU:lTTO!\. (i) Let I dx. Then,
te

1 = J1 xexdx =
[
xex] -
11
J 1,exdx = [ xex ] 1 -[ex] =(e - 0) - (e-e
1 0
) =1
m

OTTT O O O 0
fro

2
(ii) Let I = log x •
-
J 2- dx. Then,
= ..
X
d
de
oa
nl
w
do

tt/2
(iii) Let I = f x sin x dx. Then,
O I If
n/2 n/2
[ = [- X COS X] J1 X ( - COS :,:) IfX
0 ()

T = [-x cos X J' l0

12
+ sin x]n/
-0
2
( n
2
CIIS ·1r I
2
O l'OS O) I ( sln ; sin 0 )

(iv) Let I = t( xex + sin ;c:) dx. Them,

I .. 5
l 1
I = J xe" dx + J sin 1t x dx
O o 4
 19.11

1
Let J = J x log (1 + 2X) dx. Then,

m
0 11 I

o
1
x2 1 2

.c
I = -log (1 + 2x) - J -1 +2-2x x ~2 dx

du
[ 2 ]
0 0

oe
1
I= - log 3-0
)
-f1 1 +x22x dx
l
(2 ,\1 t

0
ch
1=!log3-j {(x_!)+ 1 }dx x
2
x 1 1
te

2 2 4 4 (1 + 2x) [ ·: 1 + 2x = 2-4+ 4(1+2x)

0
m

1 '1 ·
x2
=-log3-[---+-log
1
(1 +2x) _"-log3·
- {(---+
X
-log3)-o} =
1
-log
3 1 ·• 1 1 1
fro

I 3
2 44 ] 88 2 4 48
o·
d

fl'LE 14 Evaluate:
de

3
2
r 2
Sx dx
2
[NCERT, CBSE 2010} (ii) f 2
l dx lNCERTl
(x + 1)
oa

1 X +4x~3 1 'X

2
5 x2
J
nl

flOU.,7 10N (i) Let I = dx. Then,

2 + 4x+ 3

l
X
w

1
do

2 2 2
J " 5 2 J
- - x - -dx
1 x +4x+3
=5
! [ 1 -4x
2
3
-+--
x +4x+3
dx

l =5
2
f 1 dx-5
2
J
2
+ 3- dx
4x- 5 J 1 . dx - 5 J 2 ~2.r + 4) -S ,I\'
1 l X + 4X+ 3 1 I ' . f 4.1 f 0

1
2
= 5 J 1 • dx -5 f J { ; <2x + 4·) 2 5 } ,/.\'
1 I. I X + 4x I· 3 \ I 4X I 3

2 2 2
2x + 4
r = 5 J 1 . dx -10 J ifX I· 25 J d~·
1 1
2
x + 4x + 3 I "2 ·I 4-\' I- 3
2 2 2 ·1
I = 5 J 1 dx -10 f 2
2x +4
dx + 25 f dx
1 1 X + 4X ~- 3
1 (x+ 2}2-1 2
19.12

⇒
I = S [ x ]'. -10 [1og (x + 4x+ 3)
2
l'. + 25 ' 2 )1) [log :: ~:: I
⇒
r = 10 (log 15 -log 8) + ~ (log(:)- log (!)}
s (2 -1) -

I = s - 10 log ( :) + ~ log (; , ~) = 5 - 10 log ': + ~ log ( ~ l

1
⇒

1 A Bx+C
(ii) Let
X
2
(:\ + 1)
= x +1 + .x2

m
Then, 1 = Al +(Bx+ C) (x + 1)

o
PuttiJ.1g x =0, x = -1 respectively i11 (ii), we get: C = 1 and A = 1

.c
Equating coefficients of x 2 on both sides of (ii), we get: 0 = A + B

du
Substituting the values of A, Band C in (i), we obtain

oe
1 1 -x+l 1 x 1 1 1 1
? = - -+ = - - 2 +- == - - - -+ -
ch
2
x- (x + 1) x +1 x2 x +1 x x2 x+1 x x
3
te

Jl X
2
1
(x + J)
dx = fl (- +--.!_+--\)
X 1
1
x
dx
X
= [1og l x+l l -logl xl-].]
X I
m

= ( log 4 - ~ )- (log 2 - log 1 -1) = log 4 - log 3 - log 2-

fro

log 3 -

4
(3.)3 + 23
d

= log ( - - ) - .!_ + 1 = log

2x3 3
de

EX,b ' E 15 Evaluate:

oa

.[2 1t/6
(i) f ✓2 - x2 dx (ii) J (2 + 3x 2
) cos 3x dx
nl

0 0
w

.ff_
= J ✓2 - 2
do

SOLUTION (i) Let I x dx. Then,

0
F2 r -
L= J ✓<Ji./-x2 dx
0
.J2
~ [i l- 1 ~ (.j2)2 sin
;::;,
I =[ x
1t/6
2 ) co& 3x dx. Th,
I X
✓2 l 0

(ii) l.et 1 =
tr (2 3x[+
IJ I II, J (1

~
l ]rt/6 >1/6
I = (2 + 3x2 ) x -3 sin 3x J6 l
I
x ,, sm
,
3x d.x
I "'
0 0 ,,
l .. (
l jn/6 rr/6
= [ 3 (2 + 3x ) sin 3x
2
- 2 x sin 3x d •
0
J
O I l[ ~
 19.13
& IIIE1NTEGRALS

sin O} I
~2
2 2 , l 2 16)
2
-+
..
=
lt
+
..
-
(
lt +
336936936

m
IXAMPLF: lb Evaluate:

o
2 1

.c
(I) { (x + 1) (x + 2) dx

du
1 A B •.. (i)
9(1WTIO~ (1) I et-~ - = - + -
x + 2°

oe
(x + 1) (x + 2) x + 1
, .. (11)
1111n, 1 = A (x + 2) + B (x + 1)
ch
Pulling x ..- 2 = 0 or, x • - 2 in (ii), we get: B - l
Ming x .,. 1 = 0 or, x • -1 in (ii), we get· A == 1
te

Ming the \ alu<.'S of A and B in (i), we get

m

1 1 1
---- - = - - --
(x + 1) (x ➔ 2) x+1 x+2
fro

2 I 2 l 2 1
~ 2) x-:;:-1 dx - ~
- dx
I = { (x + dx = { x+2
d
de

I = (tog (X + 1)1: -[log (x + 2)]:

l
oa

/ = (log 3 - log 2) - (log 4 log 3) == 2 log 3 log 2 - log 4 = log 9 - ll1g 8 = log I :
nl

1 A Bx+C
w

Let - - ~ • - + ~2.
x(l + i) x 1 + x
do

•.. (11)
Then, 1 '"' A (1 + x2 ) .,. ( Bx + C) x
Putting x - 0 in (ii), we get A = 1.
2
C.omparing the coefficients of x and x in (ii), we get
A + B = 0 and C • 0 ~ 8 .. - 1 and C = 0
Sel.tituting the value:. of A, Band C in (i), W(' gt•I
1 1 X

;(1 + x ) = X l + x2
2

1 l
2 2

=
2 l
J _;_ 2 dx = J - dx
2 1 1 2
J 2x
l ~Y 2
dr • [ lo~ ,
I
t In~ (I + \' 2)
2
111
1 T (} + y ) 1 X 2 1
I l 3 l
= (log 2 -Jog 1) -
J
(logS - log 2) log 2 log S ➔ log 2 = - log 2 - log 5
2 2 2 2
2
19.14

EXAMPLE 17 Evaluate:
21t .
J ex sin (
0
i +
X) dx.
2
SOLUTlON Let T=
2nf e·t. sin (- + -X) dx. 7t

rr I 4 2
0

x
ecos (1t
- +-x)dx
J1 I 4 2

ex sm
· (1t4 + 2XJ dx

om
.c
⇒

du
oe
⇒
ch
1 21t ... 1
te

l+ - 1 = e • (1 - 2)
4 2-.fi
m

SI e21t+ l
- = - - --
fro

4 2,fi

⇒ I =- '7 (e2" 1) +
d
de

rt/2
EXA~PLE 18 Evaluate: J cos 2x log sin x dx
oa

rt/4
,r/2
nl

SOLUTION Let 7 = J cos 2x log sin x dx. Then,

w

,r/4 Il I
do

I rl
= - (log sin
L2
x) sin 2x
]rt/2
-
rr/4
rt/2 1
J - cot x sin 2x dx
,r/4 2

⇒ I = [o - ~ log ( /2 )~ -nj2 cos 1

2
x dx
.1 11/ 4,

1 J n/2
I =- log 2 - - J (1 I co9 2x) dx
4 2 n /4

1 .
I = -1 log 2
l [
- x -i -sm 2x
]n/2 I
--= log 2
4 2 2 n/ 4 4
00

EXAMPLE 19 £valuate: J ~-~1 -- d

0 (x2 + a2) (x2 + b2) x.
 INTEGRALS 19.15

N Let .x2 = y TI1 cn,

~--1~- I
(.i + i) (.i +b2 ) = (y -
+2a ) (y + h2)
___ 1 - ~ A B ...(i)
(y + ,?) (y + ,,2) = y + ,r2 + y + 1,2
2 ...(ii)
1 A (y -'-b ) + R(y + i)

o m
.c
du
oe
ch
te

r= 1-.. lr1 tan- t.:: ~]"o'

m

1 tan 1
a2 -b2 b b a
fro

l = - 1 ~(1 tan - 1 ;x; - -1 tan - 1 oo~ - ( 1 tan 1 1

0--;;tan
-1 o)lJ
a2
1

-b2 b a ) b
d
de

l =
1
a2 -b2 ~
,..(n2b - 2nn) -(O-O)Jl = 2ab~b)
:r
oa

2
a,IMP U: 20 If/(x) is of the form f(x) = a +bx+ cx , show tl,nt
nl

f f(x) dx = :!_6 {t(O) • 4/I.!.2 \ + / (1)}

w

O
do

I
,,
ION \\'e have, f(x) =a +bx-1- ex

/(0) ... a, f ( 1 ) = a+!!.. -r £. ;ind /(1) • a+ ti+ c

2 2 .\

: {1 (0) • 4 / f~) •J(I) f a :, \ H 4 (" • ~ < : ) < (" '" c) l

1 1
J /(x) dx = J (a + bx + ex2 ) ,h
0 0
1 ... (ii)
(ba + 3b + 2c}
6
 MATHEl'IIAT

19.16

From (i) \11\d (ii), we g~l { / ( 1 ) I f(l)l

J /(x) dx 6 .f(O) I 4 2 f
()

[ LEVEL· 1]

En1/w1tt' th,·fol/awing dc/ini/,• inh•gml~ (L 60):

9 I
1 Jb d.,
4 •
1/ ::!
~. J r-2--

m
dY
o ✓1 - x2

o
.c
~ j x~/ + 1 dx

du
2
00
1 1 8. J e-x dx

oe
7.
J - 2 dx
-1 1 + X 0
ch
n/2
'l.
1
J -x+ 1 X
dx 10. J (sin x + cos x) dx
te

0 0
rt/2 n/4
J sec x dx
m

11. J cot x dx 12.

fro

7'/4 0

13.
rt/4
Jcosec x dx [NCERT} 14. J1 l - x dx
d

1t/6
O 1+X
de

1t/ 4 1
"
J 1
16. -J, - dx
oa

--dx
0
l+sinx
- n, 4
1 + sin x
nl

1t/2 1t/2
17. ] cos2 x dx INCERT, CBSE 20021 18. J cos 3 x dx
w

0 0
do

n/6 1t/2
19. J cos x cos 2x dx 20. J sin x su, 2x dx
0 0
'lt/4 it/2
21 J(tan x + cot , / dx 22. J cos4 x dx
1t/3
0
n/2 11/2
23.
2
f 2
(a cos2 x + b i;in 2 x) dx
24. J JI I S ll1 \ ii.I
0
n
;r./ 2
n/'2
25. J ✓~1-+_c_o_s_x dx
0
21>. J ~ 2 Hin x tlx
(J
 19.17

2 t</2
x cosx dx 28. 2
J
x cos x dx
0
4 ~/2
J x2 sin., dx 30. J .i cos 2x dx
0 0
2
32. J log x dx
1

--2
log x d
X 34. Je -ex (1 -1 x log x) dx
1 (X-'-1) 1 X

m
e2J {-l - -1-} dx

o
log x dx 36
• e log x

.c
X (log x)2

du
1 '
x+ 3 dx 38. J ~2_
2
dx [NC£RT]
~ (x + 2) , 0 ,5x + l

oe
1 ,, l
2
[ 1 , dx [NCERT]
.40, ~r . ,
1
dx
ch
2
·. O ,2,X + +1.
04- x-x- ' I ,
X

,. J2l' J,r-' r:' 1'' -'

te
1 f :/ ' I
1
_2
J.Jx(1-x)dx
dx
2
. ' o,,·y3 + 2x - x
m

0
I 1• ,

•4f .·,.s•J ';/ +·12x +5

11
fro

1 I• I

dx [NCERT}
-.=== dx 1

o ✓4x-x2 ~ ,J
1
d

I '

· ~b, f x p - x)5 dx
de

!(
. 0
oa

·4~, xe 2x + sil~ 1t
2
X) dx
nl

50, 1e"•(l1-cos
- X) !;\CERT]
w

sin dx
x
1112
do

52,
2
ex f
0
cos(7t4 + ~_)l dx
1 I
(CBSE 2016) 54, J J1f .\ - .,_\T. dx lNCERTI
0
n/2
(N Cl3R1'J Sh, J Rin '1 .,. d ,\' lNCERTl
()

· jo (sin 2
~-cos2~J
2 2
dx [NCERTJ 58, 2.J

l
e2v ( 1 l---1,Jdx IN
X 2J-~
CERT, CBSE 2020]

2
,..,...__,1 (2 - x) dx
· J1 v(x-1) {NC"ERT EXEMPLAR]
19.18
61. If J 3x2dx =- 8, find the valu1• <Jf a.
II

60. If k J - 1-2 dx = - n , find the value of k. 0

2 + Bx 16
0 [ ~EVEL•2]

£;,a/1111te 1/iefollowing integrals (61-67):

~n/2 63. J
2n [,
1 +sin· dx
X

62. j .jf~-2.r dx 0 2
r.
rr, 4 65. JI x Jog (1 + 2X) dx
64. J 2
(tan x + cot x) - dx 0
1t/4
(l
rr /3 67. J («2 cos2 x + 1/ sin 2 Xj dx

m
l>b. J (tan x + cot x) 2 dx 0

o
;.16

.c
1
t>S. Jo - - ;---- dx

du
3
1 ... 2x + 2x- + 2x + i

2. log 2
oe
ch
1. :::
7l
- 1 log 2
:,. 6. -
te

2 2 ab

9. log(~)
m

••
7. - 8. 1
2
fro

10. ~ 11. ½.log 2 12. log (..fi + 1)

d

13· log b,'2 - 1) - log (2 - .J3) 14. 2 log2-1 15. 2

de

16. 2 17. 1t 2
18. -
4 3
oa

5 2 2
19. 20. - 21. - -
11 3
/3
nl

22. 3 ;r
23. ~(a2 +b2) 24. 2
w

16
do

25. 2 26. n-2 27 1t 1

2
28. - - 2
n2 29. J2 + _rr_ n2
4 2,Ji 16 .fi.-2 30, - 1t
3 4
31. 1t 7t
48 8 l2. 2 log 2- I
33· ·; log ::; lu~ 2
35. l ,2
2 36, ' - 1'
1 1 2
37 - log6 38. lni,; r, 1
2 5 +- JS tan I .Jr:, 39. _.:! I ( 21 +
✓17 ogl 4
40. ~I lan-1 ..p5 -tan
...7
I ~7}
"' 41.
71

8
42. 7E
3
o£FINITE INTEGRALS
19.19

11 44, !I 15 57 - J--J
8 5

42 - .., ..,
1
47. "
r·
2
•• 4'3
{'2
+ +
2

q ]
+-- •'\ ,,_
so. /It/ 2 1 0
4
" It

;'[

3/2 ( l) 25/2
5 l'
2 r.
+ J 53. -3 ~4.log(~:)
4 2
2 57. e -2c
56. 0
J 4

m
1t .n
I 1
2
hO, 2

o
,-..,

.c
1t
0• 8
32

du
4
;:log, 3 6~
h .fj

oe
2 21t12 2 1
(a +b )-+-(a -b) 67. - log (2e)
8 4 4
ch
_ _ _ _ _ __ _ _ _ _ HINTS TO NCERT & SELECTED PROBLEMS
te

1 I
2-dx =f x+l-1 dx= J1 ( 1 - 1-- ) dx=· [ x - log (x +l )] 1 = 1 - log2
m

l =f
X +1
0 x+l O
x+l 0 0
fro

I = J cosec dx =l log (cosec = log cosec ~ -

ltl4 /4 ( ) ( )
x - cot x]n cot !: - log cosec~ - cot~
d

n/ 1t/6 4 4 6 6
6
de

⇒ J = log("2-1)-log(2-../3) = log( -f_J3 j

oa

1
2- 1) rlx - [
nl

l = f 2-_- x dx = j ~ -(1 + x) dx = J (-1-+x 2log(x +1)-., )

w

l+x 1-t-x ll
O
do

0 O
⇒ I = (21og2-1)-(21ogl - 0)=2 log2- 1
n 1 n 1 - sin x "J 1 - sin \
I ==J ~ - dx = J -----~ -
(1 + sin x) (1 - sin x)
dx = 2 ,h
1 + sin x
0 0 0 co-. ,

⇒ I = J (sec 2 x -secx l.inx) dx =[t.m\' -6('<'

0
'J:
⇒ I= (tann -sec1t} (tanO -!>t>eO) Ot l ➔ I 2
n/2 n/2 1 . 21 sm .,
11
/2
2 [
2 l + cos 2x
7 I= f cos x dx = f- 2- dx ,+ -
2 o
0 0

= ~ ;( ~ - T)-( 0+~%0)] = ~
-----------~-----~-
19.20

J _ eI
2
_l_ .1 dx - [
e2
l 2
xl
dx :: ~g_y
]e2 ej _
I' - x (log
1
xl
x dx - ef
e
1
(log x) 2
dx

36· - log x • (log x) e

e I D e
2
e2 e _ e2 ___ e =---I!
e
⇒ r loge2-log;- 2loge loge 2 J 1

2x + 3
1
1
2 x dx +
1 3 J 1
dx =- -
10 X d
-2=--- x +
g 1
J --- J J
38• r = 2J d;1. = 2 1 5x 2 + 1 5 o 5x + 1
--2 - 2
'Sx) + 1
0 ( -v;
5
0 5X + 1 0 X + [ 0 "" ]1
1[
5
3
⇒ I = log (5x + 1) 0 + ,Js tan
- ✓5x 2 ]l
o
l1

m
, s i 1 3 - i .Js
⇒ = 5(log 6 - log 1) + .Js

o
I tan - .js = 5 log 6 + $ tan

.c
1 1 1 ,

du
2 2 2J
I "' J dx = - J dx l dx
39.
O 4_ + X-X
2
O x
2
- X -4 .
= -
-½) -(~ )
0 ( X
2

f .JW. 1] oe 2
ch
J
(!;7]' +-I('• ffi'l ogl N _,,., J
1 1 ( 2x -l
1
~
te
I • o o
m

⇒ I= 1 Jlo ../fi .,_ 3 _ lo .Jfi -1 } J

= _1_ lo 26 + 6 ,ff7 _ lo 18 - 2 .jv
fro

..jl7 L g -./17 - 3 g .f[7 + 1 ,Jf.7 g l

8 g 16

⇒ I= _1_ 1 (52 + 12 ..Jff) = J171__ log (L521s+-12✓17 x 18 + 2'117]

d

% 0
g 1s - 2,m 2N 1s + 2ffe)
de

⇒ I = ....!_ lo (1344 + 320.,/17) = _ 1 lo ( 21 + s,Jf.7)

oa

✓17 g 256 ) .f[7 g 4

nl

1 1 l 1
14.. I= J -2 - dx = J - -2 dx
w

-1 x +2x+5 _ 1 (x +1)2 + 2,
do

;::;, -
r
I - - cot 2 e
L
Xx
]" - "J - ,, cose,::2 -
,,1
2
1t/Z 2
l x X
2
llx + f 1e x cosec2
n

1t/2 2
l"

2
dx =- [ O - ,,
r./ 2]
DEFINITE INTEGRALS 19.21

1 1 ~
= J- __ dx _ f \fl + x + ✓
1
X
✓1 + x - ✓x
54. I l {
o • - o (1 + x) -x dx = f Jf+x) + Jx } dx
0
1
⇒ f =[!(l+x) 3/ 2 +3.x3/2] = 4,fi.+2_2 = 4 .J7_
3 o 3 3 3 3

55. I = J1 (x + 1)x(x + 2) dx = J2 ( -
I l l
x +1 + x +2
2 lJdx

I= [-1og(x+1)+2 log(x+2)]2 = [1og(x+2)2]2 = log16 _log 2- = log 32

1 X + ] 3 2 27

m
1

/2
~4 0f (3 sin x - sill 3x) dx = I.4 [- 3 cos x + :!..3 cos 3x]"0

o
rr/2 .r/2
J sin 3 x dx =

.c
56. I =
0

du
⇒ ~ cos \n}-(- 3 + -~)] = !
oe
l =~ [(-3cos%+
ch
J
57. 1 = " (sir? 1- cos2 ~) dx = -J cos x dx = -
1t [ sin x111 =0
te

0 0 0
m

2x ( 1 1 ) 2 2x 1 2 2x 1
58. I= 2Je ---2
dx =J e -dx - Je · - 2 dx
fro

l X 2X 1 X l 2 X
II I
2 4
l 2x]
2 2
2x 2 x 1f(1 4 1 2) e
+ - e dx- e - 2 dx = - e --e = - --2e2
1 J
d

⇒ l= - e -
[ 2x
de

1 l 2 x2 1 2x 4 2 4
2 2 1 2 1
J 1 f -======dx =f -a==== dx
oa

59 I = dx =
, ✓-,2 +3, -2 , - ((,-})' -(½l') , (½)' -(,-{]'
nl
w
do

19.3 EVALUATION OF DEFINITE INTEGRALS BY SUBSTITUTION

Consider the integral f f(g(x))g' (x) dx.

b
Let g(x) =I, then ~'(x) dx dt. Also, whert
a
x = a, t =g(a) and t "" g(}i) for x =b.
b g (b)
J _r(g(x))g'(x)dx - f f(t)dl
a g (a)
Thus, if the variable in a definite integral is changed, then the substitution in terms of new
Variable is effected at three p laces :
(i) in the integrand (ii) in the differentia l, say, dx (iii) it1 the limits
 19.22
. . . the values oft corresponding to the Val
51 1
Also, limits of the new vanable tare ~P; b putting the values of x in the su
original variable x, and so they are obtain Y
relation between x and t. .
t definite1ntegr als by makingsomes
l
We may use the following algoritlun to eva ua e U

ALGOR HM b
Obtain the definite integral and express it in the form I = f (g (x) g' (x) dx. f a

5T1T 'l Putt= g(x) which gives g' (x) dx = dt.

Puf x = a (the lower limit for x)int= g(x) to get t c: g(a) as the lower limit fort.
srrr JI
Put x =b (the upper limit for x) i11 t =g( x) to gett =g(b) as the upper limit for t.
Substitute g' (x) dx = dt and replace old limits of integration by the new limits to

m
g(b)
1= f f(t) dt

o
.c
g(n)
Evaluate this integral by using standard methods ofintegration discussed in secti

du
Following examples will illustrate the above procedure.

oe
ILLUSTRATIVE EXAMPLES
I
ch
LEVEL-1
E~ Evaluate:
te
,• ,r E ,

f x+1"xr; dx
4
(i) (") fl 2x
m

ll 2 dx
0 o Sx +l
fro

SOLUTIOK (i) Let x = t2. Then, dx = d (t2) ⇒ dx == 2t dt.

When x =O1 x -- t 2 ⇒ t 2 = o⇒ t = 0. When x = 4 12 _
d

2
4 2 I -x ⇒ t =4 ⇒ t =2
de

I I
I= I d - 2t dt
0 x+fx x - 0 t2+t =2
2 1
!2
t+ldt =2 [ log (t+l)] =2 [log3-logl]
oa

(ii) Let 5x2 + 1 =t. Then, d (5x2

10 xdx =dt
+ 1) - dt ⇒
. 0
nl

-
2
When X::0,t =5x +l ⇒ t=l. Whenx =l L·-5 2
w

1 ., - X + 1 ⇒ f - 6
2 6 -
do

Jo Sx 2x+ 1 dx = J 2xt x .!!!_ = !. J6 1dt = 1[log t]6 :: .!.. (log 6 _ log l) ..

lOx1
5
1
1 5 5
EXP. 'vi r 2 Evaluate· '.l
1 '
(i) J sin-1 x dx (ll;('J l<TJ
n/2
0
1t/2
(ii) J /cos O s in 3 0 dO
(iii) I cos e
0
1/2
0 (1 + sin 0) (2 + sin G) dO ICHSI 20041 (iv) J 2 r dx
,c/4 0 ( l + X ) \I l - X2
(v) Jo cos3 x ,J21 siri 2x dx
SOLUTION (i) Lett= sin - 1 x
. or,x =sin t. Then d .
' x = d (sin t)
= cos f dt
 19.23
oEANI 11 mTEGRALS

X 0, -::::> I sin -I O 0 ..ind \ I ~ , sin I I

n
2
l 1t 2 ~ .,2
I= J , Ill 1 ., 1h J0 II l"<)S I if/ [ I sin I J ~in I di
II Il n
0

I"'~ ,l
n lt
[ r t' n n 0 1
~ s,n
/2 !J t l ( >'i I ()';
Ism I 2 2
l (l 2 2
cos ll t. Tiwn, d (fos 0) - ,if ⇒ sm {l ,IO ,fl
(it) Let

2 0

I J , co:.- 0 sm
3
0 JO J /t s,n :1 0
1
I Sil\
- •!! ) - J ./1 sm 2 O tit

m
0 1 (l 1
0 0
l
~
2 t7/2 lo=-ro-(2 72)J= 218

o
512 2
J ,t (1 -1 2
) Jt - J (•.'f t ) d/ ~tl/

.c
ca
7 L t3
1 1

du
11) Let Sill 0-=f. Then, d (!'in 0) -= .it ⇒ c1l'< 0 d0 dt
8 0~ t sin O = 0 and, e "' '2 ~ t = sil.1 .:12 "' 1.
n/2
.:os 0 oe
ch
I = f- ----d0
(1 + :,m 0) (2 + sin 0)
0
te

1 t {1~,-2S}dt [!3y using partial fractions]

m

I= (l+t)\2+1)dt =
1
fro

1
I= f1og(l
L
+olJo - flog
~
(2+t)l
'Jo
,~
~
d

I (log 2 - log 1) - (log 3 - log 2) -= log 2 - log 3 ·I log 2 = 2 log '.2 - log 3 = lQS
.
de

V2
J - -l ~2 dx. Let x =c;m. 8. Thc-n, d:t = cos 0 clll
oa

Let/
2
O (} +x ) ,1 x
nl

x = 0 ⇒ sill O - 0 => 0 = 0 and x "" ⇒ sin O = 1;.2 ~ 0 - ~6

1
w

2
1tl r, 11/ t,
do

I= J- - _ l r==----=- cosO dO J 1 t sin ,,· 0 do

0 (] - 1,m 2 0) ✓I - bin'.! 0 0

1t/6 20
sec
I- J0 - -2
Sl'C () + tan 0
2 dO

11./ ,, 2
Sl'C (J
I= J0 1+2tan
- 2 dll
0
7t
Let tan 8 = t. Then, scc2 O d{) di Al,o, IJ l,111 0 0 ,111d II l,u\
6

I-
1/,/3
I - 1
o 1 +212
1
,11 =-
2
l/J 3
J - - -1
(l/,!2)211 2
,Jt _ , x
2
I
(I/J2)
t,m r I
/ 21
11/,, "\ = r,:;I t,1n - l f.2
n v2
-
~3
0
19.24

;r/4 1
(v) Let I = J dx. Then,
0
cos 3 x ,J2 sin 2x
4
'lt/ 4 11t/ 1
-1/2 - 7/2
T= J --,,-:==== dx = -
3 x ✓4sinx cosx 2 o
sin x cos x dx f
0 cos

We observe that the sum of the exponents of sin x and cos xis ( ~) 1-( -i) = -4, which·
4
negative even integer. So, we divide numerator and denominator by cos x to get
1
-;r;/4 - - rr./4 4 1t/4 2
I =~ f
1 cos4 x dx = .!. sec x dx = !. f
(1 + tan x) sec2 x dx f
2 sin/ 2 xcos7/ 2 x_ 2 0 ,Jtanx 2 0 ,Jtanx

m
0
cos 4 x

o
Let tan x = t. Then d (tan x) = dt ⇒ sec2 x dx = dt.

.c
du
Also, x =0 ⇒ t = tanO =0 and x =~⇒ t = tan~ =1.
4 4

oe
1 2 1 1
l=!_f 1-'-t dt=.!.J (t- l/2+t3/2)dt = .!.[2tl/2 + 3_t5/2] = .!_ x 12 = 6
20 .[t 20 2 5 0 2 5 5
ch
I:XAMPlE 3 Evaluate:
te

a 1/,/i . - 1
(i) J ~
x4
dx (n f .
") Sill X
2 3/.2 dx
m

2 2 0 (1 -
✓ a -x X )

l
0
fro

(iii) J x tan - 1 x dx (iv) J sin - 1 ( 2x dx

o (1 + x2) 3/2 o 1 + x2
d
de

SOLUTION (i) Let x =~ sin 0. Then, dx = d (a sin 0) = a cos 0 d0.

Also, X = 0 ⇒ a Sfil 0 = 0 ⇒ Sin 0 = 0 ⇒ 0 = Q
oa

1t
And, X =a =:> a Sin 0 = a=> Sin 0 = 1 ⇒ 0
2
nl

a x4 n/2 ( 4 1t/2
I = J -.====
1
dx = J a sb.: 0) a cos e d0 = n4 J sin 4 0 d9
w

2 2 /2 2.2
0 \ 1a - X O \I a - a SIU !) 0
do

⇒ I = a4 x 3 7t = 31t a4
16 16 [See Example 3 (iii) 1m f
(ii) let sin - l x = 0 or, x = sin 8. Then, dx = d (sin <➔) = cos fl d e
Now, x = 0 => sin 0 - 0 ⇒ 8 = Oand x "' 1 ~ • 1 1t
/2 5111 0 -r::: ⇒ 0=
11.fj. 1 . _ " v2 4
l
=
J0 (1-x2)32
sm x
dx
'It/ 4 (-} 1t/4
l = fo --3 cos 0d O =
cos 0
J 0soc2
1' n
OJ O
,.
[
0 tan O O
]"/4
- J I x t,Ul ft ,I 9
it/•I
0
14 0
14
⇒
I = [e tane]"o +[log cos 0]1t
o
- (re
- 4- 0
)
~ { log ( -;fi
1 )
- log l j = l -i -21 I
 FINITE IN TEGRALS 19.25

1
.. l)Lettan -' •8or.x ... tan 0.1hcn,dx =sec 2 ()JO.
x '"' 0 => tan O • 0 ~ II • ()and,,\ I ~ tm II

x tan - l ~
1
I - f(I+
- ~2r112
- ,1\
0
rt/-1
e
J0 sec 3-0 =2 OdO
O tan
l
, 4
l J Osin 8d0
o I 11

"' ]"''4 - 11/4

J(-cosO) [-8 n/4 [

m
1 - 1-fl cos O dO • cos O
]
+ sin

o
~ 0 0 O

.c
I - I - 4,2
:r - o) + (~ - o) =

du
,12 4,2

oe
= J ~m - 1 (- 2 x 2 ) dx. Then,
0 1+ X
r ,
.-
ch
1
J 2 tan - 1 x dx 2x ) 2 tan- 1

l
I (
te
I ·: sm l - x2 = :r-
0
-
m

1 = 2 f tan - l x · 1 dx = 2 x tan - rLLr 1

T-j dxl
fro

x _ 12•:r
O I 11 .JO 0 l+x J
d
de
oa

J 2 [[x tan- 1x J: -½ :log (1 + x2)J:]

nl
w

J =2~(1/tan- 1 1-0tan- 1 0)· ~(log2 logl)]= 2{(:-0)-~togl}=;-1,~:.

do

2
I.et x • tan 0. Tht:n, dx = d (tan 0) = sec 0 d O
It
:r '"' O :::) tan O = O :::) (! -= 0 11nd :i· = I ⇒ tan O 1 :::) 0

1 = J sin 1 r 2x 1,fr 2
o \lr.r)
:r./4
I sin-] (bin 2 a) a;cc2 0 do
l
I =

1 ..
::r./4
0

J 2 esi.:c2 ode
0
2 !~
11/4 2
se)~ 0 dO = 2 [ 0
l t<1ll O lo - !I x t,m Od8
11/4 rt/4
19.26

0 taJ1 e]
1e/4 [
+ log cos 0
~1e/ 4} ==2 f(~
l 2:-o) +
4
tan
4
(10g cos n - log 1)
4
⇒ {[ 0 0 .
= n + log 1 = r. - log 2
T = 2 -1t + Jog - 1 } = -1t + 2 log -;;;'2
1 2 2 2
{ 4 ff. 2 '1"-
rt/4
F..xA \tPLE 4 Evaluate· J tan 3 x dx
0
11/ 4
SOLUTION Let T = J tan 3 x dx. Then,
0 rc/4 2 n ,14
n/4 rt/4
f tan2 x J sec

m
2 = x tan x dx -
1 = tan x dx = J (sec x -1) tan x dx
o

o
0
0 0

.c
⇒
2
Let tan x = t. Then, d (tan x) = df sec x dx = dt.

du
Also, x = 0 => t = 0 and, x = 2: ⇒ t = 1
4

Ia l ·r tdt - Om X dx a [ ': i: - [ log "' xJ:1'

ch
oe
te

⇒ I =( ½- 0 ) - log sec : + log sec O= ½- 1

log ff. + log 1 = - - - log 2 =- (1 -1
2
1
2
1
2
m

EXA..1\4:PLE 5 Evaluate:
fro

11/2
(i) J::>+4cosx
_ 1 dx [CBSE 2005) (ii) f 3+2cosx
1 dx
d

0 0
de

1
SOLUTION (i) Let J = f - -=--
1t

5 + 4 cos x dx. Then,

oa

I -----,----...:.1_ _
It
nl

I =
l-tan 2 .!
w

0
2
do

5+4
l +tan 2 ~
2

lt l+tan2X
sec2 X
:::) I= J- _--2,. dx - nJ
2
_2__ tx
0 9+tan ~ 9 + tan 2 x ' ,
2 0
2
Let tan ~ =t. Then , d [ tan -x J di => I ""C2 Y /
2 2 2"" 2(.\'-r// 2 tli
S\:C2 X

Also, x =0 ~ t = tan O = 0 and x = n 2

⇒ 1 "' lM 7t
= 00
2X 2
00 sec -
1 = J 2x
2 -
2dt
O 9+t sec2 ~
2
 19.27

""
I = 2 fo -.,J-~
3 +t
= ~[tan-1.!...]""
3 3
-_ 2 ( tan -
3
1 oo -tan- l O) = 2
3
(1t - o]J = _1t
2 3
rt/2 O
ii) Let 1 ::c I0 l
3 + 2 cos x dx. Then,

I
l = fl/2 1 nr 1 + tan 2 ~ n/2 sec2 X

0
+
[1-tan2~ldx=
___ ~
o3(1+tan 2x) + 2
2(2
1-tan
2x)dx = J
2 o
S + tan; ~ dx
2
3 2
2
1+tan -
2

m
tan ; ::ct. Then.d ( tan ;) '"' dt ⇒ ½sec2 ~ dx=dt ⇒ dx 2
:\

o
sec -

.c
2
=0 ⇒ t ⇒ t =

du
X = tan O = 0, and x= 1t tan~ = 1
2 4

oe
ch
Ltan _,
te

=
2
.js 1 1
.js - tan
- 1 ]
O = .js tan
2 -1 ( 1}
l .Js
m
fro

MPt.E h Evaluate:
rt/2
• fl./2 1 s11· x
0
d [NCERT]
(1) J - - - - - dx (ii)
J0 1 + '
2
X ,
d

2 cos x + 4 sin x COS X

0
de

fl./2
(iii) J . 2 1 2 dx
oa

o 4 sm x + 5 cos x
n/ 2 l
nl

(i) Let 1 = J -2cosx

--- - - dx. TI1en,
w

+ 4sinx
0
do

rt/ 2 l
f
I =
o 2(1-tan
- - - - + - -- -
2
~J 4(2mni)
dx

2 x 2x
1 + tan - 1 + tan
2 2
2 X /2 2X
rt/2 1 + tan - n sec 2
I = J ZX
2
X dx = ? X
J X dx
o 2 - 2 tan - + 8 tan - O 2 - 2 tan - - + 8 ta:n -
2 2 2 2
tan x = t. Then, d ( tan X) = dt ⇒ 1 sec2 x dx = dt ⇒ dx =2 dt2 x
2 2 2 2 sec -
2
x = 0 ⇒ t =tan O = 0, and x = n ⇒ t = tan 2: = 1.
2 4
19.28
1 1
2 X 1 1 f
dt- - - - dt
f - J -~-x
1 sec 2 ~=J2 X
1 -t2 +4t - 0 -(t2 -4t-1)
- 2
2 - 2t + St sec - O
0 2 1 __ 1
1 l 1 = It J } dt = J fr=" 2 2 dt
⇒ I•! -(,'-41+4-4-t) o -{(1-2)2 -5 o (vS) -(t-2)

Js [ Ii-~; :~j1)
1

I = 2 log
1 [
o == 2 :./5 Jog
( Js: -1 ) - log ( Js
:./5 + 1
- 2 )]
Js + 2

($ + 2)}] == ~ log( + Js:)

3
_ ~[ {<Js -1)

m
⇒ I - 2Js log (.}s +1)($-2) 2$ 3-Js

o
2

+$)

.c
3
I =-2,..lo ( 3 +~x 3 +Js)==~lo (
✓5
3
+JsJ : : ~ lo (

du
2 .js g 3 - 3 + .js 2 ,/5 g 2 2 Js g 2

oe
T == _!_ log( 3+ $\
.Js J
ch
2
tt/2 .
te

(ii) Let I = J SID \ dx. Let cos x =t. Then, - sin x dx =dt.
0 J+cosx
m

Also, x == 0 => t = cos O= 1 and x = ~ ⇒ t =cos ~ = O

fro

2 2
I = J sinx(~l
1+t l x)
d

2
SID
1
de

I= -fi ~ -[tan-1t]O =
oa

1 + t2 1
(iii) We have,
nl

tt/2
I -- J 1
w

2 dx 2
o 4sin x+5cos x
do

n/2 2
1= f sec2x dx (Dividing n orrunator
· .
and denonunator bf.
o 4tan x+5
Let tanx=t.Then,d(tanx) "' dt ⇒ sec 2 x dx =dt
X-=O => t ta11 0 =0 and ,X--
_n ⇒ t l
2 = an 27t
=co.

r,
T = f dt I "
o 4t2 + 5= 4 0f t2- + (l ~ J2 ti/
2

1 = _ i 1tan-1( .3!.._)J"'
2
"5l l"5 = 2~(tan- 100- tan-lo) = _ l_x.'.: = ,c
0 2$ 2 4
 INTEGRALS
19.29

n/2
n>SX
3 en., \ - "in .t
dx (ii) f co,, r ,h
O I • <~ .t + SIil r
cos\' n/1
dx (I\) J () , 16♦sin 2r
111
" <.OS l" d\" [C BS F 2014, 20181
cos ' +- sin ;)3
~
0
, sm. , +co",
I b \ Ill
-
2 l
I1\ IC BSE2010, 2020J (vi) l .Jtan x t, ,ot x ),tx IC USF.2002, 2003I
n/2(
(1)Letcosx - K(J,os\+sin \)+! d (~,osx+sin \).Thl·n,
dx

m
o:;:1 K(1cosx t'.:>in x) I (- ,,m, +cos,) •.• (1)

o
~ co<'lficicnls of n1::. x and sin x, we gl't

.c
1K ' =1 md K 3L 0

du
3
the::.e tv- o equations, w(' get: /1.. = and L - ..!.
10 10

oe
ng Lue ,·alut'S 01 k and Lin (1), we get
3
sx (3 coo.; l - o.;in x) · _!_ (- 3 sin x - cos x)
ch
10 10
rr./2
f
te
COS X
= J0 (-------
3 cm, x + sin x)
.X I
m

3 1
-: 2 ( 3 cos l' + sin x) + (- 3 sin .x + cos x)
fro

J= J0 10 10 dx
3 COS X + sin X

f - 3 s m. x
d

3 J -3 -cos- --r +- -. dx x
rr./2
Sill 1
n/2
+cos x d
de

= - - x
10 3 cos x + sin x 10 O 3 cos x I sin x
0
oa

1 - -3 Jl·dx+-1 J -- 3-sin-. x-·-cos-x dx

rr./2 ~ 2
nl

10 10 3cosx+sin x
0 0
w

3 rr./ 2 1 1 ] rt/2
l x + -1 log I 3 cos x + sin x I
do

10 _o 10 ~ 0

10 2
3
( it - o) + I (log 1 - log 3)
10
3it
20
1
10
log 3

avt>,
'l'C/2
COS .l
- 0J 1 COS X + Sin X
,fr

rr./2
I - f co~ x tfx
0
(1 + cos x) + sin >.
2 X . 2 ,
co~ -sm
2 2 ,fr
"> X X X
o 2 co~- • 2 ~ill cos
2 2 2
19.30

2x
n/2 1 - tan [ Dividing numerator and denom.in1.1tor
1= f ~"< dx
o2+2tan
2

1t/Z(1 -tan;) (1 ~tan;) dx

⇒ I=~
2
f -l+tan
- x
0
2

T=
2
!
1 1t/Z(1-tan X) dx
2

m
1t/2

o
I= -1 [ x+2logcos-
X
2 2] o

.c
~ ~ + 2 log cos : )- (0 + 2 log 1)]

du
l = [(

½[~ + 2 log ~~] = ½[i + log ½] = ½(; - log 2 ) oe

ch
I=
te

(iii) We have,
rr./2
f
m

cosx d
l= o( x)3
fro

x . x
cos +sm
2 2
d
de

⇒
oa
nl

Let cos~ + sin x = t. Then,

w

2 2

-J2x'\ = dl
do

(
d cos - ~ sm
X
2
•
⇒ -1 ( - sin -X i cos · X) dx = dt = ( 2\'
CO(,,
2 2 2
Also, x =0 ⇒ I =cos O + sin O= 1 and x =~:;;,I= cos n + sin n
2 4 -l

1
12
di 2( l I) ✓2
I \ -
)

1t/ 4 .
(iv) Let I = f sm x + cos x dx
9 + 16 sin 2x
0

Here, we express the denomi11ator in term~ i,in .\ l·os y wh ich b Lnl<•gration ot nu!J\
. )2 , 2 2
Clearly, (sin x - cos x = stn x..,. cos x 2 sm x cm, x = L - -.in 2x

⇒ sin 2x = 1-(si.n x-cos x,2

FINITE INTEGRALS
19.31
n/4
I• f sin x + coi:; x
o 9 + 16 {1 -(sin x -cos x)2f dx
tt/4
l= f sinx+cosx
-dx
o 25 -16 (sin x - cos x) 2

sin x - cos x = t. Then, d (sin x - cos x) = dt ⇒ (cos x + sin x) dx =dt.

x =0 ⇒ t = sin 0 - cos O = - 1 and x = .'.: ⇒ 1 = sin .'.: - cos ! = o
4 4 4

I = fo dt 1 OJ dt 1 OJ dt

m
-1 25 -161 2 = 16
-1 1.6 - t
25 2 =-
16 -1(!) -i2 2

o
.c
I]o

du
I _ 1 1 [1 5/4 + t
- 16 x 2(5/4) og 5/4-tl _
1

4
oe
ch
I = l [logl- Log(l/ )] = l_[logl-log (~)]=l_ (logl+log9]=l_log9
40 9/4 40 9 40 40
te

)We have,
m

n./3 .
I = J sm x + cos x dx = n./3
J .
su1 x + cos x dx
fro

n/6 .Jsin 2 x n./6 ✓1 - (cos x-sin x) 2

tt =(COS x - sin x). Then, di = d (cos x - sin x):::) -(sin x + cos x) dx =dt
d

,J3-1
= cos -1t3 - sm. -1t3 = 1-,/3
de

;c 1t . 1t 1t
x=- ⇒ I = cos - - sm - = - - and, x = - ⇒ t
6 6 6 2 3 2
oa

1-,./3
1 [ -1]]-:s = - [ . - 1 ,/3 - 1]
-11-/3 - sm
nl

l -r:J dt = - sin. t ~ sin 2

11 - r :{3-1 2
w

2
do

= - sm
. -1 1-../3 . -1-./3 - 1 2 . -1./3-1
- --'-- + s!Jl -'--- = sm
·: sin- 1 (- x) = sm
. -1 x]
I 2 2 2
[

n./2
~ l = J {.Jtan x + ,.)cot x} dx. Then,
0
rr/2 .

1
= n.t{ ~in x + !cos x dx =
Vcos x ~ sir, x J
1
sin x -1 cos~ dx = n.t J2, J
O
sm :i: + cos x dx
J2 SUl X COS X
0 O ..Jsin x cos x
1t/2 .
l= ..Ji J smx+cosx ? dx
o {i-(sin x - cosx)-
ainx-cosx=t. Then,d(sin x-cosx) = dt:::) (cosx+sinx)dx=dt
1t
, _%-O ⇒ l=-1,andx=-:::)f=1
2
19,32

n/ 4 sin 2x
E>-.AMrtES Evaluate: (ii) J __4;:.:.:.:....-., 41x dx
n/2
_...,..::::::;,.__4
sin 2x dx 1cusE 2003Cl O cos X + SU\
J
(i) o sin 4 x + cos 4 x n/f 2 sin 2x 2 2 dx
n/f 2 sin 2x dx = sin 4 x + (1 - sin x)
SOLUTION Let I = 0 sin 4 X + cos4 x O . 2x dx = dt

m
. 2 x = dt ⇒ 2 sin x cos x dx-
- dt ⇒ sm

o
Let sin 2 x = t. TI1en, d (sin . ) 2 7t

.c
n t =Sin - = l
x =0 ⇒ t = sin 2 0 =0 and x =2 ⇒ 2

du
Also,
1 __l __ dt 1
---:,- dt

oe
-
lJ 1 dt = J 2t2 - 2t + 1 t2-t+ -
1
T =0 t2 + (1- t/ 0 2
ch
te
m
fro

1 1 -1
d

f = ( tan- \ 2t - l ) ] =tan- 1 - tan (-1)

de

0
2
n/ 2 sin x cos x
sin 2x n/l
J - --,-------:-- J
oa

A ETFR Let T = 4
= dx
4 4
O sin x + cos x O sin X + cos 4 X
nl

Dividing the numerator and denominator by cos 4 x, we obtain

w

rt/2 2
1 = J 2tan;sec x dx
do

0 tan x+1
Letl=tan2 x. Then, dt = d(tan 2 x) = 2 tanxsec2 xdx.
Also, x=O=;. f = tan 2 O= Oand,x =-
1t
:::>
t = tan 2 -Tt = QO

I=
,.,,
J 2
t + l2
1
dt
ltan
2 2

0
n /4 . n,/ 4
2
(ii) Let 1 = _!tn x
4
J-
. 4
dx. Then, 1 _3s in x c:os;1:
- - dx.
J
o cos x + Sm X (J cos 4 x + sin 4 X

Dividing numerator and den ominator by c:os 4 x, we ge t

n/4
1= J 2 tan x sec2 x dx
4
o l+tan x
 19.33
2 2
tan x = f. Then, d (tan x) = dt ==> 2 tan x sec2 x dx = di
Iso, x = 0 ==:> t = tan20 = 0 and x = 7t ==> t - tan 2 it
= 1
4 4
2 2
ubstituting t = tan x and 2 ta11 x sec _, dx = dt, we get
1 1
I = f
0
l + t2 dt = [ tan - It]1 =
0
(-4TC -Q) =4 lt

cos2 x
1tf2
f

m
Evaluate: --c:-----::-- dx [CBSE 20121
2 . 2
0 cos x + 4 sm x

o
.c
2
n/ cos 2 x
f

du
Let I : 2 . 2 dx. Then,
O COS x+ 4 Sill X

oe
n/Z cos 2 x
I =f dx
ch
2 ')
O COS X + 4 (1 - COS - X)
te

tt/2 2
l = f cosx dx
m

2
0 4 - 3 cos x
fro

rt/2 2
1 = _ .!_ f -
3 cos 2x dx
d

3 0 4 - 3cos x
de

1
!
n/2(4-3 cos2 x)- 4 dx
oa

l= -3 4 - 3cos2x

T{l-4-3~',l
nl

~
w

I • d,
do

11/2 4 rt/2 1
I = --
1
3 J 1 · dx + 3
fo4-3cosx 2 dx
0

n/2 2 sec2 x
4 rt/
I = !3 J 1 •dx -t- -
3
f 4 (1-- 2
+ tan x) - 3
dx [Dividing N' M\d D' by cos2 x]
0 0

I = - -l [X ]rr/2 + 4 n/2
f -J+4 tan
sec2 x
2
dx
3 0 3
0
x
00

1 =- ~(~-o)+! f0
1
1 +4t2
dt, w heret = tan x
19.34
2
«> .\ - dx
E'\;\MJ'lE t(l Er'/1/llafr; (ii) J 2 2)5/2
a 1 () (n + X
(i) f ~ 2 2 d.r
o (.r + a )
II J
SOLUTION li) Let I =- J 2 2)2 cf.\.
o (.\ + a
8) - a sec2 0 d8.
Letx=atan0.Then,ifr=d(at.ln - 0 =a ⇒ tanO"'1 ⇒ 9
=> O-Oand,x=n ⇒ ntan
Also. r. = 0 ::::> 17 ton a = 0 => tan e = 0 -
;t/4
2
I=
"J/-1
,, 2
t
, 2 asec
2 0 de _ _l_
- 3
cos 0 dO f

m
() (a- tan 8 + a-) 0
a
4 1
I= J:....3 ,.;J\1 + cos 20) dO =-\-[0 +.!_ sin 20]

o
2
n/ = 3 (: + ~) = g:3 (1t + )
2 2

.c
211
2a o a
0

du
"' 2
(ii) Let I = J ,- +X ,_):,_/,- dx.
oe
0 (a X
2
ch
Let x = a tan 0. TI,en, dx = a sec 0 d0.
0 ⇒ atan O= 0 ⇒ tan = 0 ⇒ 0 = 0 and, X e = 00 :=::, a tan e =00 :=::, e=
te
Also, X= tan <X) :=::, 8
J n /2
m

ll, :! 2tan_2Q 1
: 1= J 2 ll 2 2 512
asec 0d0=
2
3
Jo sin 2
0cos0d0=~ft2 dt,wheref=
fro

o (a + a tan 8) a a2 0

⇒ I•:, [•: 1: -
d

3: 2
de

-x:
oa

EXAMPLE ll Evaluate: J X /
1
dx
nl

o \l+x ·,_!
~

Jl
w

SOLUTION Let/ - Jl · -x2 2

do

- x dx. Let x :: f. Then d ( ,2

·' ) "' di ⇒ 2 x dx
2
o 1+x ' = dt
Also, x =~ __,ff :_ 0 and x =1~ I =l
•• I-=fx l-lxdf
o l+/ 2~

J = ~f (.11 I di
0 V~ I
⇒
I = !2 fO ✓ff:~
,-,.1
"
1 /
I II
~
l l
⇒
c=-
2
I I- t
l ~ dt
0 yl - t2
 19.35
f
I
1
I -
2 -
I
' 1// + J
I T
4 o
2,/
dt

r
\ I /4 12
', 1
1 1 I
I - I
12 \
...,
r
sin / -t 12
o 4
lo
]
l ~in l l - sin - 1 ll) + : ( 2 x n 2 >< l) I { rt
2\2
0)-1 I ( 0
l\
2) "' 4
7t
l
2
7t/2.
Et1l1wtc j - , - .-, d\ f"1C I R r EXEMPLAR!
0 (a COii 'I:+/, Slll \)2
2

:: 2

m
I= Io -(,,~c~-.1
, - , .!___2 - 2 , d\ '

o
+I• ,in .\)-

.c
\iding numt:'. rator and dl'nomin.itor by co~+ x, W<' obtain

du
:t : -l-
SC'( .\
I- JCl -.,- ., --, , dx

oe
(a-+ Ii- tan - t)-
- 2 ")
ch
• 1 + tan~ 2
l - I , .,
sec x dx ,
0 (a- +b- wn- x) 2
te

,
t = tan.1. Then, df = ~cc- x dx.
m
fro
d
de
oa
nl

I 1
I - -tan
w

b2 ab ~
do

2 2 r
o)_ + b /1~,1 f
, 1
i.,2
l ( r.
ab l 2
2 2
t JI I when• 1, =
o
7t b .•• ( i)
, _ '3+ 2Ii I1
2ab b
a t.ln 0. Titen,b di asec:.2 0 dO
It
f =O ⇒ atanO=O~tanO o ~u 1ll,111tl ..,
Otmd,/
-
_ 1 ~ II !:,l'\'2 l) d0
, 2 2 2 b
(a-+ a tan 0)
1 Isin20]n/2 = n
• cos20) dO
2a 11, [
8•
2 o 3
411 b
19.36
2 2
Substituting the value of r1 in (i.), we get 2 7t
1t 1.,2-a2 1t _ _ rr. _ (2i+b -cl)= 33(a+ b)
1 =--+ -->< - 3 - 1 1 4a b
2ab3 /J2 411'/I 4a' b"
. rr/2 l __ dx. Then,
ALITER Let / = j ;;
(11-cos2 x+b sin x)
-- 2 . 2 2
0
n/'l
-- ----d;i;
sec4 ,.. J -(1~tan 2 ·x) sec2 x dx
,_ rr/2
T=
J (a2+b 2 tan2 x}2 - 0 (a2 +b2 tan 2 x) 2
0
LRt I> tan x =a tant. Then, 2
2 2 . asec t
d(btanx)=d(a tanf) ⇒ bsec xdx=asec td!: ⇒ dx "" bsec 2 x dt

o m
Also, x=O ⇒ a.tant=btanO ⇒ tanl = 0 ⇒ t =O

.c
It It It
and, x =- ⇒ atant =b tan - ⇒ tant = oo ⇒ t,, _

du
2 2 2
2
f) 2
tl
oe
11 2
'n/ 2 [1+ b;tan 2 n/2
( 1+a b2
- tan
I= Jo -"-;;--,,.---<- x asec t dt = J _,_.......,._...,_.....£.. a 2
ch
(a2~a2tan 2 t)2 b o 4 4
a sec t
x-bsec tdt
te

n/2
I- 1 n/2
⇒
[ 2 2 2 2
- a3b3 • (b cos t +a sin t) dt =
1
3 3 J (2112 cos2 t + 2az si.n2 t) dt
m

o 2a b
n/2 0
fro

= I= 2a;b3 l {b\1+cos2t)~a2(1 - cos2t)}dt

d

n/2
de

1
⇒ I = 2a3& 3 J {a2 + b2 + (b-? - a2)cos2t} dt
0
oa

[b2 -a2) . 2tlrt/2

nl

l=-l-[(a2+b2)L
2a3b3 .,.. 2 sm
w

0
do

I = 2a'b3
_l. (a2"' b2) ( -
n -0 ) l- [ b2 _ 2) .
_ _!:_
2 2 (sm n -sin O)

I ~ (a2 -t b2) n:
4a3b:f •

EXAM PLE 13 /
f
_v11/u11te . I x (tan I x):Z tlx
(J INC.TR r EXE
l
SOLUTlON Le t / - f X (tan I x)2 rlx • 'l 'IIC't1
() '
1
I = f0 (tan JIx/ x dx = l(tan -1 x)2
l[
x2 1I -fI 2 la n--l
2
\C
,\.2
o O 1+ xix 2 ux
EFINITE INTEGRALS
19.37

I
',h

l - n.., -
3~ 0
2
fI { t,\t\ 1'

m o
.c
du
oe
ch
te
m
fro

Jf./4
If 1 =
• 11
f tan
O
11
x dx, prot•e that 111 + 111 + 2 = -!-
II+ 1
d
de

\~e ha,·e,
n/4
J tan" x ⇒
oa

111 = 111 .. 2 -
0
"I
nl

Jf./ 4 4 11 2
= J tan' x dx + J tan
w

I)
0
do

/
11
+ 1,
1
~ 2 :a f
11/ t (
tan" x t tan • - .,
? )
,fa
0
11/4 2
/
11
.,. /
11
..
2
= J tan'' ., (I 4 tan xJ d1
0

1.. + 111. 2 =
Jf./4
r t,111
11
,\ li4'(:2 \ ,11
0
t = tan x. Then, ,ti = s.:1:2 .Y .Ix
n
n ) l,111
t,m O - 0 nnd, X
x :O ⇒ t ;:
4

I 11•1
,11
I,. + 111 .. 2 = J
n
dt = - -
n+l [\ 1· .JI]
,:
n ~ 1
19.38 1 l
1t/4 1 1 +I ' I J ,... · from
+
If I11 = -f tanu x dx, +
show t/Jat _-1
2 14. ' ]3
, + ls , I4 6 5 7

F.XA.MPLf: J5
0
Find the common difference of this 11rogressiort.
SOLUTION We have, .r./4
1t/4 J tan11 + 2 x dx
1
11
= f tanu x dx ⇒ 111 + 2 =
0
0
7t ;t/4 11 2
In+ In+ 2 = J tan 11
xdx + J tan + X dx
0 ()
11/4
11 2
= tan x (1 + t,tn x) dx
I0

m
l,1 + In+ 2

o
,i/4

.c
11 2
=
⇒ J0 tan x sec x dx

du
1

oe
⇒
= J t" dt, where t = tan x.
0

•[~•::I •":
ch
, • • 2, 3, 4, 5, ....
te
⇒
1
m

1
I J1_ .,_ Iti+ 2 = n + I, n = 2 3 4 5 I'- I I I• • • •
fro

1 1
⇒ l2+J4=3,--
l3 + 15 = 4' 1 +1 l = S, - 1- - 6
Oearly 3 4 . 4 6 ls + I7 - ' .....
d

· ' ' J ' 5, 6, .....

1 1sanAPwith common
· ,
de

difference ·1
Hence,-.- - _ _ 4 1 6 ·
I 2 + I 4 ' I + ls' l "+r , ···· .IS an AP with common difference 1.
oa

3
nl

LEVEL-1 [
w

Evaluate
4 thefi //owmg °
. integrals:
do

1 J
2 x2
X
+ 1 dx IN CERT) 2
2
f 1
• X ( rix
2 I l+ logx)2
3x n/2
3· I1 - 2- dx f() ..,"-- 'I
coi;--
9 x -1 4.
a
\" -~ 3-.. dt
5. J ' 2x dx
I
,,.,· Slt1 \'

0 Va + x2 6. I
{) I ➔
,,,_ rlx
e~·'
1 ?
x-
7. I0 xe· dx
8.
;1

J c:os-(log
- __x·).
dx
l X
2x
9. I0 -.L +- 4x dx 10.
ll

J '\;
0
ra2- ·- x2. <X
j
EFINITE INTEGRALS
19.39
rc/2 n/2
11. J.fin IP COSS$ d<I} INCERTJ 12. I 1'0~ X
rh
0 () 1 I ~in 2 ,\ '
rc/2 .
Slll 0· n/ •'
t3. f J1 -icosO d0 14
·
J co:; .\
dx
0 O 3 I4 i;in 1·

l {tan -lx 2
15. f --'---2
0 1 -+ X
dx "I 6. J x ✓x 1 2 ,Ix INCf.H1 I
0

11.j tan
1 2x ldx
Il1 -x IS
.
1t/2 .
J $ 111 X COS X
4 c/;\
2
o o l + sin x

m
1t12 d n/2
J -acosx+bsinx
x n, b > 0 20. J -5+ 41si.n x clx

o
19.

.c
0 0
1
f .
It
sin x

du
1t
f
21. • --dx 22. dx
O SIJ1 .\+COS X O 3 t 2 Sm X+ cos .-r

Xp"(
oe
1/2 . -1
23.
1
f tan - 1
x dx
24.
ch j 2
dx
o 1-x
0
n/4
n/4
f (.Jtan x + ✓Cot x) dx [CBSE 2012] f 1 +ta,;tcos3 x2x
te
26. dx
25.
0
-0
m

rt/'l.
1
f-
It
28. f 2 . 2l 2 2dx
fro

27. -- dx
5 +3cosx ...o fl sm 1 + .Li cos x
0 1 - 1
ll/2
f x+ .
30. f tru.i ;°' dx [NCERTJ
d

29. sin x dx LCBSE2011]

o l +x
de

1 +cosx
0
1
rr./ 4 .
f tan - 1 X dx
oa

-i;mx+cosx [CBSE 2015] 32. X

3L
J0 _ _..:.,:.~dx
3+ sin2x 0
nl

3
1 l
33. f - -x
2 34. Jl 24· x2 4 rlx
w

4 2
dx
· 0 (1 + X )
0 X +X +1
do

1t/2

35.
12
f x( x -4) ax 113 36. f x2 sin x d~
0
4 'I
l - x2
37. f ~1 -x dx [CBSTI 20041 38. J 2 2 dx
o( I ~.\)
O 1+ X
n/2 2
l Ll- COH,\ I I CUSl". 20 151
39. J 5x ,p? + 1 dx 4 [NCW{TI 40.
J0 . 2 • I ,\
l + 3sm .1
-1
1t 4
41..
'11/4
f sin 3 2t cos 2f dt [NCliRTI
42. J 5 (5 4 cm; 0) l/ sin Od 0
0
0 2/3
(1t) 2 3/2
43.
ff/6
J cos- 3 2 9 sin 2 9 de 44. J .Jx cos x dx
0
0
 r,./ 2 5 d
19.40 cos X X
46. J
1 dx [CSSEZ0031 0
2
45, f x (1 + log x)2
]

9J ✓x - dx
47. - 3/ 2 2
(30 - X )
4
2
"
in 3 X (1 + 2 COS X) (1 + C05 X
) dx.
48, f
0
S

n/ 2 - 1 ( . ) dx
49, J 2 sin x cos x tall SJJ1. X
0 [ LEVEL-2]

m
1
f (cos- 1 x)2 dx

o
11/ 2 51.
f sin 2x ta.11- 1 (sin x) dx

.c
so. 0

. ,--
du
0 n/ 2 .,)1 + COS X
53. J (1-cos x) 3/2 dx
J sin- 1 ~___::_
oe
52. dx 1113
a+x
0
ch
~J x
1a2-x2 55. aJ ~a-x
- - dX
54. 2 d,x a+x
te
2 -a
0 ~a +x
1t/2
m

rr./ 2 tan X
sinxcosx 57. dx
56.
f cos2 x+ 3cosx+2 dx J 2 2
fro

o l+m. ta11 X
0
1/2 1
1
58. J - dx 59. J
d

o (l +x2),'1-x2
de

1/3
Jt/ 4 · 2 2 1t/2
J
sm xcos x
f vcosx-cos 3 x(sec2 x - 1)cos2
oa

60 • dx Gl.
o (sin 3 x + cos 3 x) 2
0
nl

n/'2.
62. J cos x dx
iJ
w
do

O (cos~+ sin

1, 1 tog 17J
2 r5 log 2
2. - -
log 2e 3.
1
6 (log 35 - log 8) 4.
5. a(../2 - 1) 6. tan - ·1
e 1t
7. :__- ]
4 8.
2
9. ~ lo.
7t a2
4 4 I I.
-231
64
12.
1t

13. 2(./2-J)
14. ~log( ~:✓~J 15. - 1t3/2
·1
4

12 16. ·t§_ (2 +
15
 19.41

18.
1t
8 log [" + 11 ~ Ja2 ' bl ]
,, _..,, Ja2 +bl

22. II n I
23. log 2
i 2
24 l
2-.iJj
1t
2,,. 1
27 1t
8 l

29.
1t .,c2 I
30, 11 log, 3
2 :12 4
1 35, 720
33. log 3 34. 1
2 7

m
37. - -1
lt 1
38. - 39. 4✓ 2
2 2 3

o
.n 1

.c
-12. 9../3 -1 l3. 3
8 4

du
4,:,. lo~2- 46. S 47. 19
1 + log 2 15 w
4· 1t 1 so. -1t
oe
1 51. it-2
ch
2 2

53. 1 a2(~ -½
) 55. l'tll
te
54.

{3_
m

58. ....!_ tan- 1 59. 6

.[2 V3
fro

61
8 62. -2[2-;-1] 1
d

21 2-n
de

_ _ _ _ _ HINTS TO NCERT & SELECTED PROBL;;A1S

oa

4
1
[Jog (x 2 1 1
I
1
nl

+1 == (lo~ 17 - log 5) = log

2 2
w

2
do

1t • 1t
0 ⇒ t == sin O = 0 an d ~ = 2 o I == "'° 2
1 I
I .Jio-t2>2dt - J J1<1 2, 2 .,,-'i.t1
0 0
f 2 13/2 _! ,712 ➔ 2 ,1112 l' 2 I 2 hi
231
L3 1 11 j0 :1 7 II

x .Jx+2dx.Letx+2-= 12. lhl•n,,h 2tdt

2
_- = 0 ⇒ 12 = 2 ⇒ t • Ji and, x == 2 ⇒ t == 4 ~ t = 2
19.42

T = f
2 2 2 lt22tdt = 2
(t - ) \ fi
(t
2
I
4- 2t2) dt "" 2 [~ - 2 ;'

ff.)
t-v2
32 + J6 Ji
ti ✓--'i -4- -
32 16J_ 4 -
l = 2 [(-- -3) 5 3
( li•)] "'2 (!~ 15 +
l
s;2 "' - 1s
⇒ ltJ/4 -tan~,3 .
5
dx = 1-2
1t/4
f tan 3 .t .sec2
1
x d x --
-2
f
O
t 3 dt 1 w heret = tanx
')61 -
~- - -x •0

02('10S[t~ ]1= ~ (~ - 0) = .!_

⇒ I=-
2 4 24 8

m
0

tan - dx and let tan - 1 x = t. Then,

o
1 1 .,
30. Let f = J - l+x2

.c
0

du
d (tan - 1 x) = dt ⇒

oe
Also, x = 0 ⇒ t - tan- l 0 = O and ' x=l

[l it/ 4 2
ch
rr/4t2 K
I
= .r tdt = -2 - 32
te

0 0
. x+cosx sinx+cosx dx
~ _'Jt/4
J sm 1t/J4
m

dx = - 2
.>l. Let I - 3 +sin 2x 4-(sinx-cosx)
fro

0 0

= I =41 [ log 1\22 +-sinx+

sinx-cosx!J
co.sx, o
rr/( =_!_Jog.!.. == .!..loge 3
4 3 4
d
de

1 1
3- f -Jr -.=~dx
1-x l
~~dx
=J - 2 +-1J
2 {j-2x 2 dx.
oa

I. - /---;:;
o -v1 -x- o i/1 -x O, 1- x

l 1
nl

=[sin- x
1
J: +½[ 2.f1-x
2
== ( sin -l l -sin- 10 )+ (0 - 1)"' i-1
w
do

1
4
39. Letl = J5 x
1
{x5 + l d:1: and lcl x5; I - /, Tlwn, d(. 5 . l)
1 1

Also, x = - l ⇒ t = 0 and, x = l => /· = 2,

312 2
T= J .Ji dt = 2_ [1 ] - '!: x 2 3/2,,, ±-./2
0 3 o 3 3
FINITE INtEGAAlS

.
Let I = J 1..50s
'11:/2 2
x
3,:-r-tfx.Thcn
19.43
o -t sm x '
'lr/2 2
I - I -~'"
. 2 - d:f. D1cn
1
0 + 3sin X '
'11:./ 2
I- J -sec:?- \~(, 2
sec2 y
-
ac,0 • sec x + 3 tan 2 x) dx [Diving Nt and Dr by cos4 x ]

I=J - 2~ - ,,~ it
O (1 + t ) (1 + 4t~)
1 , Wl1cret = tan "
.,

f (-1

m
4 1
I= .!_
3 l+f2· --l+ t2 ) dt
· ==-3 [ tan ~1t - 2tan 1 2L] "'
4

o
0 0

.c
I = -½(;- rr)=:

du
111 4 ,, I
f sin 3 2 t cos 2 t dt ,md le\: sin: 2t = u. Then
oe
• Let I =
0 I
I' ! I
ch
'
tl(sin 2 t ) = du ⇒ 2 cos 2 f dt = dti ⇒ ~os 2·t ,d t •~ !_
• 2)!
au
te

Also, t = 0 ⇒ 11 = sin O = Oand,t \ .

m

!2 f 11 3 du [u
1
!.
fro

4
.. I = = I_ ] =
8 o 8
0
d

9
,.Ix
I d
x. Let 30 - x
3/'1.· ' ' 3 r:c •
= t. Tb,ep , ,,- ": x dx . F dt.,
de

.I =
(30-x 312) 2 , 2 '
4
Now, r = 4 ⇒ I = 22 and, x ,⇒
oa

= 9 f = 3'

1.!.. I[!.]3 _!._) "'

nl

l-= - 2 dt = = 2 ( t, - 19
3 22 t 2 3 f 22 3 3 22 99
w
do

4 PROPERTIES OF DEFINlfE INTEGftALS '

~ ' we will study some fundamental properties of definite in tegrals which ,ut:' \ e ry
an evaluating integrals.
A.1 PROPERTY I
ATEUENT fb f(x) dx = fb /(I) df i. 11.1 integ i-atio11 is i111/epe11ck11t of f/11• clwng<' of vc1rhil 1le.
a 11

Let c!>(x) be a primitive of f( x). TI«' n,

1
"'X
{c!>(x)} = f(x) ⇒ d {q,(t) } "" /(I).
dt
'1 b
,J /(x)dx = [<1> (x)] c 4> (b) - ,j> (a)
...(i)
« a
h b
...(ii)
/ f(t) dt "' <I> (x) c <j>(b) - <j> (a)
a
19.44

. and (li) ' we obtain

Frain (J)
I n
{ f(x) d., = _f(I) dt J
/1
a n
19 4 2 PROPERTY II /1 .r / (.Y) dx
.. .
(Orrfrrof,11/c):nl/lt>II ·
. )· J f(i)· d.\ 1, . .
STATEMENT ' 11 ., . /111! c/1n11gPs by 1/ltn /.Jr, sign
.' t/"''/// 1l/'(' //1/.(1, .(·l1n111;<'
" ' C
d then , s va
i.e., if thr limit;, o/ ,1 d,'/1111'.i' ''.'.
f f(r) Then, s,
.·
rR(.l(lf L12 tJ.'x)bea
~'\• pnm1tivc o . · ·
1,
J ((x) dx 4>(b) - <l>(n)
,1

m
and, -r ((x) dx = - l(<l>(a) -4>(/i)I = 4>(b) - c~(n)

o
.c
(,
b a
f f (x) dx J f(x} dx

du
=-
n 1,

oe
19 .4.3 PROPERTY Ill b c b

J f(x) dx = j. f(X) ';{X + J f(x)·

ch
s~ATEMENT (Additivity): . dx, where a< c < b.
a C
te

PRllOi- Let ch (x) be a primitive of f(x) . Then,

m

b
J.f(x) dx = <!>(b) - cl>(a)
fro

a
C f>

and, [ f(x) dx + J /(x) dx = [<!>(c) - cl>(a)] + (cl>(b) - <l>(c)] = <l>(b) - <!>(a)

d
de

a C

From (i) and (ii), we get

b c b
oa

J f(x) dx = J f (x) dx + J f(x) dx

a a c
nl

The above property can be generalized into Nu! following form

w

GENERAUU>- TION
~ '1 '2 b
do

J f(x) dx =J /(x) dx + Jc f(x) dx + ... + Jc /(x) dx,

" 11
where r1 < cl < r < c_1 ....... ,·,,
2
1 11

ILLUSTRATIVE EXAMPLES

EXAMPLE l Eva/11n/1•:
l LEVEL-1J
J
0) J /(x) dx, wl,ere f(x) { 11 I
- I ( II) J /(1) riY, wit,,,.., I ( 1)
I
SOLUTION (i) We have,
l - 2x, :r < 0
f(x) ={ l 2x, x ~ O
HEMATI
olflNITE INTEGRAL.$
19.45
1 0 I
J .f(x) dx = I f c~·) dx Jf(x) dx I

- 1 1 O
1 0 l
J / (x) dx = J (l - 2x) dx + J(1 ~ 2:,;) dx [By using the definition off(x)l
- I - I 0

J / (x)dx=[x~x2 ]0 l +[r+x2.1: = [0 - (-1 - 1)1t[(l+l) (0)1=4

(ii) Using additivity of i\1tegration, we obtain

4 2 4
J f (x) dx =Jf (x) dx + J .f(x) dx

m
1 1 2

o
-t 2 4
J/ (x) dx =J (2x ~ 8) dx + J 6x dx

.c
[Using the definition of f (x)]
I 1 2

du
j f (x)dx =[x2 +sxJ2 + [ 3.,.:2r = [(4+1,6)-(1+8)]+[48-12] = 47
oe
I 1 2

l!ICAMl'LE 2 Evaluate:
ch
1[ I, ,5 I
l
ti) JIsx - 3 I dx (ii) JI cos X l ~;\1 (iii) J Ix - 2 1dx (iv) J e1xi dx
te

Q I fl - l
0
2 4
m

(v) JI x 2
+ 2x - 31dx (vi) J(lx-1 \+lx-21 +lx - 3l)d.t [NCERT]
fro

0 1
2
(¥Ji) f I.r3 - xi dx [NCERT, CBSE 2012, 2013, 2016]
d

-1
de

SOWTTON (i) Clearly,

3

l
-(Sx-3) when 5x-3<0 i.e., x<
oa

Q. 5
ISx - 31=
Sx - 3 whe11 Sx - 3;:;: 0 1.e.,
' x ~ ·3
nl

5
w

tlit graphof y = ISx - 31 is shown in Fig. 19.1.

do

j
l
,!.. - -
(l , ll} X
X'

y•
Fig. 19.1 Graph of !I ~ 15-" - 31
 MATHEMA

[U~ing additive pr

13

m
10

o
.c
(ii) \Ve know that

du
n
cosx when 05x~
r 2

oe
1 cos xi
= 1cos n
x when - :,; x
2
~ n
ch
y
te

(0, 1) =
y= !cos.,! (r-, 1)
m
fro

X' 0 (it/i, 0) (rr, 0) X

d

= •
de
oa

y·
Fig.19.2Graph of y = cos xi
nl

1t

JI cos x I dx. Using additive property, we obtalt1

w

Let/ =
0
do

n/2 n
I = JI cos x I dx + J I ens ~- I clx
fJ 71/2
lt/2
I J<¼x dx ◄
!/
f(
•ii
('Oh X) Jx

,r/2 IJ
I fsin .1 Ju lf,111 xJ
,l/2
I I I

(iii) Clearly, Ix 21 J x 2 w lwu 1 .!. u

l ( 1 2) w l11•11 .!. ()
The graph of y =Ix 21is shuwn in l 'ig l':1..l,
 y

(-5, ?)
y ~-(.,· 2)
'

x· (-5, o)

m
Y'
Fig.i9.3 Graph of)l= I~ 21

o
5

.c
I= JI .x - 2 I dx.

du
Usil'lg additive property, w e obtain

oe
2 5
I== J [.x-2\dx+ f \ .x - 2ldx ch
-5 2
2 5
1 = J -(x - 2) dx + (:x - 2) dx f
te

-5 2

•·[,,-,:L •l': -,'],

m
fro

I=(2+ ~J+(~+ 2)==29

d
de

x, when x2 0
oa

(iv) Clearly, I x \ = { - x, wen

h .x< 0
nl

The graph ofy ==elx\ is shown in Fig. 19.4.

w

l
let l = J elx\ dx.
do

-] y

Y'
1 1
F,g .19.4 Graph ol e ·'
 MATHEMA

19 .48

. additive proper·tv
, , we obtain 2
Usmg o
11·1 dx + [
1
e
1xl
dx
j / (x)
I"' J e • I

-01 0 ]1 1) + (e 1 - 1)
2
= 2e - 2
1 [ ]0 [
-x + e X == (-l+e I (-x
I=f e- -'dx+J idx= -e -1 O I
⇒
-1 0
x2
--2
_2f I t2 + 2x - 3 I dx
(v) Let I - for different values
. s of x 2 + 2x- 3
2
Clearly, X + 2
o
3 = (x x- + 3) (x - l ). The sign

m
shown in Fig. l 9.5.

o
00

.c
-«> -3 0 z1
Fig. 19.5 Signs of x + 2x - 3

du
1 O<x<l
-(x2 +2x-3) , ·f
rl
oe
·
I x2 + 2x - 3 I = 2 , if 1 s x s 2
(x + 2x - 3)
ch
Using additive property, we obtain
te

1 2
1 = J I x2 + 2x - 3 I dx + J I x2 + 2x - 3 I dx
m

0 l
=
1 2 2
fro

1 = J- (x2 + 2x - 3) dx + J (x + 2x - 3) dx

0 1 l 12
d

=-lx: +x2-3xl+(x: 2
de

::::, I +x - 3x
1
oa

r= -{(½ +1-3)-o}+{(l+4-6)-(~+1 -3)} 5 2 5

= -+ - + - =4
3 3 3
nl

j (1 x - 11 + I x - 2 I + I x - 31) dx and f (x)

w

(vi) Let I = = I x - 1 I + I x - 2 I + I x - 3I
1
do

Here, we have three critical points, namely x = 1, 2, 3. When these points are m,uked
line, it is divided into four par~ as shown in Fig. 19.6. 'Therefore, to remove the modul
we consider the following four cases:

1 2 3
Pig. 19.6 "'
(i) X <1
(i i) l S x < 2 (iii) 2 s, < 3 c1nd (iv) y 2:: 3.
-(x-1) (x-2) (x 3) -~x , 6 ii'\' 1

f (x) = (x - 1) - (x 2) (X 3) -x l· 4 , lf l s x < 2
Le
(x - 1) + (x - 2) (x 3) - x if 2 $ x <
3
(x -J) + (x - 2) +(x -3)
= 3x-6 , if x~g
Using additive property, we obtain
 19A9

(U ng (1J

o m
.

.c
3
y f tx) = x - x = x (x -1) (x + 1)

du
stgns off (x) for different values of x are shown in Fig. 19.7.

oe
+ +
00
-1 0 1
ch
Fig. 19.7 Signs of j{x) for different values of x
te

f (x) > 0 for all x e(- 1, 0) v (1, 2) and, f (x) <0 for all x e(O, 1)
m

f(x) I = { f (x) , x e ( -1, O) v (1, 2)

fro

- j(x), x e(O, 1)

x3 -x i = J x:-x,xE(-1, O) v(l, 2)
d

1-<x
de

-x), xe(0,1)
0 1 2
= J !x 3 - xldx+ J !x 3 - xj dx+ Jl x3 - xl
oa

1 dx [Usingadditivepropeny}
- 1 0 1
nl

O 1 2
I = f (x 3 -x)dx-J (x 3 -x)dx+ J(x3 -x) dx
w

-1 0 1
do

,.[{-':L-[': ,:i: •[': -':I

1 = (14_IJ
' -rl 1)+(16 4)-(1 I)=3 +(~ 2),. ll
2 42 44 42 4

[ LEVEL· 2l
3~
If a > 0, find f Ix 2 2
- a I dx.
0
LUTION Let f(x) = x2 -a2. Then, J(x) = (x ti) (x r 11). The sign o[ /( t) for diffownt Yalues of x

shown in Fig. 19.8.

 •
19.50
____. . ~o~ 3a

- -a fig. 19.8

m
Ev

o
We

.c
du
oe
ch
1,
1
te

n/4 J
e-x !sin 2xl dx = ~ (4 + en/ -e
4
l:XAMPLE4 Find Je-x sin 2x dx.Hence show that f
m

-n/4
fro

SOLUTION Lctl = J e-x sin 2x dx. Then,

I U

r = - 1 e-x cos2x - J(-1) e- x x -

d

l cos2xdx
2 2
de

1 =--1 e-x cos2x - -1 f e-x· cos 2xdx

oa

2 2 I II
=> 1 -x cos2x--
/ ---e 1 -x sin 2x -
1 { -e J (- 1) -x l . }
2 2 · e x s in 2x dx
nl

2
2
Je-x sin 2x dx
w

=-!.e x C(JS2X - l e-x stn

I
2 4 · 2x - l
4
do

_ 1 -x
I -- 1 -r . 1
e cos2x--e sm2x -- 1
2
5 1 X 4 4
1 e (il L
4 =4 (2cos2x 1-si.n 2x)

I 1 -x
= e (hin2:r ➔ 2cos2:r) 1-C
5
..• .
I sm2xl= 1-sin2:r,if - n:4 ""x< ()
sin2x, iJO s:r., n
usmg
' additive property, we obtain 4
lt/4- 0
I= f e-xlsin2xl dx = f ,,-, I . R/4
-1t/4 sm2:rld~·+ J -x
" 14 0
e · Isin 2xl dX
 19.51
O a 4
J " ' ,m 2x dx + J e .t
sm2rd.T
% 4 0

1,; e " ("-m 2x .. 2cns2.l.) I

e 2cos2t
11 1 l S

E ·• J log xldx.

m
l t
\\ e knm, that logrx <0 for x E(O, I) and kig, z.0 for x ~1.

o
T

.c
-log, 'I'. , til e<x<l
X ""

du
(

oe
- f log x dx Then.
1 r
ch
1 r [Usmg additive property}
- J logr x dx • I Jl logr xi dx
te

• t 1
1
m

t
- j - logt x dx + f log, ;r dx
fro

t 1
1 t
- - f loge ::r d:r - J logt x dx
d
de

1/t 1
.r(logtx-1) 1 -1-[x(logr x-l)lrr__l' f IOSr' J:r=x(iog:x-1)
oa

,. 1
--lit - 1
1 -L-1..-; r "'] ..,
nl

- l(O-J)--;(-1-1) .,.. ..e(1-1)- 1(0- 1) ,. T(O+l) .a 2-:

w
do

2
dx (ti) f (::r2I dx I)
0
l1 U ing add1t1vc proper!), we obtain
1 2 1
j f
Ix! dx .. [x1 d.r t- (xi dx .. Ix! ,11 f f
0 I Z
1 2 'J (2 t) t(b 4l
f f
Ix) dx = Odx + 1 . dx ..- 2. dl o " J J
0 1 2
Using additive property, we obtain
2 I 2 ,:J 2 ~
f (:r2J dx = f ti)dx-. "JI:?! ,h " J ti1 dx,. f tx-·J dr
0 o 1 ,2 "')
19.52
13 2
1 Ji 'f 2dx+ r 3dx
f [x2] dx = f Odx + f 1. dx + Ji h
0 0 1
[ ]2
/3·+3X
[x l
2 ,J2
J rx21 dx "' 0 + + 2 [ x J.,12 ./3
0

⇒ J2 [x2Jdx "' 0 + ( li. -1) + 2 (v;;,f"ii3 - ",1.Fi) + 3 (2 -,/3)

0

r;:; 6 31'J=5-.fi-.f3
=> I rx2ittx = ft.-1+2fi-2v2+
2
- ";;,

m
0

o
(iii) Using additive proper ty, we obtain

.c
15 1 ../2 1.5 2
J [x2] dx = J [x2] dx + f [x2] dx + J [x ] dx

du
0 0 1 ../2

oe
15 1 fi 15
⇒ J 2
[x Jdx = J Odx + J 1. dx + J 2 dx
ch
0 0 1 /'i.
]fi +2 [x ]15
te

⇒ 15J [x ]dx
2
= 0+ [x _ = 0+(..fi_ - 1) + 2(15 - ,fi) = 2 - ,J2 -
1
O 1 v2
m
fro

E>..AM.PLE 7 If[· ) denotes the greatest integer function, then find the value of J2 [3x] dx.
1
d

SOLUTION We observe that 3x e[3, 6) whenx e[l , 2].

de

l
3, if 3 :,; 3x < 4
[3x] = 4, if 4 $ 3x <5
oa

E.
5, i£5$3x<6
nl

3, ifl $x < 4/3

w

[3x] = 4, if 4/3 $ x <SI 3 '

1
do

5, ifS/3 ~ x <2
2 4/ 3 5/3 6/3
{ [3x] dx = J [3x) dx + J [3x] dx + J [3x] dx
J 4/ 3 5/3 [By usi ng ,ldditive
2 4/ 3 5/ 3 6
/3
I [3x] dx = I 3 dx 1 J 4 dx 1 J 5 dx
1 1 4/ '3 !j/ 3
2
⇒
J [3x]dx = 3(i - 1) +4( 5 4) 1 s(63
EXAMPLE 8
1 3
Let f (x) = x - [x], for every real n111nbe
3 3
3
!)- J +4+
3
9- = 4.
1 r x, where [x] 18
•
equal to x. Then, evaluate J f (x) dx. · the greatest integer I
- 1
 19.53
L!JflON Clearly, x-[x] ={X-(-l) =x+ 1, if-1 $x <O
l
.,-O=x 1'fO ' S.,1:'<l

J (x -[x])dx"' -11 (x -[Yl)dx + ft (.·' _ [x]) dx

- 1 (By using the additive property]
0
0
= f (x + l) dx + J0 (x -0) dx = [(x +21)2.] o + [x2]
-] 2
1=2+2=
l 1
1
- I 0
lt is evident from the graph of y = f (x) that

om
A(O, 1) B(l, 1)

.c
du
X' X

oe
(-2, 0) ch
te

Y'
m

Fig. 19.9 Graph of f(x} =x- [x)

fro

•,
1
Jf (x) dx = Area of 6 OLA+ Area oft. 0MB
d

-1
1
de

=2 (Area oft.OMB) = 2x - (OM x MB) =1 x 1 =l .

2 .
oa

EXAMPU: 9 Evaluate:

'J 1[ 2] 1J (3x - .3]

tan
- 1 ~
dx
nl

(1) tan - 2x dx (ii) 2·

1 -3x
1 - x 0
w

0
do

2 tan - 1 x , if - l <x<1

- 1t +
2 tan - l X , if X > 1
1t+2 tan - 1 x , ifx <-1

I == f tan- 1 (-2x- 2Jdx

../3

I
0
O

= f1 tan - 1 (
1-X

2
1-X
2
l
dx + /3 tan
x
t
f 1(
1 -x
2x ] dx
2
lUsing additive prnperty]

l / 3 1
I = f 2 tan - 1 x dx + J (- 1t + 2 tan - x) dx
0 l
 MATHEMATl

19.54
1 .[3 .[Jj - 1 d1
J 2 tan - 1 x dx + J - 1t dx +
l
I == 2 tan x .

0\ 1 1
1.f-3 2
,1 -
,
1 ,. dx - 1t
/Jf
1 . 'iX
1 = i2tan- xdx+ l tan ., 1

o m
⇒

.c
du
= 2(~J3-½(log4-log l)1 -n(J3- l) oe
ch
1
te

r = 2 .t..}3-log4 -1t (J3- 1) == n(l _ ~-l -log 4

= 3 ~ -.i3)
m

3tan- 1 x if- l /J3 <x < l .f3

fro

(u.. } 0 eary,tan
l -1l3x-x3J-
J - 3x-
? -
- n+3tan- 1 x, ifx >l/.j3
..., = -
•
if X <- 1/-!3
d
de

l =f tan-1( 3x-x:\dx
l
oa

o 1- 3x J
1/J -•
nl

= I= tan-1 J3x-x: \ dx + [ tan _i( sx-x3\ [Using additive pro

J l J
w

O ( 1 - 3x 11f'5 1 _ 3 x2
do

1/ ✓3 1
I= f
0
3tan 1 xdx+
Jr (-n+ 3 tan 1 X·) ,a
_,
1/.fj

⇒
I = 11/
l
J
O
3 tan 1 x dx +

I
1/ ../3
f 3 tan l x 1/x l 1 JI _Tt dx
11 ✓ 1

⇒ l = f3tan lx 1t f dx
0

rLx an
1/..j?,

. ta: 4
"') J j{.t) d.t, w

l
1 - 3
- 1
-1
x-21 log(l + x2) Jo1, 1t X
1
~
J I
x d.Y I
L11/ ,/3
- x tan 1 x _.!_ log (1
2
 19.55
I =
1) 3 log 2
4 2
tan x
.,,,.,..,.,LE 10 Prove that f di I
tut r
J I dt
l/ r I ~ ,2 2 I /or all x for w/Jirh tan x and cot x r1re
1/r t (1 I 1 )

tan .1·
f CtJt .'(
Let 11 = f ~ dt and 12 = J _.!_
1/e l + t 1/el(l+t2) dt.
1 1
Hlno t = -
•--o II and dt = - -.2 du in ! , we get
2
u

m
tanx 3 tanx
Ie u x-~
f ~ du

o
12 = 2 2 du= -
1 +u u

.c
e 1 +u
tan x

du
t
12 = - I - - dt
2
] +t
[·: Integration is independenf of change of variable!

oe
ch
te
m
fro
d
de
oa

1
= - x-2 1oge = l
nl

2
w

_ _ _EXERCISE 19.3
do

- -- - - - -----;==
LE
~v==e=== 17
L-=:= _ __

&aliu,re tlie following integrals: ,

4 { 4x+3, if 1 ~ x ~ 2 (iv) j I·',·. I dx
l. (1) J J(x) dx, where f(x) = 3x + 5, if 2 < X < 4 I
1
f sin x , 0 S x < rr.l 2

(ii) r
9
f(x) dx, where f(x)
J l
lex - 3
, 1t/2:::1•.;3
3 :s X <; 9

4 7 x+ 3 , if 1 sx!.3
(iii) f f (x) dx, where f (x) = { Sx if 3 ~ x ~ 4
l
19.56 3
4
3, J1x+ll dx
2. J Jx+2.ldx - 3
!I,
-4 2
l 5- J I2x + 31dx
4. Jl2x + 11 dx -2
10.
l 3
li
2 7 J 12x - ll dx
0• J Ix2 - 3x + 2j dx 1
0 2
6 9. J Jx + 11 dx
s. f J.,+2jdx -2
- (, rd2
2
u. J Jcos 211 dx

m
10. J Ix - S: dx 0

o
1
n/4

.c
2n:
11. J sin .i.j dx 13. J Jsinxldx

du
0 -rt/ 4
n/2

oe
-!
14. flx - 5(dx [CBSE2020) 15. J (sin Jxl+ coslxl)dx
ch
1 - n/2
4 4
J{lx -ll+Jx-2l+lx - 41} dx
te

16 J x-1 I dx
J [NCERTJ 17.
0 1
m

0
1S. J / (x) dx,where/ (x) =IxJ +Ix+ 21 +Ix+5 I
fro

-5

!(1 xl..-J x-21+I x-41)dx

4
d

1':l. - b
de

2
oa

2.C. f(xTll +lxl+lx-lJ) dx

-1
nl

2
: ~ Jx eIxi dx 1t/2
w

2 22. f sin x I sin xi dx

do

- 1t/4
~

23 - f cosx I CQS xJ dx Tt/2

0 24. f
- rr./ 2
(2sinJ x l +cos .xJ) tfa
.
ll
25• J sin -I (sin x) dx lt/ 2
-n/2
26.
f
2
27. J 2x [x] dx
0

1. (i) 37
(iii) 62 ('lV) I
2. 20
-- INTEGRALS 19.57

-52 5. 25
2
6 1 6 40
3
5 10. -"l 11 1 12 4 I 2-,..2
le;
1S. 4 16. S 17
n ft-1
2 2 2
19 - l
20 20.
...., 21 0
8
+
4
0

6
. , __2
2.-- •'- 26
,I
27. 3 2 f'.2
s J
HINTS TO NCERT & SELECTED P OSLE

l;-2

m
4 - 4 r 2
2
1,
f f/x-:ldr=- 1+-2r
1

o
r ... :!. dx- j -(x + ~) dr + +- - +2.xl
J-2

.c
-4 -4 ._ -4

du
oe
ch
-11 : l
J 2r - 11 dx - j - (2.r ... 1) dx + J (2x + 1) dx
te

-1 -1/2
m

2 - J/: "'t

J 2x - 3j dx = J - (2.x: + 3) dx + j c2x + 3) dx
fro

- 2 - 3/2
3 V3 3
d

J 3r -11 dx = J - (3.x: -1) dx - J (3x - 1) dx

de

0 1/3
oa

: 6
.x:+ 2 d.x:s J -(x - 2) dx .. f (x ~ 2) dx
nl

6 -2
6
w

1 2
J .x:..-1 dr-a J -(x- l)dx~ J (x-l)dx
do

2 -2 - I
2 2
J lx-3ldr = f -(r -3)dr
1 l
K/2
~ 2
[ lcos 2.x1 dx=
r./4
fI C05 2xl dx +
f (-cos 2x) dx
0 O R/1
21t
2!t
J Isin x1 dx =
1l

{
~in r dx ... J (- sm r) dx
It
0 0
"''4 0 J _ .x dx + " ' ' sin x dx J
51n
J lsinxj dx=
- -::./-1 - ,i/4 0
 19.58

1'
4
- (x - 1) dx + f (x -1) dx
1

19.4.4 PROPEPTV V
STATEMENT if f (x) is II contiiluous function defi11ed on [a, b], then
I> b
J f(x)dx = J f(a+b-x)dx

m
o
" a

.c
["'.,.•(' Let x = a + b - t. Then, dx = - dt
AL<;0, x =a ⇒ t =b and x ::b ⇒ t = a

du
b a
J f(x) dx = -f f(a+b-t)dt
oe
a l•
b b
ch
Jf (x) dx = f f (a+ b - t) dt
te
a a
1, b
J f (x) dx = J f (a+ b - x) d.x
m

a a
fro

b b
Hence, f f(x) dx = J f (a + I, - x) dx
a a
d
de

ILLUSTRATIVE EXAMPLES

I LEVEL-1 I
oa

Prove that: J _ __f (x) _ _ b

nl

TY./U,1l'Lf l
a f (x) + f (a + b - x) dx == ~.
w

b 2
= J __f::-:(x-·)'-_ _
do

SOLUTION LetJ
a f(x) + f(a+b-x) d-1:
Then,
b
1 - J- - - L(a + b - x)
a
b 1 (a + b x) ~ (n-----
f +b - (ai- b -x))dx
I =f L(a +b x)
. a f (a+ b X) I f (x ) rlx
Adding (i) andb(Li) , we gel
27 =J f (x) !J (a :!:_ b _ x) 1,
a f (x) I· f (a+ b - x) d.t = f I dx -
l "' b - a a - (b - a)
⇒
2
DEFINITE INTEGRAI..S
19,59
E)(AMl'l.E ~ Evaluate:

Jx
.1 J2 - - - '
,-dx 'lt/0
() 1 ,./3 -x+,/x
. ' (li). J- 2_ dx INCERT, CBSE2014]
n/1, I + ,[cot x

...(i)

[Using Prop. IVj

2 13 -x
l -
- I V
✓X+ [3-X
d' ... (ii)

m
X
1 "

o
Adding (i) and (ii), we get

.c
2 ,Jx +.,j3-X 2' r ' ]2 I
~ ~ d,-c = J 1 · dx = I x
I1 '\,x+ ✓ 3-x

du
2l = = 2 -; l =1
1 L 1 ',

oe
I = 1/2.
Here, a=l, b =2 andf (x) =$..
ch
1
' f(x)
Oearly, 1 = J J(x) + J(11 ~b -x) dx.
te

,1

Using Example 1, we obtain

m

b -a 2-1 1
I = -- = - = -
fro

2 2 2
,,;3 1 ,r/3 ' •;~/sin 'I
...(i)
r = J --.- dx= J6\ISID
7;7~~
d

n/ 6 1 + .,.,/cot X
,r / , X -;t- \1
·
de
oa
nl
w
do

...(in

f(

I = -7t-.
12
Here a = !: I b = !: and f (x) -:c ✓ sin 'X,
6 3
1
0ear J' f~"<l-- dx.
ly, 1 = J(x)-"- J(a+b-x)
a
 19.60

Using Example l , we obtain

b-11
I =- 2=2 3 6
l(1t-~)=1t12
Q EVEL-2]
n/2 .
.nan1 1h
E"\Al\'fl'LE 3 Ev,1/rrnt,•: •f ,•·' ~
1
-n/2
71/2 .
.1"5111 .\
SOLUTION let 1 = f 7 - dx
-n/2 r +1
Then,

m
r. / 2
(-lt/2 + it/2 -x) sin (-;c/2 +rt/ 2 -x)
,= I

o
-;,./2 (-"-+~-x)+1

.c
e' 2 2

du
-;r/? ,r./2 . X
I= 's - -xsin (-x) dx = j. x s~ ~ dx
e-x + 1

l
ex+ l

oe
-n/2 - r,./ 2 ch
Adding (i) an~J~),(we _obtain . x
xsm x xsm x e d
-I - + -- -
te
2]= X
-n/
X
e +1 X
e + 1
2

l)
m

tt/2 [ rt/ 2
21 = J xsinx \
X
~ f f sin x dx
fro

dx=
-'lt/2 e .,.l -n/ 2 IT
d

= [
21 = -xcosx l rt/2
+
1t/2
f [
cosx dx = - xcosx
],r./2 [
+ sin x]
1t/ 2
de

- -n/2 -n/2 -n/2 - n/2

::;, I =1
oa

n/4 2
EXAMrt.t ~ Evaluate. J se<:2 dx
nl

X
- n/ 4 l +e
w

n/4 2
f
do

SOLUTI(JN Let 1 = ~ ~ dx
11/ 4 1; ex
Then,

IL !:, ill~

II f ti)
n/4 2
[ -_ f e
\
X
&t;'C l .t
1,,.r
-n/4 1 +e·
 19.61
dding (i) ru,d (ii), we get
rr/4 [ s e 2 c·,.)
21 = J · -'
+ t'.- sec- .,• it/4

- 1t /4 1+/< ·1
.f('
- , ldY= f 2
sec x dx l,,n y ]
it/4
= 1-(-1) : 2.
11/ ., nO
I = 1.

EXERCISE 19.4
IQ}uate ,,f eacl.7 of Iii<' followi11g integmls (1-·ts)· -
2« .,su,.i: '- •
2n
1. J" •in,'.·+ e- ·~in.,· dx 2• J log (secx + tan x) dx
0 ()
ll J 1·
,,tanx rr/3 {sinJ

m
3.
f ,.-- -d:i.
\ltanx + .Jcotx 4. J -~inx
.Jsmx Jc,osx
- dx

o
• f, 1t/C, ·I

.c
71./~ 2
tan X 17 1
5. f 4 l+e x dx f

du
6. - -x dx , a > O
-11 -a 1 +a

oe
1l 3 n/2 2
l d
I
-"If./ 3 1 + e
lanx X 8. - f-X dx
COS X
ch
.• ;/2 l+e
•/4 11 9 7 r-
te
x -3:x +5x -x:> +1 . .bl. xl/n
f4
ll COS X
2
dx 10.
a x 111
F - - - dx neN,n?:.2
+(a+b-x) 11" '
m

a/Z a ~

f (2 log cos x - log sin 2x) dx ✓x

fro

d
12
· J · Ix + la -x x [NCERTJ
0 0 ... "
d

u. soJ(x~4+~
'{/x+4
7
J
3fx
de

14. 3.,_{-x+"7
" 3~ dx
-x dx 0
-x
oa

a/ 3
15. f !,__ dx
T./ 1 + ✓tan
[CBSF 1007, 2011 I
nl

6 X
w

b +b b
16, If /(a+ b -x) =J(x), then prove that J x f(x) dx =~2 JJ(x) d.r.
do

a a
ANS\\'ERS
-----
7t It
L 3. - 4.
,i 2. 0 l2 l:!
It 7t
s. 2-? 7 s. ~
6.
2
(I
::i
·11
,,
fJ -ii II . In~ ''. 12. ')
10. ")
2
it
t3. ~ 14. 7 15.
12
1 2 2
l.4.5 PROPERTY V II 11

iftlEMENT If f (x) is a continuous function defined on 10, ll] ' !lien f f (.,) ,fr = f0 f (a
0
x) dx.

[CBSE 2019]
19.62
f) -? dx =- - df
,/ (II
Ll'I' \ I, "Jhen, d:r
r1
I 'R(.)OJ'
(1 :> / 0
/\ l;.o, r () > I 11 mid 1·
(l

f l (x) ,fl -i' j (II

ti
I) di

l)
a
,, ti

⇒
f / (x) dx J ((11 - l)dl
(1
0
n ti

⇒
J0 ( (1) ,h I I (!1 r) d.,·
0
,, (1

Hencl', f f (x) ,h = J f ((l - ;r) dx

0 O

om
ILLUSTRATIVE EXAMPLES

.c
I LEVEL-1

du
n/2
Prove f11at: f ___sin x dx = n

oe
0:AMl'LEl
ll sin X .,_ COS .t 4
n/2
ch
SOLUTION Let I = J . sin x dx.
smx .. cosx i Let
te
0
Then,
m

= it/2 sin (; - x) a a
J f (x) dx = J f
fro

1
J0 Sin (ii: )-+ ( Idx
2- X COS 11: - X
[
Using:
0 0
d

,i/2 2 /
J
de

I = COS X
o cos x + sin x dx
oa

Adding (i) and (ii), we o-et

Jt/2
f 0
nl

2/ = sin x
-.- - - - - dx +
n/2
cos x J
o sm x + cos x . -- - - rlx
w

n/2 0 Sm X 1 cos X ,
J
do

21 c sin x + cos x 1t/2

O sin~ cos-;; dx - 1 . rlx J = [x ]"/2 - -7t
- 0 7t
I - 1t 0 O 2- =-
4 2

EXAMPLI 1

SOLU I !OJ\ Let I

Then,

Io _
,r/ 2
T = ) sin (n/2 )
r;;. - - \'
vsm (it/2 - r) J.cos(n/ . n a
f f(x) dx = J
I
2 -~t) dx
~

Using:
- () 0
 12 19.63
" ..jcos x
Io --le===
Jcos x +\sin ,\
d.t
... (ii)
and (ii), we get
~
_J
-
r</2

1-
ySll'\ X
~dx+
tr./2
J r;:;,,
vcosx
0 vcos ~-+,.)sin x r;,-:-~- dx
12
- 0 ySm X + .Jcos X
" [sin x + ,)cos x 1t/2
= JO ✓5fn x + ~ d.x = J l ·dx =[xln/2 =~-0
0 -0 2
1t

m
lgjn X
_ v~ - - - l h = ~
✓sil, x + .Jcos x

o
4

.c
Evaluate:

du
[CBSE '.1007]

oe [NCF.'.RT, CBSE 2002C,03,04,11, 131

ch
n/2 ,
te

(i) Let I = J log tan x·~ ... (i)

0
m

-
a a I
fro

=l\ogtan(i -x)dx Using: J f(x) dx = f f(a - x) J:c

[ 0 0 ~
0
d

rr./2 ... (u.. \

de

= J log cot x dx
0
oa

(i) and (ii), we get

tt/2
21 = J (log tan x + log cot x) dx
nl
w

0 rt/2
12
"' f log (tan x >< cot x) dx = J log 1 dx = J O dx cO
n/2 ""
do

0
0 0
I=O .. ,(i)
,r./ 4
I = f log (1 + tan x) dx
0

rc/4 { (1t )} llMng: j' /"{ 1') ,t.,·

J= f Jog 1 + tan
4-X rlx o
0
,r./4 {
}
tan. 1t/4- tan,:_ rlx ,,
nJ/1 \ I- 1 tan
log 1 · 1 + tan .,
.,·1 ,/_\
I = J log 1 + 1 + tan re/ 4 tan x o ·
0
19,64 rr/4 { 2 } dx
1t/4 l+tanx+ l - tan -X} dx := Jo log 1 + tan X 1t/2
.. J log sin
rJ log { 1 + tan x
= 1t/4 _ "j14
log (1 + tan x) dx 0
x/2
o } J log2dX o
"/ 4{ I (1 + tan
I = f log2-og
x) dx -- o c J log sin
0
0
,./4 "' -(log2) (
I = (log 2) [x ] o -1
%
Evaluate: J
2 1 = ~log 2 ~ J=.'.:.Jog2
8

m
=> 4 0
EXA:'vll'U, 4 £valuate: •

o
12 LetI=J
" sin x - cos x dx o e

.c
(i) J0 1 + sin X COS X

du
2x) dx
(ii) J (2 log sin
tt/2
X-
.
log Slll
e
cos(x-

oe
0
0
,: e-
1t/2 sin x -cos x dx
SOLUTION (i) Let 1 = f0 "' Io e
ch
1 + sin x cos x -COSl

Then, ) and (ii), we

te
•
sin (n -x) - cos(;-x) a a = Jldx
dx
•/2 2
f f(
m

•.· J J(x) dx = 0
f = J0 1 + sin (1t2 - x) cos (1t2 - x ) [
0 0
fro

12 2,
" cosx-sin x dx
= J l +cosxsinx Letl=J
d

=> f
0 0
de

Adding (i) and (ii), we obtain n/ . - X :/,

2
,.72 sin x - cos x cos x - sin x d f =0 => t=
I -:---:----:-:-::-::
oa

dx + .,-- - - - - r 0
2f=
0
l+sin xcosx 0
l +sinxcos x = I t12a-
2a
nl

=>
21
= l
,/2 ( sin X - cos X cos X - sin X ) d
1 + sin x cos x + 1 + sin x cos x x
2,
= J ((211-
w

•/2 . 0
do

sinx-cosx+cosx-s1nxd
=> 2/ = I
0
.
1 +sin xcosx
X=
0
I =
211
J_{(2a
=> I =0 0
(ii) We have,
•/2
f
I= (2 log sin x - log sin 2x) dx
0
1</2 Evaluat
=> I "' f {2 log sin x
0
log (2 sin x cos x)} dx
rt/2

=> I = !{ 2 lcJg sin x - log 2 - log sin l. - log cos x} dx

1t/2 •/2 •/2
=> J = f log sin x dx - f lcJg 2 dx - f log cos x dx
0 0 O
 19.65

=
,c/2
Jo log sin x dx -(log 2) Jo 1 · dx _ nJ
1t/2 /2
(it )
1ogcos2 - xdx (Using pcoperty V)

) dx
= !
rt/2
log sin x dx - (log 2) [ x J: _J
/2 n/~
log sin x dx

= - (log 2) ( i -0) =- ~ log 2 °

1t ecosx
Evaluate: f e'~osx
. + e-cosx dx ICBSE 20091
0

Let I= "r _ ___:_

ecosx

m
1 _ _~ dx ... (i)
ecos·~+e-cosx
0

o
l 1

.c
I

du
COS(7<-.t) -cos(',c-x) dx (Using: f(x) dx = f(a-x) dx
e +e
e-cosx

oe
n.
I= f e- - cosx
- - - d x,
+ e cos x
0
ch
(i) and (ii), we ge t
1t 'It_
f0 1 dx =
te

2I = 1t ⇒ 1 = --:
•2.,
2a n ' 2a
m

1
Prove that: Jf (x) dx 7, J/(2.a·- x) dx.:. [CBSE 2002C]
fro

0 0 .
2a
Let I= J/(x) dx · ' '
d

0
2a -x =t. Then, d(2a -x) '~ dt ⇒ - ,dx·=dt =,,dx =- dt
de

x =0 ⇒ t =2a - 0 = 2a and ;ic =a ⇒ t = 2a - a =a

0 0
oa

I = J/(2a-t) (-dt) = - f / (2a-t) dt

2a 2~
nl

2a
= f f (2a - [·: I dx1
w

I t) dt f(x) dx=-I /(x)

[-.- !
do

I =
2a
J f(2a - x) dx
0
/(x) dx = I/(t) dt1
21< 2a
f f (x) dx = f f(2a - x) dx
0 0
1
LE7 Evaluate: J x(l - x)" dx INCERT]
0
1.
ON Let T = f x (1 - x)" .dx. Then,
1
0

1 =
1
f (1 -x) (1 - (1-x)}" dx
0
[·: I J(x) dx l
= f (a - x) dx
19.66
1 1

r=
1
d·· -
(1-,•\ .,- .) /I I (,\ ,,/i+
_11 _ ) dx
J
o lr { 1 ____ 1 _
I=:..---
i+ l
[ 11+1
/~ 2

11+2
-
. ""

(l
_ l_ --- - .
11+1 11+2 1
- (0 - 0) - (n+ 1)(11+ 2)

F.\. \MrLL" s I'ro,,('//1al:

0
.)(1 rtJ/ C.·u·, -·
~
-i 1 \o" tan x dx =
"
(1
SCll LlT!ON (i) Let
. 1(/2
J sin 2x log tan x dx

m
1=
a a

o
0

l\in 2 ( ~ - x) log tan(; - X) dx ·: J j(x) dx == J / (a -

.c
1 = [
0 0
==>

du
0
rr/2

oe
==> J = f sin 2x log cot x dx ch
0
Adding (i) and {ii), we get
,./2 n/2
te

21 = J
sin 2x {log tan x + log cot x} dx = sin 2x log (tan x cot x) dx f
m

0 0
rt/2
fro

==> 21 = f (sin 2x) (log 1) dx =O

0
d

I = Q
de

J/og ~ ~ -1
1 ( ) 1 ·1
Iii!' I = dx = J/og ( : x) dx
oa
nl

1 -- [11og {1-(1-x)} n a
•
- - - dx
1 -x
[ Using: J f (x) dx = f f {a - .
w

0
0 0
l
do

:::;.
l = log(- x -]dx
0
1-x
Adding (iJ and (ii), we get

21 = ( { log (1 : ➔
X)
I
flog I •cf\ 0
()
⇒ T =0
EXAMl'LE 9 Llla/uate·
n/2 , 2 ··
(i) J --~~
· -- dx
0 5 inx+ cosx
rt/2 ICMSE')oo2
- I 2003, 2016 NCERT
{ii) J -~ ' EXEMP
0 1 + sin x c~;; dx
 19.67
1'/ 2. ,
(i) Let l = J -_::---l>in, ~- -
l·
dt Th ... (i)
0 Slt\ ' 1 cos x . en,

n(2 sii,i(; -~I

j0 sin
(1t- - x ) + cos-( - it _ )
dx
• [
Using: ff
()
(x) dx - r
(I
f (a x) dJ
j
2 ~2 X
n/2 2 ·
=f ccsx
cosx+sin x
dx · ... /ii)
0
) and (ii), we get

m
r./2 2
1 = J ·
si.n x <:osl!' x . n/'J.
+ - - - - h - fsin -~
1
-d

o
o sm x + cos -X sin x -1 cos x • - x + cos x x

.c
0
rrJ 2
1
J 2- tan
- ----=--~-,--

du
ilx
0 x/2 1-fan 2 x/2
2 ~ + - 2- -

oe
1 + tan x/2 i + tan, ·1•~)2· ch
te
m
fro
d
de
oa
nl
w
do

I = - ~ log(..{i - 1)
✓2
n/2 . 2 ... (i)
I = J sm x di. Then,
O 1 + sin X COS X

7
r sin'(i -x) dx fUsing : J f (x) dx = f f (a-x)1
L O 0
= o l+sm(;-x)cos(1-x)
.....,,,..,....- - - -- ~ - - - - - - - - - - - MATHEM~
-- -
19.68

1t/2 '
l _ J cos- x dx
⇒
- 1 + cos x sill r [
0
Adding (i) and (ii), we get
n/2 . 2 .2 . 1t/ Z 1 /
21
__ J
sUl X + cos ,\ dx =
l + sin .1 cos X
c .\' f -----
l~sinxcosx o
0 d denomina to r by
:r./2 2 [Dividing numerator a n
21 = J 2
sec x dx I
0
sec x + tan x
l .et tan x = t. 111<:m, d (tan x) = di ⇒ sec2 x dx = di
1t 7t
= 2 ⇒ t = tan 2 =
00

m
Also, x= O f = tan O= 0 and x = •

o
"' dt

.c
21 =J
1

du
:
J - - - - --
+ t2 + tdt
di = -
l [
- t an
-1 [t + 1/2Jloo
oe
21 =
0 (t + 1/2)2 + (.J312)2 ,f!,/2 ch /312 0
0

21 = -./3 tan
2 [ -1(2 t +l
- ./7j IJo
'JJ 2 [
= Ji -1
tan c,:; - tan
·1(~/31)] =
2 [re 7t]
J3 2rc
2 - 6 = 3 .J 3
te

I =
m

A
fro

1
1
EXAMPLE 10 Emluate: cot- J (1 - x + x2) dx
0
d

1
de

SOLCTION Let I = J cot- 1 (1 - x + .i:2) dx. Then,

0
oa

l= jtan-
1
(
1
1dx
nl

2
0 1- X·IX )
w

= I = f tan -1 Jll l } dx
do

0 x(l - x)

= f =J tan - 1 { x ~ (l - x) }
0 l - x (l - x) dx
1
I = f !tan -I x+ tan-1{1-x)I dx
()
1 I
= J tan I :r dx , f Lan I (I
- "C) rlx
0 u
l I J
I ,= I tan - X clx i- J tan
0 0

[-.- j f (x) clx -

0 -
1f 1
f (a - x)
0
 TEGRALS
19.69
I
l
J tan
(1
-l
., dv ~ J t,)1) I
,\ d\
ll
l
" In l,ln I
,\ ,h
l
~
• J0 t.1n 1 l .\' I 11.\
11

o m
0)' (log 2 log 1) - ~ - log 2

.c
n

du
oe ... (i)
ch
Let I
te

,e/ 2 cos (n/2 - x) dx

m

= J0 l+cos(n/2-x)+sin(n/2-x)
fro

:r/ 2 sin X ... \i.i)

""J - - --dx
l+sinx+ cosx
d

0
de

i) and (ii), we get

,r/Z COS X + Sin X dx
oa

= f l+sinx+cosx
0
nl

= 1t/J2 (1 + sin x + cos x) - l tlx

w

1 + sin X - COS X
do

.:fl, -,..~ :.,~ ,l,,

1t/2 n/2 l
~
J0 l . d:i - J() 1 + hi n :r + cos , "'
1
n/2 11/2 I l,111 ,/ l ,/\
2T =
[1 X

0
- IO l + l.UI
2 ,/'J
· -
-i 2 l,lll 1/';.
l
t l [,Ill ' I I'-
.\

7t 1t/J2 sec2 -c/2 - ,t~ = 11 _J 2t/l I

wlwn: I . t,,n 2
2 (1 21-21
21 = 2 2 + 2 tan x/2
0
19.70

2, =
rt
~-ling (1,
r ' I} lo! - _2rc - log 2
⇒

)l 1 2
= - lo~
4 :!

a_UTER I
[DividingNr and D'
"( ' t-tan~x/2
' t
I= 0 2(1--;ta~ 'X
n/2 ·Jn/2
= 2~ J (1-tar.-':)

m
T
2
=!2 [x-2 log sec~2 0

o
0

.c
E'\AMl'LE 12 Iff and g are continuous on [O, al and satisfy f (x) =f (a - x) and g (X) + g

du
,hou• that
"
f f (x) g (x) dx = Ja f (x) dx
oe
0 0
SOLL 110N Using Property \I, we obtain
ch
• a
JJ(x)g(x)dx = Jf(a-x)g(a-x)dx
te

0 0

!
a
l
a
m

f (x) g(x} dx = f (x) { 2 - g(x)} dx [·: g(a - x)

fro

a n
Jo f (x) g (x) dx = 2 Jf (x) dx -J" -f (x) g (x) dx
d

a O 0
de

= 2 Jf (x} g (x) dx = 2 Jf (x) dx

a O 0
oa

0
:::;, f f(x)g(x)dx Jf (x) dx
nl

0 0
w

19.4.6 PROPERTY VI
61.lilEMENT Tf f(x) - conl11r11ous
.
do

2'1 "a f11nctio11 de"'

" ; •~~002
Ju f(x)dx f /(r) d~·•fi f a , ,1], then
2u
o +
O
(2n - x} dx = J{! (x) ➔ f(2a - x)}
O 1
JJ(x) 11:\. I tw, ' \'
Ci'' X, I, l /
,, . 1

1l
2
J fix/ ,ix
0
I j" ,/(,) '~I
"
ii

J/(1),/, 2u ,, I' I'(.,) th • III

u Jf{1)t11 a
Let 2n I x, I lien, 111 II

>I
ti
" a,,'I ~
 ;.
ll ; j 't•'' ' = f !a - II\ dtl -J' f(::., - 11 di J '12'1 I di j 1(2.a- x) ,t:i

. .•
tt!ttllRf !ht?\ a]ue ,"I! It ·--: \I' \\ ~ l;<'t

:•-tl.it=f
-
'(tl.f(~- ., \ d,
0 0 0
:•
.f fl.):)d.x = .I (, - •,:~ -,11 d,.

o m
ILLUSTRATIVE EXAMPLES

.c
LEVEL-1

du
1
oe
ch
I.et I = •i -, - - - .fr. Then~
0 -
_$-.C.l'.
-
-
ll,,;~t f lX) ~ 1v
te

l l Jx (X) - • (1a - x)l d:c \

_ 0 11 j
m
fro

:t -;.
d

dx ~j 1 Jx =-'
0 - _t)
de
oa
nl
w
do

'I .- C
. ,'?/dt-1''1
• , XJ dr
19.72

a
PROOF Let I = x J(x) dx. Then, J II a
0
a
Lfsin/~ : J f(x) dx J f (a
I = J (a - x) f (11 - x) d.r [ () 0
0
a I·: f (u x)
⇒ I = J (a -x) f (x) dx
0
a a
I =a J f (x) dx - J xf (x) dx

m
0 0
a

o
= I =a J f(x)dx -1

.c
0

du
a
2I =a J f (x) dx

oe
0
a
=~ f f (x) dx.
ch
= I
2 0
te

ILLUSTRATIVE EXAMPLES
m

LEVEL-1
fro

O.:..._\fl'LE 1 Evaluate:
,t X
=
(i) J a2-cos2 x+b 2 sin2 x dx
d

0
de

(ii) r" .
sm ; dx
X
[NCERT, CBSE 2003, OS, 08, 11, 12, 13, 14,
=
1 +cos x
oa

•0

SOLt.'TION (i) Let

."
nl

l:af2
" 2x dx
w

o a cos x + b2 siri2 x
do

n
I =f (n - x)
a a
22 2 dx
o a cos (n-x) +b sin 2 (n-x)
[ ·: J f (x) dx = Jf (a -
•
=> I _
-
J
lt

2 2
n- x
2 - dx
0 O
I • :
•
Jl
0
o a cos x + b sin 2 x
Ad ding (i) and (ii), we get a ,.~
n
21 = J x+n-x x-o~ r•
I
2 2 2 - dx
O II cos x + b sin 2 x
n I • ; fI l
21 = n f 1
2a
Using J f(x) dx
o a2 cos 2 x b2 • 2 - dx
= r• Jf
. + sin x
}
I = -i(
Ol (x) + f( «-x) dx, We obtain
O 2
 19.73

lt rrf 2{ I
2 2 , t
l1 n rus x+r,• 4111 2 l 2 2
II <O~ (71
ll/2
2n f -2 2
I
1 dx
o a cos x;I, sit,2~
1t/2
sec2. x
2n J - -2 dx [Dividing numerator and denominator by coi xi
0 a2 i· o t,m 2 X

m
n/2 2
sec x
Jo II2 + b2 tan .;,,. x dx.

o
n

.c
t. Then, d (tan x) = dt· => sec2 x dx = dt.

du
⇒ I = tan O = 0 and x =~2 => I = tan ~2 ="'.

dt 11
., dt
oe
rt l
[tan-1 (albI J1"'o
ch
a2 + b2 t2 = b2 Jo (a/b)2 + 12 = b2 x (a//J)
te
m
fro

_ J" x sin
'
x dx
... (i)
-
0 l+C'OS X
2

· ·
d

l
_ 1"
l Using: f / (x) dx = J f (a - x) d.,:
de

(1t-X) sin (1t-X) dx

0 0 _
- 2
1.,. cos (n - x)
0
oa

...(li)
_ Jl't (;t-X) sin X d
- 2 X
nl

0
l+cos x
w

(i) and (ii), we get

do

sin x
=f
" ) , 1t
2I (x + 1t - x sin x dx = 1t J 2 x d:r
l+cos2x ol+cos ,
0

[ -_ -n J1t '
SJ.I\ x
2
dx.
2 0 1+c06 x
811
x = t. Then, d (cos x) rft ::=:> - ' ulx • cit
- / cOSll I
X = 0 => t

I = -
7t
--J di f
= cos O"' 1 and X
-l - l ·1t JI
it '

I rll II l 1,111
2 l 1 12 2
2 11+t l

1
I= -il™.,_-1(-1) - tan COJ "
 \''
\
\

do
w
nl
oa
de
d
fro
m
te
ch
oe
du
.c
o m
 lii&JJ~-----
19.74

EXAMPLE 2 Evalwlle: (NCf.RT, CBSE 2001, 2004,

X
(i) f -1 +-sindx x
11

0 iNCERT,CBSE 2008,
" xtanx
(ii) f --dx
sec x + tan x
0
n/2
(iii) J si.n X
X + COS X
dx
0
it X

m
SOLUTION (i) Let I = J . dx
1 +sm x

o
0
a a
·: f f (x) dx = f j

.c
1t 7t-X
I = J---dx [
1 + sin(n: - x)

du
0 0
0
1t
1i -X

oe
1-f--dx
- l+si11x
0
ch
Adding (i) and (il), we get
= "J x + 1t - x dx
te
2I
1 + sin x
0
m

1t 1
21 = 11J - - -dx
fro

1 +sinx

21 = rr
0
~2{J 1
_ + 1
'l .
d

o 1 + sm x l + sin (n-x) j dx
de

1t/2
21 = 2rr J l+sinx
_l__ dx
oa

0
1t/2 l •
nl

21= 2 n r - smx
J 2 dx
1 - su,x
w

0
1t/2
do

= 21 = 21t J (sec2 x - tan x se } [

0 c x dx = 2n t ] rr/2 [
_s_ii;-◊-:-.~-1-
~~J"
21 =2rr [ - sin:2..:.::2_ ·jn/ 2 [ an x -secx O :: 2n
12
cos x (sin x 1 i) = 2n
I = Itrr O + su, x
O
= 2n (O
+ 1) =21t
(ii) Let I = J __:_ ta,~x
0
sec :r.+tanx dx = f
11
X Sill ~.
"
" ( () I ;~1;- cl.t
J = .2:..:_x)_~:_~) X
0 1 ,;- sin ( n _ x) dx

J = J
1t ( rr/2
n-x) sinx
o l+sin;-ctx J~
0
 19 7r,

r 1f ,-.in ' \ sin \

:, 1+ sin , d
.md {1i) \W ~• t
n
7t ~1ll '
[ ,h
ii 1+ Sll\ '

J (r r
-
"r-/ '
sm .\ ~111 l
o~ 1 + sin .x • 1

m
sm x ifx

o
it/'2
J sin, Jx

.c
~ = ln 1.o.sinx ·

du
0
n/2 .
~m x (1-Sill x)
21 - 2;;: .[

oe
' lb
0 1 - ~in - X
n/2 . • 2
ch
smx-~
2l = 2r. r
• 2 dx
0 COS
te
X

1</2
= 2rr J (tan x sec x - tan 2 x)
m

21 dx
0
fro

n/2
I - J (tan X!;ecx-tan2 x) ,fa
d

0
de

1t/2
1 = rr J !tan x sec x -(sel t -1\I dx
oa

0
n/2
=r J (sec x t3'll
nl

0
w

1~,i
r l
do

I - n ,i;c x tMI it + 1
(J

1f I I'..~J ~ I I (/2 111 \

/}.
L..t J d
Jti ,U

n/2
II
J
7.
J
l2 J
11
I 1 I(
o :,ill
(u)
n/2 n .\
I = r t cos ~
1· t su1 1
1h
19.76
/2
Adding (j) a11d (ii), v,e get ,
n/2 x+
1t
-X 1t
;u2
J -----~· ·
l dr I0
21 = [ 2 dx = - ~in x + cos X 11/l
• si.Jl , -' COS X z 0
I0
2I = ~ j
0
1t/2

0
2tan 2
X
1
1-tan 2
-- 2 X 1<(2

I0
x + -- 2 X
2 1.,_tan
1-1-tan 2 2 2X X.
x n/2 2 sec
1t/2 1 + tan
2
. _~ J -----~--;:;--- dx
21 =~ J _:t ___22x ,h - 2
0
_ tai/ :: + 2 tan ·:.2 + 1

m
2
2 0 2tan -+I-tan -
2
l 2

2r) "'
.
2 x dx - dt => dx
2 dt

o
1·
1 = ·
tan~ "'t. Then, d tan dt :::::, 2 sec 2 - sec2 ::

.c
Let
2

du
1t 1t
Also, x = O⇒ / = tan O= O and x = - ⇒ t =tan 4 =L
2 1
21 = ~ J - ? 2 dt _ = 1t [ dt = rr J
oe di 2
- 21
ch
2 t- +2t.,.l
O
-(t2- 2t - J) o -((f-1)
0 1
1
te

J - -dt- - - = rc x- 1 - [ log\f'i~~/-1
-- ]
= 2T = n
O<.Jil -u-1>2 2.J2 1 ✓2 - t + 1 0
m

Lei
r r1 jl logj,~2-1 )= - n-lo [.fi.+lJ
fro

21 =~r,, Jlogl - loglv~ - 1 n~

2 ✓2 · ,✓2 + 1 2 ✓·2
·
1 - -

2v2L ., 2+ 1 _ g J2 -1
l
d

= 2f = r
n - lo l~(✓ 2 .,. lf -- , '} - rc ,- 2 n ~
de

2 ✓2 g ()2 -1) (-}2 + l) - 2 , 12 log (✓ 2 + 1) == -,= log (✓ 2 + 1)

' ✓2
oa

I
;;:::,
2
Ji log (✓~2 + 1).
:t

n/2
nl

l.)(k\.fPLE 3 Evaluate: f X sin X cos x

. 4 4- dx
w

0 Sill X+ COS X [C BSE 2010, 20 11,

do

r./2
SOUJTION Let 1 = J x sin x cos x
~ :r--- 4 dx. f hcn
osmx.,.cosx '
n/2
/ - J (n/2-x)
-:-4 -sin- (n /2 ::_xl_cos (n/ 2 - x)
o sin (n/2-x);cos4( -;,--- dx 21 -
s/2 1 x)
1t - -
J - [ <n/2-x) sin x cos x
;, cQS4 x , , 4 dx
T btn l.
n/2
I='.:
2
f -, sin;i- - XC()<JX
- - rJx
11/2
I ,. X
Yi11 .\ ('OS.\ tan-
u sm x + COij4 •
d.x 2
~/2 X 0 ~·,,,,1
· X+cos4 x
"n [ ~nxcosx X e

2 o sin 4- x + cos4 x d~: - l

 19.77
rc/ 2 .
it J-. Stn X COS X
y - -- ii,;
2 O Sm 4X + ens x
n/2 ,
J1"
.
f-~
tan .A Sl'C
dx
4
·'

2 !Dividing numerator ,md d<'nominator by cos* x)

0 1 ~- tan· ,
;r;/2 ,
n J 2 tan.....::._ sec~_2 dx
4 o 1 ,. (tan 2 x)~1
2
11 x. TI,en, dt = d (tan 2 x) = 2 tan x scc2 x dx.
0 ⇒ t = tai·? 0 and, X = ~ ⇒ t :: tan 2 11 = ~
2 2 ~--

m
=~ j' l dt

o
4 0 l +t2

.c
du
oe
=
16
ch
te
re X 1t_(1t -' 0:)
Prove that: J -1 --cos- a- s-in -.'< dx, 1

sin· o.,
m

0
1t
X
J0 J - cos a sin x dx.11le~,
fro

Let 1 =

1t (n - x)
d

' I
dx
= J 1 - cos a sin (n - x)
de

0 1t
re X

J 1 - cos a sin x dx -Jo

11 dx
oa

•
= 1-cosasinX
0
nl

1
= 1t
1t
.....::..-- dx-1
1 J 1 - COS 0. Sin X
w

0
do

re 1 - dx
21 =n J l -cosa:sin x
0
" 1 + tan 2 x/2 ___ dx
21 =n J (l+tan 2
12) _2 cos a ta n x/2
x
0 2
rc sec x/2 _ tl r
2r = 1t J ---z--;2-::.2
tan x
('06 (j tan y/2 I I
0 2
re sec x/2 rlr
r=~ J tan2 x/2-2COSCl li111 x/2 I l
O (
x ) - di ⇒ i;ec:zx
2 rfa·. -2dt,
tan x =f. Then, d tan
2
2 - re ~-
1 = tan 2 - •M

o = Oand :x = 1t ⇒
x = O ⇒ t=t~
 MATHEMATI

19.78

:t ' ~ dt
=
-' J ,-' 21 ,,,~ a
1
1
11
~

, di
=-- I
' J {/ -,,i, <'()- qi
0
'
,<>S" IX)

"' di
:::!- I - :t
f
[\ sin·' a ~ (I l'<>,; ll)"'

= 1
,in o.
L " _
tan 1 - . -
sm a
l, c,1~a1J~

m
0

0,)]

o
:t I tan - 1 (- cot
~ 1 - tan oc
,in a

.c
du
;. 1
1 = [;~tan (cot a)]
sin u

oe
re (n - a)
= 1 = si: a[1+ tan - l { tan (; - a))}] 1t
= --
ch [
-1t +
2
1t
2
-a ] =
_ sjn a.

IT 2
te

EXAMPLE 5 Prove tllat: X2 2) 1t

J - -2 2 . 2 2 dx = 3°3"(a +b
0 (a2cos x-b s111. x) 4a b
m

Let J , , x 2 dx. Then,

fro

SOLLTIO:'\
o (a" cos- x +b2
sin x)2
,:
d
de

lt
rr-x
J0 _2_°J_2--2-dx
oa

I-
(a cos- x +b sin x)2
nl
w
do

=
..
= 21 'It fI2. 2
l
2 dx
a "'" x,. /, sin 2 xJ2 I

(f)
•1

</2
21=27t f
IJ (,,2 u,i \

I
lh, iug : f(r) d.r
2.u
0
ti / ( 211 .r) "'/(x)

[Seo example 12
on page 19 11/
 19.79
ERTY VIII

- Iff (x) is a conHnucus f1111ctio11 d""i d I

'J' ne 011 - 11, al, lh<'II
11

f (x) dx = J {t (x) + / (-x)l dx

a O r
sing additive property m, we obtain
" 0 •
J f (x) dx = J / (x) dx J / (x) dx "t
.. .(l)
,, -a 0
1. Then, dx =- dt.
-a ⇒ t = a and x = 0 ⇒ t = O.

m
0 0 0 a
JJ (x) dx J J (-t) (-dt) = - JJ (-t) dt = J/ (-t) dt

o
= [By property JII

.c
a a 0
O a

du
Jf (x) = Jf (- x) dx (By property II ... (ii)
a 0

oe
nd (ii), ·we get
a a a
ch
Jf (x) dx = f f (- x) dx + Jf (x) dx
- a O 0
te
l

Jf (x) dx = j {t (- x) + f (x)} _d:i; · Q.E.D .

m

-,? 0 '
fro

ILLUSTR~ VE

I
d

n/2 l
de

Evaluate: f sin x dx
- n/2 1 + e
oa

l
1ti2 l
= J
nl

Let I sinx

-~~' • '..,.-,} ,,
1+e
l·j/''',,.. ! 1~/2
w

r • '/' f {1r,1 •! <- a} "'

do

o
l 1+e l +e
. x } "/2 1 i- ~~in ; dx =
n/2 r

f I dx "' lx .u =2
re

J =
n/ 2{
J 1 lm _
.
X + - s jnX
dx = JO ~iln:i'
1 ~e 0
1 + esin. 1+e
O ICU~t· .!0151
2
:t1 ~ dx
Evaluate: J
-1ti2
1
+ ex

Let J 2 cos: dx. 'I11(lll,

, 1+e
- 1tJ 2
f
·.• J J(x)dx= {f(x) + / (- .\')} dx]
1t12{- cos x co5(-x)}ax
- - +- - '
[ -a 0
1-
- I
0
1 +/'
-x
1~-e
 MATHEMATIC

19.80
rc/ 2 {
⇒
T= "f{~ + cos_:x} dx = J
l
1 + ex +
ex } cos x dx
-i:-7
0 l+ex) l +e Jt/2 0 [. ·]rc/2 -sin~-sin O=l
"12 (1 + e'. cosxx-
d - J cosx dx = s m 2 .t -
⇒ 1= J - O X 0
o 1 +e

19.4.9 PROPERTY IX . fa ction defined on [- a, a) ' /hen

STATEMENT Iff (x) is II contmuous m
• 12 f f (x) dx , if f (x) is an even function
f f (x) dx = 0

m
:.
0 0 , if f (x) is an odd f unction

o
Using property vm, we obtain

.c
PROOF

1f(x) dx =j {t(x) + J(-x)} dx

du
-a 0

oe
a

a
J {f(x) +f (x)} dx , iff (-x) =f (x)
ch
⇒ ff (x) dx = 0
a
-a J{J(x) - f (x)} dx if f (- x) f (x)
te
= -
0
m

r2 j f (x) dx
a
f f (x) dx -1
_ I if f (- x) = f (x) , if f (x) is an even function
fro

=>
O , if f (- x) = - f (x), if f (x) is an odd function
-a 0
d

j f (x) dx if f (x) is an even function

de

•J \2 ,
⇒
f (x) dx = o if f (x) is an odd function.
-a 0
oa

L "· RK The graph of an even function is symmetric about y-axis that is the curve on the left side
y-axis is exactly identical to curve on its right side.
nl

a 0
So, Jf(x) dx = Jf (x) dx
w

(SeeFig. 19.1 • 1- x'

0 - a s..,. f(.'I:) is an e,
do

y It :

• 0
f J<x) d"' - f f (x) rl.t
0 •

I = j
0

n
l = I
0
• Adding (ii) and
X' d/(:r) lb
(- a, 0) O
y,
(/1, 0) - X I+ I

Fig. 19.10 Graph of an even I unction

 INTEGRALS

19.81
f an odd functior1 the curv .
0 e is S)'n'= . .11' o
uuetr1c
• ,
ppositc quadrants. Consequently
J f (x) dx = -J
-a
f(x) d
X (See Fig. 19.11)
0

y O Q
j j<x) dx ~ ~ f x) dx
0

X' 0 X

m
0

o
f /(x) dx

.c
~

du
Y'

oe
Fig. 19. 11 Graph of an odd function

tLLUSTRATlfE E~A~l!Es ' ,. ~ ·•1

ch
I LEVEt~;<'..\. ' il
te

PL£ 1 Evaluate:
n/2 1t/ 2·•
m

2 [NCERT]
J sill x dx [NCERTJ (ii) J sin x dx
fro

- n/ 2
-a/2 /.'
7
ON (i) Let f (x) = sin x. Then, 11
d

/(-x} = sm.7 (-x) = lsin (- x)l7 = (, sinx{ = - sill x = - f(x)

de

(x) is an odd function.

1t/2 ,r,/2 .
7
J J (x) dx = O ⇒
oa

f sin x dx =0
- rr/2 - it/2
nl

t f (x) = sin 2 x. Then, 2

w

f (- x) = sin 2 (- x) = {sin (- x)} 2 = (- sin xl = sin x = f (x}

do

(x) is an even function.

n/2 ,r,/2 ...(i)
2
J sin2 x dx = 2 f sin x dx
- n/2 0 ... (ii)
n/2
I= J sin2 xdx
0
... (m}

I = f sin2 (!: -
n/2 )
X dx =
1t/2 2
f
cos X dx
0 2 0
ding (ii) and (iii), we obtain 12 ic/2
,r,/2 " 2 J (sin2 x+ cos2 ~.·) dx.
I+I = f
sin 2 x dx + f
cos x dx = o
0 0
19.82

;c/2 1t
⇒ 2I = J 1. dx = -
2
0
1t
⇒ 1=-
4
tr./2
.2 d n
⇒
J0 SU\ ~'. x=-4 71,/2
· 2 " dx = 2 x -4
1t
= -
lt
rr./2
2 =2 SU1·• 2
From (i) aild (iv), we obtain f sin x dx
J
0
-n/2
n/2
EU MPLE z Enaluate: f I sin x I dx
-n/2

m
SOLUTION Letf(x) .:\ sinx[. Then,

o
f(-x) =I sin(-x) I :c[-sinxl =[sinx\ =f(x).

.c
So, f(x) is an1t/2
even function. "/2 71,/ 2

du
[·.- sin x 2:: 0 for OS:
l= f 1sin x I d.x = 2 J I sin x I dx = 2 f s in x dx

oe
,r/2 0 0

r= 2 (- cosxJ:12 2 {- cos i + o} 2
ch
= cos =
te
EXAMPLE 3 Evaluate:
rr./4
(ii)
,J, ~
~ dx T =
(i) J x 3 sin: x dx
m

a+x
-a
- n/4
fro

4
SOLUTION (i) Let f(x) = x 3 sin x. Then,
f(- x) = (-x)3 sin4(-x) =-x3(sin(-x)l4=-x3(-sinx4- 3. 4
) - - x sin x =- f(x).
d

So, f(x) is ail odd function.

de

7ti 4 n/4
Hence, f_J(x)dx=O ⇒ J x3sin4xdx=O.
oa

-,t1 4 -ni4
I =
a ,--
1--
a-x
nl

(ii) Let T = J ,1a+ x dx. Then '

w

- a \
I -
do

[ -- aJ 1a---xx -a-x
-dx =
-a ,a.+x a-x

r
Let J(x) = 1
v~ -x2
1,;2
Tl INTEGRALS
19.83
1

om
.c
du
oe
ch
r./4 X + n

let l = f + dx. Then,

_ ;N 2 -cos 2x
te

x/ 4 11 / 4
1
J .,- - cos2x J --
m

I = x dx ... !!. - dx ... (1

_ ,.. 4 _ 2-cos2x
fro

14 1114
1
~._.r. t: tha t x is an odd function and-- - - is an even function .
2 - cos 2x 2 - cos 2x
d

11 / 4 11/4 ~ 4 l
J J .
de

f
·
_x_ - dx=O and,
2- cos2x
_!__dx = 2
2-cos2x
dx.
- :r./ 4 -11/4
0 '.! -co,-'.!x
oa

tutmg the,e values in (i), we obtain

r./ 4 1
nl

I '"' 0 + 2 ( :,: dxJO 2--COS

- -2X
w

4)
do

rt/ 4 2
I = !!. J l + tan ~ dx
2
2 0 1 + 3 tan X
n/4 2
1 == n J -2 sec x 2 dx
2 1 +(f3 ton 1)
0
,3tanx = I. Thtn,dl d(J'J t,w x)
II " I
Xc O ~ I J.11.111 o-0,1nd,, 4
I ft t.:11\ I, I t.,n IO )
t.111 I .!, ii
II
 19.84
,,I " 2x (1 ~- sin x) d .
6XAM1't.E5 Evaluate: f 2- x
}' -n
t+cos x

SOLUTION
Let l = f 2x (1 + si1_:~¥~ dx. Then,
-"
1+ cos2 X
I =
" -- 2.\ Jn 2x sin x ,,,.

- n
I 1 + cos2 x dx+ - -~ , , . ,
_ ~. 1 + cos x
'"
" fn 2xsinx
⇒ I = 11 - I2 'where I1 = f 2
__x~1
1 + cos X
dx and 12 = _ n -1 + COS2 X dx
-Jr

m
V•.' b 2x ( ) 2x sin2x 1s
e o serve that _f (x) =- - - is an odd function and g X -
.
an even

o
l+cos2x l+cos x

.c
1t " " l't2·
11 = JJ(x)dx=Oand,12 = J g(x)dx=2 f g(x)dx =2f xsm2x dx.

du
-ir -:i o o l+cos x

oe
It
!\:ow, I2 = 4 J x sin; dx
0 1 + COS X
ch
= z = 4 ..J (n -x) sin (it - x) dx
te
2
o 1 +cos 2 (n -x)
m

f2=4"J(n-x)sinxdx= 4~ Jn sin x "

o l+cos2 " 2 dx-4J xsinx dx
fro

" . x o l+cos x o l + cos2x

12 =4ir J sm\ dx -l2
o 1 -cos x
d

;, =
de

212 = 4n J 1
o 1 + cos2 x COS X
oa

⇒ 212 = -4n[tan- 11r· 1

nl

2 1
I2 = it
w

Hence, I =I1 +1 2 = 0 + Jt2 = rc2.

do

1 3
EXAMPLE 6 Evaluate: J x
2
+Ix 1+1
-dx
-IX +21.t l +l INCERTEXE
l 3
SOUJTION Let/_ J x +Lx j1-J
x2 I 2I I 1-1 dx. Then
1
3
I =f { 2
x ~. I XI I l ,} E,
-t
J
x +2 l xl I J x2 12 1
xI I 1
,1\-
3 1 w
l= f _2_x_
l - th- J l xI I I
] <X
-J x +2 xI I l x2
I 3 -l • 2 1,r l 11
[ -- JI 2
X
dx + J - Ix_
I +_
1
l
-1 X 1 2lxl +11 2 ()<.\.
-1 l xl +21xl1~·lc/x
 I 19.85
J I., I I l
I(1\I~ If
l
ri.t

1 -- O 1 2 J1 ( Ix- I + 1 dx ·: -
_,.3
' _ and I ~:i + 12
0 I xi~- l)2 (I I )2
,x+l.
(1-tl+l ) arc•ndd and
_ Jl _ 1 even functions respectively
1
I -2
o
-
I A I+
dx=2 J - l
I "
[ L
L
o x +~1 dx= 2 lo~ (x+1) = 2 (log2- log1) =2 log2

m
J (cosax - sin

o
Evaluate: . 1,x)2 d.x.
[CB SE 2015}

.c
-n
n
f (cosnx- sinbx)

du
Let I= 2 dx. Then,

"
oe
7t
l =f (cos2 ax+sin 2bx- 2 cos ax sinbx) dx
ch
-"
lt ? 1t ':'f >

f J cos ax stn,bx a)
te
I= f
-;r
cos- ax dx +
-n
s in 2 bx dx -2
-1t
m

I= 2 J cos2 ax dx--;- 2 J sin 2bx dx-2x O

1t 7t [ ... C()S 2 ax and si:t~ 2
bx are even functions
J
l
fro

O o and cos ax '>111 bx 1s an odd ftmction

J
7t
1 = (1 +cos 2 ax) dx+ J (1 - cos 2bx) dx
"
d

0 0
de

1
I = x+-sin2ax r 1
]" +lx--sin2bx ]"
[ 2a 2b
oa

O 0
( 1 ) / 1 1
l = ir+ - sin2air-0 +l n - -si112bn-OJ
nl

\. 2a 211
w

r =2n+ l-sm2an-2-sin2bn
211 2b
do

2n +-1- xo--1-xo =21t, if11andb areintegors

I =

l 2an
2
2bn
Zn+ ~sin a1t-..!. sin2b1t, ifa,b arc not Integer~
2a

3/2
2b
1-~~EL-2]
lNCl'Rl'l
Evaluate f I x :;in 11 x I dx
-1
ION Wehave
- 1 <x< ~ ~ -n< n x<-f
sln it x > (} ==> I x sln re x\ = .1· sin rt x
X
· nx < 0 ;:::, '
-l< x<O
==> - n < 1t X < 0 :::;, s lJ'\ 1
sin It ~- > O ::::> I .t sin 1t xi = x sin "x
:::;, sJn it x > O :::> x
 • 7t x < 0 ;:::;, I x sin it xi - - .i
19,86 ~ 0::;:,X5111
3 ,, 1i ;::> sin 1t:\' <
1 <X <- ⇒ lt < 1i X < -;,- •
2 ·r1•y<l
· ,~ 1i I i - <- -
_, $µ, ., '
r
Thus, jx sin :n\ =~ ~ ,.
l- x. sin, "~, if 1 <X <I
2

using additi\-c p;opertr, "e obtain 3/2

3,, j1 . Id . f \xsin itx\ dx
1= [ jxsinnx\dx = [xsm.ix l+
• -1 1 3/2

m
J
1 =
l
Jx ,in x dx +
3/2
f (-X sill Jt x) dx
1 .
= Jx s111 1t x dx - x sin 7t x dx

o
71
-1 l -1 l

.c
l 3/2 [·: x sin 1t xis an even

du
J f xsin 11xdx
= 2 Jxsin rrxdx-
a 1

oe
J x sm "x x =-- cos 1t x + -lj cos 11 x dx == - -x
r . d x cos n x + 2l .
SID n x
lll 1i 7t 1t 1t
ch
te
m
fro

=
d
de
oa

2
!.OLL'!ION Let f -- f1
-2 XCOS1tx[ dxand/(x) -
- 1x cos re x I,Then
nl

f(-X) =1-xcos (- . I- '

w

So,ffx)isan;ven furu:tio:J -1-x cosnx\=\ x cosnx\ =f(x)

do

I- j I x ,os
. n I dx - 2 2JI
..· " f (x) dx ~
-2
0
xcosnx l dX
[ J" a
2 J f(x) dx, lf f(- y)
0 •
 TEGR'<LS
\I.U

' ..
l
w,..

1 1
.-o - 3

•
-. '
0
:It •' '
..'.s;

l
,_-
l
,•'_+_!_ I -' = ..
\ - --
-.;- ~

' ~ i:i2 -:!lt -:

o m
.c
=

du
if J (.:U - x) (xl

oe
0

!-"sinS rroperty \"L we ootain

i .'
ch
f f(J-1 J1 : " (x) ~ I (2.Q -
->-
.r)}d.,:
te

~ 1 j {x) - f(x)} d.r. if f (la - .r) =f(x)

m

"
fro

j tJ(.r) - _r{.r)J dx, iff(2a-x) =- f(xl

0
d

if f (2a - .r) = f (x)

de

if f (2a - X) ~ - / (X)
QE.D.
oa

0
if f (2a- x) =fl x), !hen the graph off (x) is ,-ym11wtrkul .11\:1111 x ~o a., - :n
nl

9.T2.
w

2A a
J f(:r.) dx = J f (XJ dx
do

a 0

-
x· n
' LI

Y'
Ft(l 19 12
 ·g. 19,13.
19.88 f (x) is as shown in Fl
the graph of.
If f (2a - x) =-f (x), the~"
f f (x) dx = - J f (x) dx.
0 n
y
i '}JI
f/(X) dx
I
X
X'
/,,, OJ (2.a, 0)
0

m
•

o
Jf(x) dx !x=a x= 2a

.c
0
j

du
Y'
Fig. 19.1 3

oe
ILLUSTRATIVE EXAMPLES
ch
LEVEL-1 f
2n
te

EXA.\1PLt 1 Evaluate: J cos5 X dx.

m

SOLL710N Let f (x) • ~; x Theo,[ (2, - x) • { cos (2 , - ,i)' •oos5 ,

fro

2n ~
J cos5 X dx = 2 J cos5 X dx
d

0 0
"-Jow,
de

J(rr.-x) = { COS{1t-x)}5 "'- cos5 x = - f (x)

oa

!t

Jcos5 x dx = o.
nl

0 =
2
w

2,r 5 n
fieru.-e, J cos x dx = 2 J,.,,,.5 d
do

0 "'"XX=2xQ=, Q l
l:.XA MPLE 2 Pro~e that , 0 •
n/2 . •12
J log ~in :r. dx = J Jo •
~
1t
g cos X clx = - - log 2 11 "" III
•n o 2 ·
SOl.UTION Let / J lCJg &in X tfx, lNCER f , CBS 1 I in (
n/2 (I
21 "'l
Then, 1
= J log sirJ 11 _ )
0 \2 X d:r.
•12
J " J log cos x dx
0
 INITE IN TEGRALS

·ng (i) and (ii), we get 19.89

rt/2
21 = J log sin x dx ➔ 11/2
f log ~s,b
o 0
rt/2
2I = J (log sin x + log cos x) dx
0
rt/2
21 = f log (sin x cos x) dx
0
rt/2
2I = Jo log ( 2 sin ; cos x) d.x

m
y 2

o
2I = 1og ( - 2x)
sm- - d.x

.c
0 2

du
rt/2 Tt/ 2
2! = f log sin 2x dx - f log 2 dx

oe
0 0
rt/2
ch
21 = J log sin 2x d.x - ~ (log 2)
0 2
te

ri/2
2I = J log sin 2x dx - ~ log 2 ...(m)
m

0 2
fro

rt/2
11 = J log sin 2x dx.
0
d

utting 2x =t, we get

de

" dt
11 = J log sin t -
oa

o 2
1 rt
nl

I1 = - J Jog sin t dt
w

2 0
rt/2
do

lUsil\g- property\.]
11 = ~x 2 J logsintdt
2 0
tt/2 ( U~ing Property ll
I1 = J log sin x dx = I
0
Putting I 1 = I in (ill), we get
21 =I - ~ log 2 ⇒ l ""
!! log 2
2
2

H:ence,
rr/2
J log cos X dx
J log sin ;x. dx = 11/2 - - -2
It log 'l
~

0 0
n/2 t x I dx == 1t loge 2
EXAMPLE 3 Prove that: J log I tan X + co
0
- - - - - - - - - - - - - - - - - - - - - - - - - - - - --,V1:r-r-~•--- ~ ~ --

19.90

SOLUTION Let T =
1t/2
f log I tan ;r + cot x I dx. Then,
0
dt/2
sin x cos xi dx
1
I= flog cos x - +-::1
sin .,· I
0
Tt/2 1
1-~ - - t dx
1= f
0
log sin x cos XI

[·: sin X > 0, cos X > 0 for ail XE (0,

) = rrJ/ 2Jo.,. (- _J __) dx
° sinxcosx
0

m
rr/2
J log (sin x cos x)

o
=- dx

.c
0
1t/2 n/2

du
= I =- J log sin x dx - f log cos x dx

oe
0 0

⇒ I= -(-~loge2)-(-~loge2)
ch
⇒ [ = rr loge 2
te

It
t:XAMPLE-1 Evaluate: J log (1 + cos x) dx
m

0
fro

1t

SOLUTION Let J = J log (1 + cos x) dx.

0
d

Then,
de

"
=I
I log {J + cos (it - x)} dx
[ 1/ a
oa

0 (x) dx • J f(a - x)
nl

11 0
= I = J log (l - cos x) dx
w

')
do

Ad ding (i) and (ii), we get

!
I!

21 = !log (1 i- cos x) + log (1 - cos x)I dx

7t

2J ~ J log (1 - cos2 X) dx
0
1t
2T = J log sin 2 x dx
0
It

2f = 2 J log sin x dx
0
1t/2
l = 2Jtog sin x dx
0
 19.91
1l
= 2x - -2 log ,, .,<.
•12

= -1rlog,, 2
11
[
·: J l()g sm .x ,Ix
I)
" l
2 lr,g, 2 l
[pn/uatc: J x log sin x dx
0
~

Let I =J x log sin .t d.t. Then

0 '
11
f= J (1r-.~} log sin (,c-x) dx a a

m
0 ·: f JCx.)dx=f J(a - xJdx
[

o
0 O
"
J (rr-x) log sin x dx

.c
I=
0

du
n: :t

l = rr J log sin x dx - J x log sin x dx

oe
0 0
rr/2
I =2rr J logsinxdx-T
ch
[Using Property X]
0
lt
{See Example 2J
te
J =2ro:x - log 2 - 1
2
2
21--,r 1og2
m

2
lt
fro

I =--log2
2
LEVEL-2 I
;t/4
d

[NCERT EXEMPLAR]
Evaluate: J log (sin x + cos.x) dx
de

-n/4
:t 4
oa

Let I= J log (sin x+ cosx} dx. Thc11,

-~/4
nl

4
f log~f✓-2 ( ..2.-<;i.n x+ ~-cos.x
w

1= "' J )~ dx
2
.fi .J
do

-n/4

T= "l log { Ji ( sin xcm,i- cosxsin )} dx ~

I=
-:~
'J
4
log { fi. sin (x t- ~)
}
dx =
n/4 {
f log /i.., log ..in
(
i ;
n)lj ,It
4
lt/4 1t/+
1l4 1ti
4
( 11)
1= f log fi. dx.,. J log ~i n , • 4 ,/1
- ,r./4 -n/4
1
I "' ( -log2
)
LX
]"/4 +
1t/2
J lo"si.n/tf/,wllL'll't 'I "·I
n ·

l ~ log 2
2 -n/4 O n/2 ]
•: [, log sin t di -
19.92

⇒ T= 1t Iog2 - 1t
log2=- 1t log 2
4 2 4
;r/4
ALITER Let I == J log (sin x + cosx) dx 11,en,
-n/4
ri14 (
-x-) } dx·
I= f l( log(sin x+ cosx) + log (sm• (-x)+cos
0

1= nt { lug (sin x+cosx) + log (cos -sin x)} dx

0
n/4 n/4
1= [ log (cos2 x-sin 2
x) dx = f log cos 2x dx

m
⇒

0 0

o
til2
f

.c
⇒ 1 =~ log cost dt , whcret =2X
2 0

du
1 1t it
l = - x - - log 2 => I = - - log 2
2 2 4

EXAMPLE 7 For x > 0, let f (x) =f loge I dt. Find the function f
oe (x) + f (x1 J and s
ch
1 +t
1
te

=
m

SOLtJIION We hav e,
X I t
f J
fro

(x) = oge dt
1+ I

lr
1 =

(±-) =
d

:::;, f lo~ dt
de

X l 1+t
1 1
Lett = - . Then, dt = - du.
oa

u u2
Also, t =1 1 ⇒ ] 1
=> u -= 1 and, t = _
nl

-= - ⇒ u = x
X U X

fr .!.'x ) = lf' log,-1-x--

w

(1/u) - 1
du
u2
do

, 1 1+ -
u
f [1 J= J log~ du
X l (1 + U) U

f ( 1J
= I x.
xJ
I
k,g, /
('J
+-)
I I
dt
Adding (i) and (h), we get

f (x) ~ f(~J' - J { log, I

'I J ~ I
101,, 1 '
('I I I) I
ldt
21

I =
= f(x.) + /(~)
x.
"' jI Jo~".../ ( I ·I
·1 ~ I
/J
I - di· n/ 2
f f(
f (X) + f ( ~) = J ~o~,: dt ()

t
 TEGRALS
19.93

<', we get
) + f ( 1\ J = (loge e)~ _ 1
\e "- - -2
-:t/2
Show that: J f (sin 2x) sin x dx =v2 r;:; nJ/4.f (cos 2x) cos x dx.

m
0
n/2 0

o
Let I = J f (sin 2x) sin x dx

.c
... (iJ
0

du
n/2 r II r, -

= lf{sm2(; -x)}sin (~ -x)dx ·: Jf (x) dx =Jf(a - :t) dx

oe
[
0 0 J
rr./2
JJ{sin(1t-2x)} cosxdx
ch
=
0 l
te
rt/2
= J f (sin 2x) cos x dx
...(ii)
m

0
i) and (ii), we get
fro

n/ 2
1 = J f (sin 2x) (sin x + cos x) dx
d

0
r./4 [Using Propeth .\1
de

= 2 J/ (sin 2x) (sill x + cos x) dx

0
oa

= 2]2 ntf (sin 2x) ( -3--2 sin x+ .Ji cos x} dx

nl
w

f f (&in 2x) sill ( x + ~ ) dx

n/4
do

ntf{sin2(~-x)}sin (~ x"~)dx
21 = 2✓2 "ft{sin(i 2xJ}sin(; x}h
n/4
21 = 2.../2 J / (coi; '].X) cos x ,Ix
0
n/4
l = Ji. J f (cos 2X) COH X dx
1C/2 O n/4
. d r,;2 J
f (cos 2X) cos x ,tx.
I f (Sill
0
2x) sin x x = ,u.
0
&,lM%.JMQ hM & W4.
MATHEM

19.94
2r:
•)11 ~2
., sin"' l __ dx == ,.
EXAMl'LE9 Prove tlml: j ~; +-;os 211 X
0 Sill ·
.,_rr '
x sin ·' _ _ d.r
:.,, '

SOLUTION Let I .,. Jo ::,m

.. -2-;: .. cos 2" y
., ii ,?
n/ 2
I
2:t (2n x) si1/"__F1t -~'1- dx [
•: f f(x) dx= f f(a = 21t

1t/10 '
0 0
Then, = J0 sm
:-:-2"(.,-;
_,. _-;.)
· + cos2" (2n· .x) = 1t I0 2
(a
"n ) . 211 ,
-[ (2n - x sm · dx 1da2
⇒ ''I 211 ,
. sin"' X+C'OS ;\
0

m
Add.inn- (i) and (ii), we get . 211 ,
"' 2n , . 21J x (2n - r) sm x dx

o
xs1n + - - - - - --
21 = JO sin ~- -2,1, . 2,1 \' .L COS 2II X

.c
X + COS ;\ Sill ' '

du
2 21
2 n sin ' x ,
21 =
"
J -2.,-,---2,-,
sin x cos x
dx

oe
4
0
.,,,... . 2n
]- -"
J --
Sill
-X --2n,
dx
ch
-1t .2!1
0 sin X ,- COS X
n sin 211 X
te

J = 2;i J -sm. -211 x, cos2" x. dx

0
m

r:/2 Sin 2.,, X

= 4n J Sl!l.. 2n [Using Property Xf
fro

l -n
- dx
Q X + COS2 X

rr/2 , 2n ( /:2 ) ,, n
,
I -- 4it J dX
·: J f(x) dx = J f (a -
d

Sill 7t - X
z11 2'1 -
de

i:--,,l-
o sin (n/2 -x) + cos (n/2 - x) [
0 0
1t/2 2n lo,:: l -
1 = 4n J- _x__ ---,
oa

cos dx
2ll . 2n ~ T \;-
o cos x+sm x
nl

Adding (iii) and (iv), we get

1t/2 . 2n 2.ti
--- 1 t.\J\
w

27 = 4rr J __s~x cos x n/2

• 2n 21,--211-- - dx=4n J =
1t
do

o sin x+ cos 2 71 x sin x ~cos x 1.dx 4nx -

2.
[ = n2 ()

1t
EXAMPll'. 10 Proue that; f _2_ 2
_x_ _ _ _ dx
o (a cos x I· ti2 i,,in 2 x) 2
11
SOl.lJTION Let I
J -(" 2cos2 x..-/.1
!)
x 2 · dx
~in 2 Jt./
'fhen, n/2

J0 7
tan
n
IUsing Propert Jx~i.n X
0
 19.95

[Using Property X]

[Sec fa amp''
I 12 "n page 19.3!,j

o m
.c
du
EXERC/ ,f" 19.4
1.EVEL-1
,/ua(( 1/r, fol/m1•111g mtegral 1

oe
2 I
J It J
ch 1
t.m
2 ,,,,.1 :t "
te

r
' ,«•I I 1 l,1n l
dr ,/\ ll'l;CEIU I
\'
m

2 II
( 111 I
J ,, " d, I l.ll'>L 20151
fro

~ ~In I 1 n,s \

l
d
de

I Jr
oa

l
I
r \ l,ln \ d,
nl

.
0 "'\" ) o..,,,,,,,. \
w

•
do

j 1 ,m 1 , 11\
I

•
{ tsin t
, l
+ sin 1
ru
•
J.t C\i,. ~ ~ d\

tan :r -
- tb:
tan .1 + rot7 ::r
•
f
 MATHEMA

19.96 rr/2
s in 4 xdx rif (x) is
rr/2
f xcos2 x [CBSE 2020]
24, f
- n/2
23.
(ii) f (1 - x2) sin x cos2 x dx
25. ~;/} log(~=: )dx - n Provet
- 1
rr/4
27,
f,
rr log (1 - cos x) dx
26 J sir? x dx
0
Prove
- ;;/+
zx (1 + sin x) dx
lo ---x) dx,
2 rr
f
"{ ( 2 -sin l +cos 2 .x·
..~ S,
29.
J g 2 + Sin X -n

m
- n/2
2 3x3 + 2 Ix I+1 dx
J log (a-sin8)de f

o
30 a
- -.- , a> 0 31. - 2 x2 + Ixi + 1

.c
· a+sm8
-a

f1 tan - 1 lr 1 +1 -x-x
Zx J

du
1'/2
32. (i) f {sin2(3r.+x)+(n+x) 3} dx (ii) 2 dx
Ct
-3n/2 0 j1;

oe
--
6 . sin

34.f 1og(~-1Jax
2 a
33. f x ✓2 - X dx [NCERT, CBSE 2007]
ch
~ 2 x + co/ xJ dx
0
1
JI x cos J(l+sm
te

35. 1t x I dx [NCERT EXEMPLAR] 36.

-1 0
m

38. f . 100 x cos101 . d

2n
37 "J -,------,---:--
x dx [CBSE 2016] Sil'\ X X
fro

· 1 + sin a. sin x o
0
1r/2 . b 3/2 30. -') .,
39 J0 asmx+ cosxdx 40. f lx cosnx l dx -, -
d

• sin X+ CO$X
. 0 .n. -l
de

;r
1 3/2
u .flxsin nxldx [CBSE2017l 42. f l xsinnxldx
oa

0 0
43. IT f is an integrable function such thatf(2a-x) = f(x), then prove that
nl

2a a -t Let I
f f(x) dx = 2 J f(x) dx
w

0 0
do

2a Then,
44. If f(2a - x) = -f(x), prove that J J(x) dx = 0.
() Addin
45 If f i~ an integrable function, show that
a a
(i)
2
J
f(x ) dx = 2 J f(x 2) dx
(ii)
n
J x f(.l) rlx 0
2
-a 0
a ==> I
46. If f (x) is a continuous functit1n defint•d on 10, 2111 . TI1e11, prov,, th,,t

f f (x) dx Jn { f (x)
2a
1 f (2 ,1 y)
}
,h 12.. Let
0 (J
47. If /(a+ b -x) =f(x), then prove tha t
l'he
!
b x f(x) dx =( 11
b)''!.f(x) r/x.
;-
Add
 !TE INTEGRALS

lf/ (x) is a continuous function defined on [- l 19.97

11
•a , then prove that
Jf (x) dx,,, JO jf(x) ~ f(-x)} d.>1
11 • [

- a l
n 1t

Prm·e that: f x f (sin x) dx = -'2.'.. Jf (sin x) dx

0 O
a II 1
f
!'rove that J(x) dx = Jf (a - x) dx, and henceevaluat~J x 2 (1 - x)" dx
O O 0 [CBSE2019]
- - -- -------- -~-- 1t
ANSWERS
•~ . -4 rr 1t
4. 4

m
5. 4
8. 0 Jt
9. - log 2
1t

o
8 10. 4

.c
1t
1,,~- -5 13.
2 1t
3 14. - i 2
!og 2 15.n(%-1)

du
2
It (I. It lt 1t
1:-. 4 18. 12 19. 20. 3

oe
-in 0. 4
It 1? 31t
ch
"2 -
~ . 16
23. 0 24. 8 25. (i) 0 (ii) 0
3
" 1 2
27. - log 2 28. 0 29. 30. 0
te
1t 1t
b 4 - '..!
16 ✓2 2
3,~- (i) ~ (ii} 0 34. 0 35. -
----is
m

I. 2 log,7 ~
33. 1t
~

~ (a.+b) ;(%-;)
fro

38. 0 39. 40.

2
d

50. (11 +1) (n + 2) (11 + 3) PROBLEMS

de

HINTS TO NCERT & SELECTED

- - -- - ... (i)
oa

;t/2 sin 3/2 x dx

= J . 3/ 2 x+cos 3/2 X
nl

SID · n/2 cos 3/2 X d., ...('u')

0 . . 3/2(iti2 - x)
f -cos3/2 x-1- sm. 3/2 ·
w

X
;,;•2 S111 3/2 12 dx I =
Then, / = f . 3/ 2 ( ,7 x) + cos. ( '1li x) o
do

() S111 ~- n/2 it
(11
Adding (i) and .. ), we get 3/2 x
n/2 . 3/-7 + cos -th =
it/2
1 . dx "' [ x ] f =-
2
O
2I =
sin
J 3/2
. 3/2 x + cos· x
0 SU\
o

1t
= - ... (i)
4
4 x dx.
12. Let T ="Jxsin xcos ... (ii)
0
-l (n- r) ch -
- "f (n-X') .~,., ) . t dx
• ~·os4

1t • It _ T) cos ll
Then, l= J(n-x)sJll(
0
Adding (i) and (ii), we get
19.98
-1
Jt 4 dt, wheref =co«\'.
x sin x dl = - it
if .t
.. 21 = J11: cos I n/2 \ 1t -
0 = f 2- 4
rt51-
::::) 21 =-1t15 I
l -ir ( - J _ 1 ) = 21t
5 5 5
~ )= -
1t
5
0
ing (iJ and !it
cos
s/2

n
;f , coo;
s
1:,. Let I = J., sin J .,· dx. 1
(1
rr
= T\ f-
3 ➔ 0 (1
it
Then. J = J( 1t - .t) sin 3 (1t - x) dx = J (n x) sili x dx

m
0 0
1t

f sin 3 x Jx

o
Adding (i) and (ii), we obtain 21 "' n

.c
0

du
1;3 Let I = •J -x sin x dx
1 + sin x

oe
0
Then I = j (n-x) sil, (n-x) dx J.og• -
.,
ch
1 + sln (it - x) ·
0
te
:r l -
(n-x)
= I -1-
Slll X
I - - - dx
+ Sin X
m

0
rr
1t SUl X l = ~ ~::
fro

Adding (i) and (ii), we get 2 / = I ---dx

l+sil1x
0
r= .. 41-"i: :
f (1 1.,.
lt
d

2I rr + sin. x) -1 dx =nf
sm x
rr ( 1- .1
. x dx
1 ~
t,:
de

0 0
1 +sm =
" ,r./2 / .
= 1t f 1 · dx - J_
oa

~
2I 2n l . - dx
1 ~ =
o O
+smx
nl

2I - n[x]" 21t "r 1 -sin

12
x dx
'
w

----
0 o 2
cos x J
'
do

n/2
21 ,<-2n f0 (se·2
· c x Lanxsecx)d
• X
~ I
,l

•
21 - "2 2n( lan x -sec x J:/2

2
2nlsin x
co,'
11•12
0
:I.
It I ?11 \'O~.\
(l1~in,)
l-
n/''

I)
I{
,
I ::'. JL(ll I)
,\h.1,.,11 ..\sr!!d l li
t•nh ,1tion w-.1:,. k.
l -= it
1t ( \') l>l' ., c,,ntint
2 t'd into II eq
22. Wehave, ,... ,n + (11- l)h b
rr/2
I ~ J ~:tcosx
.4- -<Ix
11/1 = [,- a or,
0 Sil) X + CQS4 :,
 1 = "j2 (~~)cosx sm .t 19.99

0 COS4.t+. 4 dx
dding (i) and (ii) sin x
tt/2 'we get ...(ii)
21 = ~
2
J sin .t co
4 ~x d
ocosx+•4 x
1 SJ.11 X

21 = ~ f 1
4 0 (1-t)2 +12 dt ,wheret= Sm
. 2x

m
1t
fI
~ 2 [tan-l
21 =- 1
(t _! )

o
2
S O + ( -1 ) 2 dt = 8 X (2t-l) ] l

.c
~
2 2

du
l = ( 1t + 1t) n:2 o
8 4 4 = 16

oe
It
= of log (1 - cos x) dx -_ oJ" log ( 2 s·m 2x)
-2 dx
· -f
- " log2 dx+ 2J• l
ch
. x
rt/2
0
og sm -dx
te

1 = rr log 2 + 4 f log sm. t dt , wberet-x o 2

0 -2
m

-.··r'2 1
fro

I = n log 2 + 4 x - ~ log 2 1
2 [ 0 og sin/ di= - ~2 lo<>" 21
2
d

l ; n log 2 - 2 it log 2 = -id og

de

= f x✓2 - x dx. Then,

oa

1
2
= f (2 -x) .j2 - (2 - x) dx [
Using Jj(x) dx = j f (a -x) dx]
nl

,O 0
0
w

2 2
= f (2 - x) Jx dx = f (2 ✓X -x./x) dx
do

~
0 0
f = [± 3
;x3/2 _ 3_ XS/2]
s
2
[±
= s x 2 3/2 _ _3.s x zS/2] "'
3
_£_,/2 _ 16 /2
s - LS-
O

hitNTEGRATION AS THE LIMIT OF A SUM

ens~ection, we shall consider integration as the limi t of th!? swn of certain nurnber of terms
_ ; n"mbe, of ,~m.s
rend> to ~fhUty ,nd Oods to mm. .,.cl,"""
of foe< th• As•=•"'
•ten,;:•."""' of d,fini te fnn,g,M
is ,nore fw,d,mOO to it wM f" bef= "' I•"' In""""
3h
tion was known.
-(~ b, • continuous real v,l"'d t;,,,ctfon d, ff o-' ~" tb• ,to~d i
mto n "'l"al P""
eo<n of width h b/ """""' (• - 1)
fo, bf wWch fs
porn• , . "•" 2h.
0
•='
1
, ...,a + (n- l)h between a and bas sho,~"fl in Fig. J9. l4. T1 cn,

1111 =b-a or, 11==~

n
do
w
nl
oa
de
d

T
fro

=
m
te
ch
oe
::;.

du
_._
.c
o m
 19.101

/J .,_ sin (/J + 11) + sin (a+ 211) +. .. su1 {a+(" 1

+ sin (a+ (n - 1) Ii)=--'-- 2 J
'\ht) sm. (nh2 \J
sin (~J
{a+(11-2!l1zl sin(nh' I

m
a+ cos(a+ h) + cos (n+ 2h) + ( cos
•·· +cos a+ (n-1) h) =--'-~-,c..J_.,.,f'----__..::..2 _!_
. (h)2

o
.c
$111

du
ILLUSTRATIVE EXAMPLES
\ LEVEL-1 j
Evaluate the following integrals as limit of sums: ,,
oe
ch
2
(ii) J(2; : 1) dx
te

0
ON (i) We have,
m

b . .
f f(x) dx = lim h
h ➔O
[t (a) + f(a + h) + f (a + 2h) + .. . + f (a + (n -1) h)-1 , where 11 =b -a
fro

a .
' J n
2- 0 2
a=O,b =2,f(x) =x +4and11 = ~
n =-
d

ri
de

2
r= J (x + 4) dx
oa

0
I = Um 11[f(O)+f(O+ h)+f(0+2h)+ ... +"f (O+(n -1) Ii)]
nl

II➔ 0
4
w

= 1im h [ (0 + 4) + (h + 4) + (2fr + 4) + · ·· + ((n - l) l1 + )]

do

Ii ➔ O
J = lim h [ 4n + 11 (1 + 2 + 3 + · · · + (n - t))]
It➔ 0

<;-:.-11}
11

~ ""j
I = hl~1 h { 4n + /!
0
{
2
rJ (II -1)} [·: It =;, ,1nd /1 -) 0 ., II
I = lim 2- 411 + - >< - ~
11 ➔ 00 11 n
2

.
I= lim
{ ( 1
8+2 1--
)r = 8 + 2 (1 _ 0) == H)
n ➔ oc, 11
W
e have, f( + (n - 1) ·]
Ir) , where Ii = ~
b a
b a + Zh) + . " + II n
J f(x) dx = 1irn /1 [t<a) + f(a + h) + /( '
n 11 ➔ 0
19.102 2-0 2 : lin,2j
a= O,b =2,f(X) =2X t 1 and 11 =---- "' -- 11 11 n ➔ r.;\3n+
Here
2 -= 1im 6 4("
( +
J(2.t + 1) dx ti->"~I
I =
I • ,"'!'o ,, r((0) a J(O + l•H f(O + 21,) <f(O < SI•) + .. · + f (0 + (• - 1) h)
0 l We have
"J f lx) dx= Jim Ir ➔ O
h
1 = _lin1 Ir [f(O) + f(II) + f(Zlz) .,_ j(31t) + ... + j((n - 1) h)] a

,,--o . 11 =Z,b- 4,f\x) =

4
-= ! f(x) dx = ~
1 = h]j~10/1 G, .. 2/l(l + 2+ 3 + ... +(11- l))) 2

m
-
1 = fan h t2(2)
~

o
k -.Q
I = lim Ir {lH 2h )( (n - _]}

.c
'1 -➔ 0 L 2 I=hrnh3•1

du
h ~ C

1 = ,~o lz~1-,.11J1(11-l)]
I = fun h lll-
~ (11 -1)1
oe
[· " = ;;2 and h ➔ o
n- .J

==> I = 11->00
Jim J11 {11 + 11 X II , • 1.
ch
. I= n~O {2+4 · ➔ ao {2+4(1 -~)}-2
(11:111)J = nJim
te
- +4 = 6 11

.) 3[ ( 2 Evalllate the follmmng inteoral

..,. s as• hmrt
. . of swns·
m

rxA"fl'LE
2x+l)d·X ~f (
(ii) 4 2x· - 1) dx
fro

(L •
1
SOU.JTION (i) We have 2
d

b '

➔ O h[f(a)+f(a+ h)+f(a+Zh
fa f(x) dx = hJim
de

~
+
Here, a=1,b- ) .. . +/(a) + (n- ]
;) ""'h,l
oa

' - 3, J(~ • 2H l '"d h • 3 - -2 . h) ' where h

'
nl

\~) J.T -
I = f (2x+ l) dx n --;:; J h
a
w

1
(I !, -'
do

= I = lim hr
h ~ 0 l_.f(l) + f {l -,. /)
1 + f(l + 2h)

= 1
-- 1lim [ (ZY 1
,~o 11
+ ··· + J(·1 +(n-1) h)]
J' (\'; . .
I = lim h . r
+ 1) +12(1
~
+ /J) + 1) + 12 (1 + 2/r 0

."
+
h - , o 3 i ( 3 + 2/i) . ( 3 + 2 2 } l} + ... + [2 (1 -t (n l) I
= - lit11
1: .
11, ➔o
lll\ l1
J _
I
311 •Zh (1 -I 2 -I 3 j-
x /r)~( 3
+ 2 X '31l) +, .. , (3
' -I 2 (11 - 1) I)
1-t
I I,

,,lin1:.0
(\

J = r { ... I (n l)} I

im h
h-+O 3n - 1 2 //1(~ I)\

= f =lim h { 2
h-,o 3n+h11(It - 1)1f
J
 TE INTEGRALS

19,103
I = lill, ~11 { 311 + 2 X 11 (11 - 1)1
U-HQ H f ,, , "L

,, ➔ "°
1 .,, Jim { 6 + 4 (
11

n
1
1}J ,, li;n,.,, { 4l
h 1 I I )\
t, ' (I 0) IO
II
We have l
/1
[ f(;,.: ) d~· = lim Ii f(n) + J(a
·,
1
Ir ~ 0 L
r · I)l + }(11
I .
+ 21,) + ... + /( 11 + ( 11 l
1) hJ , where Ir
b ii

,i = 2,l, =4,f(x) = 2x-1 and 11 =i-2 =~

4 11 /I

m
r = 2_f(x) dx = hlim
➔O.)7 \ f('l) ..., f( 2~h)+f(2+21!)+ .. +/(2i(11-l)h1)

o
h~\

.c
I = . h [12 (2) -1)1 + (2 (2 + Ii) -11 + 12 (2 + 2/l) - ll ' · .. + I2 (2 + (ll 1) It) - II lJ

du
l = l1Jun
➔ 0 h[3+(3+2h) -'-(3+ 4:/)
1 + (3 +611) L .. +!3+2(11-1) /1}]

oe
I= ~a71[3n+2h(1+2+3+ ... + (n - t)]
11
ch
11
I = lirn h { 3n + 2h x ' (n - l)}
te
h➔O 2
f = lim ~{3n ... 2x3_ ><11.{11.-lt}
m

n ➔ ,o n 11 2
fro

I = lim {6+
tt ➔ '1". ;
4(/t-lJ'} n
= lirn {6 +
n➔ rJJ
4(1-!)lJ= 6 ➔ 11
1
4(1-0) = 10
d

1PLE 3 Evaluate the. fol/awing integrals as limit of sums:

de

2 3 [Cl3..-,£ :?010
[ (x2 + 3) dx [CBSE 2001C} (ii) J (2x2 + 5) dx
oa

0 1
nl

L'TION (i) We have,

J~ f(x) dx = [ 1 ~ 11 I
w

lim h f(a) + f(a + h) + f<:a + 2/r) + .. + f(,1 + (11 1) /1 , when.> h

do

a /, '0

a 0, b = 2I f(x) = x2 + 3 and h =I::1'l O - ~

,l

• /lll (II I) 111 I

, t l(11 .. l
l) ,,· t I I
 rE ,NrEG
MATHEMA per'N'
110N (i)
sai.v /J
19.104
• ? n(n-1)(211- 1)}
f f(X)
I = Jim 311+ 11- 6 /l
⇒

J =
lt ➔ O

lim
{

2
-
l 4 ,r(n-1)(2n-1)}
311+7:x 6
a"" 1, b
3
n~oo 11 11 I"" J1 (
8 (n-1) (2n-1)}
I= fun { 6+- r- .
11 ➔.,

r
6

= Jim
11
2

J6 '- ~ (i -!) ( ~)lf 2- = 6 + { (1- 0) (2 - 0)

8 26
= 6 t- - = -
3 3
-fr 1
r ::: h~
l:iJ.1
n➔"' l

m
6 n

o
(ii) We have,
l

.c
r
"1 J(x) dx = Jim 11 /(a)+ /(a+ h) + f(a + 2h) + ... + f /a+ (n - 1) hi where h =

du
a
h-->0 l
3-1 2

oe
Here, a=l , b=3 ' /(x)=2x2+ 5 and lz=- n- = -n

3
ch
= f (2x- +5)dx
?
J
l
te

T = liln I! IJ(1) + f(l +Ii)+ f(l + 2ft) + ... + f \1 + (n - l)h)]

= lr ➔ O l I =
m

⇒
2
1 = lim hlr\2(1/ +51 + \2(1 + h/ +5} + \2(1 + 2h)2 +s)+ ... + {2 (1 + (n- 1) h) +
=
fro

11 ➔ 0

I = h~O h l
(212 (1 + Ji/ -'- (1
+ + 2h)2 .,_ (1 + 3h)2 + ... + ( + (n l -1) hf} 5n] +
d
de

[ ==
2 2
= I= h~O hl 2{n + 2h(1+2+3+ ... +(n - l))+ii2(1 +2 + ... +(n-1i2)}+s
oa

I =
,JL2 {n 2h
nl

I = + x 11 ( 11 - 1) + 11 2 n (n - 1) (211 -1)} ]
= fun
1,-,0 2 x 6 +Sn
w

l ==
;,nfl
do

1 = lim h { 2n + 2h n(n -1) + 2h2 n (n - 1)(2n -1) + -

h->0 -- 6

l {7n + 2 x 2 11 (n
= 1 = lim
n ➔ ocll
-
n
l) + 2 x j
11
2
:x 11 (n 1) (2n 1)}
6 [·.- /1 ..

J = lim J 14 +
n~ ,:.
s(~ ~J n
1 8 (11_!2 (2n
3 2
1) I
"'
- II
~
1
1~ 11
~
00
1141-8(1
L
1
n
)\ 8'l,11 1Jf2 'J 11 , 11
"'
(i.i) b
⇒ l = 14 +8(1-0)+ 8
3
(1-0)(2
3
OJ 14 1111 16 112
3
Jf(
Q
ExAM:Ll, 4
Evaluate tile following integrals as limit of sums: l'let<i!
a ""l
(i) f Ci + x) dx lCBSE 2000CI (ii) j (x 2 + Sx) dx ICBSE 20101 (iii)
1,
Ji· dx ..
1 I ,1 1
"'-
 f lNTEGRALS

N (i) We have, 19.105

b
J f(x) dx =hlimO h [t(a) + f (a -t h) -i JC
a
➔ r, + 2hJ t . . ' / (11 ;. (n 1) h1, whereh=-
b- a
2 n
a==l,b =3,f(x) =x + XMc,l 1, ,, 3 -1 =- 2 .,
3 n n
J(
2
I == x + x) dx
1

I= hl~o h [.rc1)+ /(1 + h)-t j(l + 2h) + j(] + 3h)+ . .+J (1 + (n 1) h)l

1= ,A~·10 It [ { 1 2+ 1 } + { (1 + h)2 + (1 +It) } + { (1 + 21i>2 ~ (1 + 2h)}

m
J

o
1i/ +h.-cn-l) h)r

.c
+.. .+{(1+(n- 1)

du
,~o"(\
0

I•
2 2
1 + (1 +h) +(1 +2h)
2
+( l +(•-11 h J)
oe
, .. .
ch
+ { 1 +(1 + h) + (1 + 2h)+ ...+ (1 +(n - 1) h)}-1
\,+ 2h(
te

I • ,,":;', I, [ 1 + 2+ 3 + . . + (•-ll)+h' (1' d+ ... +(n - 11' )} -

m

j
fro

+ { n + h ( 1 + 2 + 3 + ... + (n-1))}

n(n-1) 2 n(n - 1)(2n-l) h n(n - 1) ]

d

I = lim h n + 2h x _..:..-- + h x ~-.:....:....--'- + n + x ?

de

h ➔O [ 2 6 -
n(n-1) 2 n(n - 1){2n-1)°j
oa

I = lim h 2n + 3h x ~--'- + h x 6
k➔O [ 2
nl

2( 6 n(n-1) 4 n(n-1)(2n-l))
I = lim - 2n + - x ~-_..:..+ 2 x 6
w

n ➔ «> n n 2 n
do

n-1) 4 (n-1)(2n-1)]
l= lim 4+6 ( - +- 2
n ➔ «> [ n 3 n

I= lim
11 ➔ <»
[4+6(1-1)+{(1-~)( 2-~~)] 8
11 18
I = 4 + 6 (1 - 0) + 4 (1 - 0) (2 - 0) = 4 -r 6 + 3
3
~ 3
(
)j I> - 11
I) II , where h =- -
11
b [ f( + 2/1) I .• ' / II I \11
f f(x) dx = Jim h f(a) +- f(a + Ii) + a
h ➔O
a
3-1 2
re a=l, b = 3, f(x) =XZ +5x and h ==--;- ~ n.
3
I = f (x2 + Sx) dx
l
 flNITE INTEGRAL

~ (11 - J) 11 )]
19.106

f( l l 2h) + ... I !( ) I = Jim

lt ➔ O
. 1 /(1 + 1,)4 }
= \~, If [ /(ll • } { Zh/ + 5 (1 + 211)
h O } {. 1)2 + 5 (1 + /1) + (1 +
I = lim I, rI (l 2 I· .'i 1) I (I I I 2 (

j(,
X

,, ~ ll ,......., +(n-l) h) ,s 1 ,(,

J "' { (b -
(b -a)
I= -
3

1 = lim rI(l~ + (1 + 1,)2 + (1 + 21,)2 + ... +( 1 + (n - 1) h l} l =-

(b - "

m
3
lr ➔O 1 ( Evalu

o
j;.XA,WPLE ;
._ +5{1+(1 + h)+(l+2h) + ...+ l+(n-1

.c
2
(i) J e:r dx

du
0
, = lim h -~ 1H 2/, (1 ➔ 2 + 3 + •·· + (11 - 1) + 112 (12 + 2'2 + ... + (n - 1/)}
(1) ~
h ➔ O lt
SQL1.J710N
{"
oe
!,
+5 n +h ( l+2 + ...+(11 j JC-t) dx "'
ch
a

[611 + 7h (1 + 2 + 3 + ... + ◊1 -1))+ 112 ( 12 + 22 + ... + ...+ (n - 1)2)]

te
I = Jim Ir
~➔ O

fe
m

n(n-1) 2 n(n - 1)(2n - l)} I =

I = lim h { 6n+7h.x -'-----'- + h x
h->0 2 6
fro

= J - Jim 2{ 611+- n(n-1)

14 X -- + -
4 X
n(n -1) (2n-1)}
=
n ➔ "" n n 2 112 6
d

= lim J12+14(n- 1) ~~!n -1)(211-1)}

de

I l -=
n ➔ "' n 6 2
11
oa

I= lim f12+a(1-.!.)+j(1-!:.)(2-~)}=12+14
,,...., "' , \ n 3 n n
+.!x2=12+14+~"'
3 ., = l
nl

II •

! ~(a) ➔ r(II+ (n - l
w

[(Y) dx = h~lo h /(a+ h) + f(a + 2/i) + ... + 1) 1,) where /1

do

4
Here, {(X) aaX
b
I - Jx2 dx
a

= 1
f - ltlu:o h u +(1i+ hi°'' -t-(t1 I 211)2 I (a+ 11!)2 1 ,. I (11 1 (11 1)1, r
'
r Ii

~I
Jr
J = }l~o h [11a2 211h { l I 2 ' 3 t ••• I (11 I)/ ' 1,2 { I ' ' 2' ' . ' I \II l).l

=>
l =Jim 1,{11i+2al,x 11 (11 J)J 1 2. n(,1 l)(''u I)} \ii)
l,

⇒
II ➔ 0 2 ' I ><
6 J /(x)
a
-r
- h ~no { (nh) a2 + n (n/1) (11/1 - h) + I (n/1) (11/i h) (211/1 /1) }
6 1-fere,
a"'- 1,
 ITE JNTEGRAIS

1 = lim [<b - a) a2 + a (b _ ) (/, 19.107

h➔O ,1 - a- Ji) + 1 (b
a) (b
(i 11
h) {2(b a) hJ]
r. , h b a ·. nll-b-a]
n

o m
[NCERT I

.c
1710~ lD ,\·e ha...-e,

du
•x)Jx = 1lm
• ➔O r
h f(a)+f(a+ h)+J(11+2h)+ ... ➔ f a+(n -l}'h 1J,whereh= b-- a ( ) n

oe
_

X 2-Q 2
a=O, b = 2,f<x) = e and Ii = - - = -
ch
n n
te

n
m

i = ,lim_ h .../(0) - f(h) - f (2h) +.. · + f t (n - l) h)]

fro

n~ u - -
J= lim he -e - e
- 0 h 2h
+ ... + e
(n- J)lrJ
d

h-0
de

: = lim h -e° jci/1 iI

11-
ar
l
=11
r Pl -1 '
-
oa

-1 ( r-1
~ .... o e -1 JJ
nl
w

~ '
do

, nh ·: 1, = - c:- 11h ~ 2
e -1 II
I = lim h{ h ~
11-.fJ e -1 J

e1 - 1 e2 1
,·2
f=lim-r- 1
1r ~o i-1
h ) I• -,1
•Jl 11 ' (11 l) Ii) j' where Ii,_ II •
f( ~ /t) j (11 I 2/J) I
jC j(J;J dx = lim [
h J(a) +- a
:, h-,0 1 - / - l)_Z
4==-1,b =1,j(xJ =e X and h-
- 11
- 11
 MATHEM

19.108

1
I= limh
1 = f , .... (/.r
h ➔O
-J ~(n-J)h)]
r = e2 (e -

·tuting these v
/J
Jf(x)dx~

m
a
4
~
J 2.r dx =

o
.c
2
4

du
J2 dx =
2
=
oe
ch
te

·: Jim
[ lt ➔ O
m

EXAMPLE6
Ei~duate the foUowing integrals as a limit of sums.
fro

l
(i) Je2- 3x <ix INCERT]
(ii)
4
J 2x dx (iii)
J
J (e 2- 3x + x 2 + 1) dx
0
d

2 1
SOLUTJON We have,
de

b -

! f (x) dx ~ ~j\ h V(a) ; f (a+ h) +f (a+ 2ft) + ... + / (a+ (n -1) h)], where Here, W= t !.;
oa

2 :bstituting thc•,c \
(i) Hcre,a~O,b=l andf(x)=e - 3x.Therefore, ft=~~ nh = ].
nl

t,
J

Jn e2 - 3, d~-
11 f F(:c) J:c = IIln-'
w

I - •
"
do

I= J(<'~- .3, ~
I
I = lim /1 [(,_,2
1, -0

+ It' 2-3(1+
 19.109
I = fuJl0 he2 [::.e-- 3nh
h ->
__:_!]
3h _ l =
2
]" <' - ::i 1
(~-::3h~ ~l X - ~
P

I,~

2 3
1
l -31, J
I "' e (e - -1) x - - = _ ~ (e- 1 2
3 3 - C ) = 1_ (r2 _ -1)
3 P.

ere, 11 = 2, b = 4 and ·f {x) = 2X · Therefore h _ 4 2

, - --- ⇒ 11I1
tituting these values in 11 =2
~
Ja f(x)dx= ,,li~o 11[/(a)+J (a+li) ➔ f(,1+211) +···-'-!( n + ( 11 - l) h )] , we get

m
4
f2

o
2x dx = 11lim
➔ 0 h [f(?)
- + .f (2 +II) + /(2+211) 1- ••• +/(2 l•(n-l)lijl

.c
4 , ) I

du
= 1t{ 22 +2 21 11+ 22+21,
2[ 2.rdx lim
/r - >0 + ... + 2
2~(, l)'1}
I
-

oe
4
J
ch
te

2''/t - l
m

h~)
fro

[ 2h

1
d

4"J 2xdx=4x[22-ll =J2_ [·: nh .o

de

= 2 and hm -21, -1
- = lug 2
log 2 log 2 11 -> h
oa

3
1Here, a =1, b = 3 and /(x) = e 2 - 3 x + x2 + 1. Therefore, /1 = -l 3.n and nh = 2.
i-1
s=>
nl

Slituting these values in

w

b
J f(x) dx= l,
do

Hm h lrJ(a) + J(a + h) + J(a + 2/t) +... +/(a+(n - 1) It we obtain

• h➔O J
1= J (e2- 3x + x2 + 1) dx = lirn h [f(l) + f(1 + h) + /(1 + 2h) +... 1-f( I+ (11 - 1) It)]
1 h➔0

l = hli~ h[(e2 3i<l + l 2 + l) +{e2-3('1+ Ii) +(l + J,)2 + 1)

2
1e2- 111-( - IJ//1 .,. [I
11
+ [e2-3(1T2h) +(l + )2 + l) T, . . 1- 1-(11 l)h\ -f ·11
211
l-3h . -1-3(211)
+e
,-J- 3(n
....... + t
l)lt}
/)2 (1+2h}2 ➔ (1 +3h/+ ... + f1+(n - 1)hl }+n]
2
2
+ 1 T (1 + + 1
{
,---,- -·- •·- - - - - - - - - - - - - - - - - - - -

19.110
-3/1 -3(21!) ,.e-3(3h)+ ...e
-3(n--1)11} + {n +211(1 + 2 + 3+" .+
-l l+e -t-e
⇒ I=limhe
[
h->0
{
+/,2
{2] 1-22+321·... + (n-

- -1 {( -~'')"-l} + lr211 - 2,I 11(11-1) +~~11(11-1)(211-1))] I - lim

⇒ /= ,:~n le 0
h
t' - - -
--;-3/J _ 1
2 6
. h->O

I = ros a

= . [ 1 (e.!.,'-3/i" -1l I+I 2,,h +nlr(nl, - /t) + I6 nit (11h - h) (2nh -h)
1 = lim hr
fi-40 {' - J .. 1
uate the follrr.ci

m
3
J (x • 4J dx

o
6

.c
-1 (e -1_)- + 4 +2(2-h)+I(2 - II) (4-h)
= 1 lim
i,40 e x- ( e-311 - l) 6
a

du
3
-3 -
~ -3/z
J (3x -2)dx
1

oe
-1 6 1 1 -7 5
=> 1= -e- (e- - 1) 1-4+ 2 (2 - 0)+-(2 - 0) (4 - 0)= - - (e J (x - 1 dx
ch
3 3 , - - - --, ,3
I LEVEL-2 I C
te
5

CXA\fPI E -
r,
Evaluate J sin x dx as limit of sums.
r l 2-.t}dx
m

3
a 2
J x·' ±c
fro

SOLUTION We have,
l
Jf(x) dx = /Jlim h jf(a) + j(a + h) 1 /(a+ 2h) + .. . + f(a ;- (n - 1) h)], where h '
J
d

a ➔O L (x: -!\d.t
de

Here, f (x) =sin x l

+
b J (x: -x)J
oa

J J sinxdx !
•
j' i!x,fr
nl
w

(l
~
do

. r h) sm. rzh J ..:~>S \' .t.,

. swla+ (n-1}
2 sin ( 11 + nit - h ) "!
= I = Jim
" -, 0
h - - -----
. Ir
2
--. = lim h -
\ 2 2
It

J costd.r
l sm 2 11 -➔ o
~11,. h (1

2 -f' (3x 1)- 2

rsin ( It 1 L)
(I j b (I
"J
hi
l /
2J . 1'' 112 6 1J) ~
- Lim l ,. '"I. J(.\ +eh)'
h-,.Q
Sill
•
l
I
,0
J (x 2
+ 2x +
0
!TE t,iTEGRALS
19.111

I -= cosa-cosb [·.. 2 sm
. A sin 8 = cos (A - 8) - cos (A + B)]
EXERCISE 19.5
. . I LEVEL-1 [
uate the following integrals ns limit of sums:

m
2
3
J (x + 4) dx 2. J (x + 3) dx

o
0

.c
(I
1
3
J (x + 3) dx

du
f(3x-2)dx 4.
- '1
l

oe
3
5
f (x + 1) dx 6. J (2x ,;- 3) dx
ch
1
0
2
5
s. J (x2 +1) dx
te

f (2 -x) dx
0
3
m

3
2 :JO. J (2x1 + 1) dx
f x2 dx
fro

2
l .2
2 12. J(x2 + ~) a,x
d

{(x2 -1) dx
de

0
1 1 2 '
4
j (x2 -x) dx t~. J (3x + 5x) dx
oa

,CERT, CBSE 2010, 121

0
1 b
f
nl

2 16. e:" dx
fexdx INC ERT]
w

/I
0 11/2
do

b 18. J si.nxdx
f cos x dx ()
a 4
1li2
10. J
f cosx dx 1
0 2 [CBSl 21.lOOl

l,
2
! (3x2 - 2) dx V I
0
0 2 l C llS F ~005I
4
24, J(r-1 1 x) ,II
2, j (x+ e2x) dx l '\JCllfl I ii
[CUSE 20071
2,
0
2
j (x2 + 2x + 1) dx 26
_ f1 (2x2 t J.x + 5) dx
· [CBSF ;!.007]
0
0
 MATHEMAT
19.112
5

27.
b
Jxdx [N CERT]
28, J (x + 1) dx
0
a 4 I = lim h

19.
3
f x. 2
dx
[NCERT] 30. J(x 2
-x) dx IN Cf.RT, CBS I, ➔ O

1
2 3

31.
2
J (.? -x) dx [CBSE20111 32. J(2x2 + Sx) dx
1 l=limh:
0 h-.o
3
33.
3
J (3x2 +1) dx [CBSE2014) 34. J (x 2
+ 3x+ e"') dx
I = lim h
1

m
1 lt-,O

o
5. -35 8. 14

.c
1. 33 2, 8 3. 8 4. 6 6. 14 7. -4
2 2 3

du
-ll
10. - 1L ! 12. 32 13. 27 14. -7 15. e2 - 1 16. e b -e17 17. sinb
3 3 3 2 2
23. 1S+e8
oe
18. 1 19. 1 20. 78 21. 4 22. 20 24. 14 We kr.o·v L'iat
ch
3 2 3 I>
2
27. b -a
2 ff (x JI -:
28. 35 29. 19 30. 38 31. 2 32. 112
te
a
2 2 3 3 3 3
,. 62 3
= - +e -e
m

3
----- - - --- - - --HINTS TO NCERT & SELECTED PROB
fro

4
13. I = J (x2 - x) dx
d

0
de

We know that

h)]
oa

!f(x) dx = h~ h[f<a)+f(a +h)+f(a+2h)+ .. . +f (a+(n - 1)

nl

Here a= l,b = 4,h =~and f(x) = x 2 - x.

"!' " '(
w

n , I =j (x • . - )

~ (1) + f (1 + h) + f (1 + 2h) + ... + f ( 1 -~ (n -1) h)

do

0
I= { (x2 - x) dx = hli_!-O h \\., kn0w that
~
J;
⇒
d.\ •
[\(1' -t)) ((1 -(1
(X)

I - ."'!', h + +h)' + h)l + [(I + 2h)' - (1 + 2/,)l + ..

H.-re, ,1 ,. 0, /, -

~ {(1+(n l) 11 )2 . f4 (x t ,.~., \ ) ,Ix

⇒
0

I - )"!', h [( 1' + (I + hJ' • (I •· 2h)1 • .... ,-( 1 ' (, • 1) 1, )') = lin:t 11

h--, 0
l<O + e°

= lilim
➔O
1r[{o+l1
- { 1+(1 + h) + (1 + 2/r) + .. .+(1+ (11-l)
 19.113

I= lim
It ➔ O

I= lim
h➔O

m
I= lim
n ➔ OO

o
.c
I = lim

du
n➔ oo

oe
ch
te
m
fro
d
de
oa
nl

4
w

I = f (x + /x) dx
do

0
We know that
b
f f(x)dx = lim h[f(a)+f(a+h)+J(11+2h)+ , .. +f(a+(11 - l) 1r)]
a h .... O
Here' a =O, b = 4, f (x) 4
= x +- e2x and h == -;:;

:. f (x + e2 x) dx
O
= lim
lt ➔ O
h [1 (0) + f (h) + f (2 h) +, ..
1
f ( (/1 I}
11
J]
" .'-'!', "[<o «'l +(h +,''i +(2h ,,'''") +(" ••""'J' · ·'(<• ll/<+ ,, ,,. _ " ' ) ]

" Um (J }{ 2h + e2<2h) + e3(3l1) + ... 4 e2 <11 - J)l1}]

h-.oh l0 +h+ 2h + 3h+ ... + (n - 1)h + 1+e
 11'41TE IN'TE
19.11 4
2/1 4/I + e611 + •·• + e
2 (n- J) h}] 5
1+e +e J (x +
= Jim h [1, {1 2+ ·.. + (11 - l)} + {
1· I 0
h ➔O e2"'-J
2 Ir 11 16 11(11 - 1) 1·..-. -2-h - -
+ i,.,
. { Ir 11 (11 -1) (e l_-:!_
= 1in1 2 x
---
2 /J -> O e -1 We have,
= hlim.
➔O
lz - - - - + 21,
2 e -1 } 11-➔"' 11 h
8 8 Bere,a =
8
es _ 1 :...-1 = 15::..:_
1) . e -l = 8 (1 - 0) + - = 8+ 2 2 3
= "lim 81
( - + hlim e2'I - 1 ) 2
~"' 11 ➔O [
2 --
J ,T
2

Zh L
'

m
b h
f x dx = h-,
Ii.In

o
li. Let I = D

.c
We know that = lim h

du
f f (x) d.t" =h~o h ~ (a) + f (a+ h) + f (a + 2h) + ... + f ( a + (n -1) h) J w h e re h =
h--0

oe
•
Here, f (x) = x
ch
! = h~oh[a+ (a +h) +(a +2h) + ... +( a+(n - 1) 1r)] = ;,I.in::.
➔0
.!:
te
xdx

!
b
m

= xdx = h~oh[n a + {h + 2h + 3 h + ... +(n-l)h}]

fro

1
= J.!ll1
j xdx hlim➔ O
= [nha+ h2 n(n - 1)1 = fun [(nlz)a+(nh) (nh - h) ]
n _,,. x- n
d

a 2 - h➔O 2
de

=
⇒
j xdx = hJim➔ O
a
[(b- a) a + (b - a) (b-a ~ h) ] _ (b- a)2 b2 - al
- (b-a)a + -- = -
fun
oa

2
2 2
5
2b. \tic have, I = f (x.,. l) d.x W tha t
nl

,..
· e N.10W
0
Eruluat,· th,:.;
w

b -I

! +f( + ( - 1) h )] , where h r
do

l. X -1
f(x) dx = h~o h [!(a)+ f(a -l h) + f (a + 2h) 1 ...
11
\
11 0

Here,a = O,b ~ 5 /(x) _

' -x+Jand lr =- 5
~.sI t
5 /1 \ ,;,
~.\:
f
(x .- 1/ dx ,
hli:no hr
f (O) 1 f (h) I f (2h)
1

0
s
1
• I t ((11 l) li)J :i.
1
J tan
= 0f (x + 1) dx = ,,lim>0 /, [1 ( 0
. I l + /J/ 1 (l 121s) I (I 1 :I/)
I I .. . I
(
I . (II l) /J )
J
5 1
⇒ J (x + J) dx = I tan -
0
Um
II ; 0
11 111
l1i (1 , 213 1 ••· I (n I ) )]
7,
0
ililfE INTEGRALS

19.115
5
J (x i J) dx Ii m
II , ,
i
n
11 ; 5 ,. 11 (11
n 2
J)
(J
3
,, have, / = J x2 dx
2

mo
.c
du
oe
ch
te
m
fro
d
de
oa

------
nl
w
do

:ialuate the foll011Jing integrals:

2

L
4
f :r. ✓4-x dx i. I x (3x --2 dx
'I,
1
fl
I
s I 'I' 11,
3. r ✓2x
' - 1 dxX
4 J cos
l)
1
I 1
I (I : }/.I
1 J ('( )I,
I I t
s. I tan - 1 x dx /j ,
0
{J
u.J3 1( 3x -x; )dx
'·I 0
tan-1 l 2x 2
l - x
Jax
jj.
J
0
tan 1 - 3X
 MATHEM

19.116
n/3 cos__:__ dx
1
l-X d
10, J 3 1• 4 sin x
9. J --- X
l+X
0
()
it/2 · X
sin.:..-~ dx
11.
n/2 sin2 x
JO -(J -+ - ) 2
dx 12
· J /1 + COS X
0 '
COS X
2
nJ sin 3 x (1 + 2 cos x) (1 + cos x) dx
1t/ 2 COS X 14.
13. - , dx
J- 0
0 1+sin"x
it/4
16. J siJ.1 2x s il.1 3x dx

m
"' x dx
15. [ 2) 0
tl

o
ll + x) (1 + X

.c
-2 -
1 --. 2 1 sm
11-.t dX lS. J2

du
1- .
J() ~l+X
--
1 X

oe
4 3 /! + COS
,r./2. X 6 - CC-, X

19.
4,

.f COS
4
X sin x dx 20. J ' 5/2 dx
ch
n/ (1- cos x)
3
11
te

-:t/2 1
1L f x2cos2xdx 22. J log (1 + x) dx
m

0 0
X d:.
fro

I0 x (tan -, x)2 dx
.f;. x2+x l
23 _ dx 24.
✓2x + 1
i ..::.x
d

2
de

l 2
2.5. J (cos 1
x)2 dx 26. J Xx(X,
+--
3 d •
.X 3 ,
- -'
i
oa

2) l
0 1
-:t/4 n/4
nl

l
27. J e'" sin x dx 28. J tan 4 xdx 29. J I 2x - 1 I dx
w

0 0 0
do

3 ,r./2
30. J I x2 2xl dx 31 . JI sin x - cos x I dx 32.
1
J I sin 2n x i ,fr
1 0 0
3 n/2 l/2
Jr;/ 2 Sm. 9 X dX
2
33.
f1 [x -41 dx 34,
35, J \"OS I
1/ ...
a 2
x ex n/2
36. j ---i- dx 37. J l 7 ilx
2H
-a 1 +X () 1 ; cot x
311,
Ill l'n/ .1 d.1

39. j fi
.Jx + ,Ja - x x
d
40.
n/2
f 1 3 - dx
n
.1· s i.11 1·
0
o 1 + tan x
41.
J0 l + cos2 x dx
 19.117
r.
·U . _[ ~ n/4
11 11" co,; 2 , + 12 . 2 rlt
> s 1n ,.
44 J Jtan xJd%
•II
\
1--.;_ en~
= ..~ 8U1
. -~ dx t7,
~12
j X •t X co,-, X dx
o sin x + rQ<; 4 x
r.
dx cos 2x log sin x dx 50. i•
:-,in .t ~ cos x 2 X z dx, a,.,J
o tr-cos x

m
n./2 . 2
. x tan x d • J sm x
" 3.

o
·' ,,ec ., . + tan .\ X -
sin-x+---
cos -x dx
0

.c
i- -- f x JO sm. 7 x dx
;r. 1
f coC 1(l -x+x )dx 2

du
X
-,----,-dx :,::,. 56.
0 ~in- X ~ t.'OS- X -;r. 0

oe
~,2 1 _
n/2
J cosec x c~ x dx
58. J - - -'---- dx 59
ch
2 cos x + 4 sin x ;r./6 1 + cosec x
0
te
. .-' dx
i ~cosx• 2sinx
m

~a,e :h.:_fo//owing definite integrals as limit of sums: (61- 69)

fro

4
J (x2 + x} d.x
2
2
62, J (2x + 3) dx 63.
1
d

0
3
de

?
1
65, J e
-x d
3
X
66. f1 (2x· + Sx) dx
J e"x dx
oa

2
l 3 ?
, 2
J (i .,. 2) dx (>9, f (x· + 1) dx
nl

68.
(x2 + 3x) dx 0
0 ANSWERS
w

l
l I ,
do

1t
5 -- og-
4 - . -1 .:.
326
2.135 'I, :!lo~:!- I
1 4
lo~ 'l
'1t 2
7. - Jog2
2 R
II I I,
Jt 12 ,?(,.!
II 2
10. } 1og (I+ 2 f3 ) 2
\ 3 ) ",.
17
fl IS ,.
'
[(, 5 [2 '
"-7
Js
3
n - lu~(~)
p
lJ,
5

20 . -2
 FINITE I
19.118
1
2
2s. 1t- 2 26. ~ log 6 27 -
· 2 TbeV
1
1t
?4 ---+-log 2 1t
~ . 16 4 2
2 (a) 0
31, 2(J2 - l) 32. -
1t 2 1 30. 2 'It
28. 4-3 1Q -
... 2 Toev
1t
36. 0 37 -
33 4 34. 0 35, 0 ' 4 (a) 0
c,:,

7t2
a
'.'IQ - 40.
1t
41. 2 42• 5-
1t
J0 1
38. 0 . 2 4
(a) l

m
2
2 1t2/4
7(0, 7(
4 -Jt 44 log 2 45. 2 -.Jz 46. 47. 16

o
2 ab sin a 7 J

.c
0
2

du
1 log (✓-2 + 1) 1t 1t (a) 2
4H. ,J2 49. -2 50.
2a ✓a 2 -1
oe
5.

53. 1 2
ch
-, 1 1t
.:,~. 2 log (.fi + 1) 54. 8 55. 0
te

,: /2
•=-- ~ 1 l { $+- 12) } -1(1)
m

J35 SB. 2$ og 2(,Js 59. tan

3
9. f
0
2
fro

6" - 1 Iog {$+1} (a ) ~

$ $ -1 61. 8 62. 34 27 ~
d

63. -
3 2
de

b"
r 2 -e -
-1 \e 2)
65 e·2 -e- 3 66. 112
10. J 1
oa

2
3
67. 62 68. 20 0 \
,, 12 3 ~
.:,
(a)~
nl

'>
w

- - - - - - - - - ~ _MULTIPLE CHOICE QUESTIONS (M 7t

Mark the cortect alternative in each of the following: t l. I0

do

1 11 +
1. J .Jx (1 - x) dx equals (a)
0
(a) n/2 (b) it/ 4
1 (c) rr./6
7t (d) rr./8 It/ J
2 f - - - dx equals
· l +sinx
0
l:?,
I
lt / 6
1
(a) 0 (b) l /2
(c:) 2 (a) 7t
rr x-tan x (d) Cl/:1
3. The value of J- - - dx ts
secx+co$x 13,
0 Give

(a)
1t2
4 (b) T ·2
(c) ~'it
2
(d)
7(2
thev-n
2 3
 value off2"0V,I1 +sin!_--;.2"1S
1 'it...
19.119

(b) 2
1t/2 (c) 8 ·
(d) 4
value of the integral J -=.lP,s""'x
o .fco~ x +~sin !· 4x is.
., 0 (b) rc/2 ( 1
1 c) 11/4 (d) none of these
:, ,.,.,,,.. --,:- dx equals
1➔ e

m
) log2-J (b) tog 2
(d) - log2
• '2/4

o
sin ,Jx
J Jx dx equals

.c
0

du
(b) l ( (ct) i/8 I

oe
I
cosx
(2 + sin x) (1 + sin x) dx equ;al
ch
of:1,,rn (b) ... rn . .11,,m
te

f .,- + 1COSX dxequals

m

' ,,,
1,
fro

~a) ~ tan - 1 ( -1- ) (b)

3 ./3
~ tan - 1 [ ·
[3 ' I
1
.J.itan- J3
•
d
de

"J /yl~
~+dx=
x
oa

lO
(a) !:. " ' (d.) it + i
2 (b) 2-1
nl
w
do

(d) (a ~ b) re
(a) ir (b) ~
Ja2 -b2 ab
11/3
J 1 dxis
n/61 + [cotx (J) 1t/2
(a) n/ 3 (b) tt/6
a, 2 ~
• Given that J __ ----3__ 2 - ~) dx .. 2 (a 1 /!) (/, ; r) (c I a) '
o (x2 + a2) (x2 + ti2) (x +c

the value of J dx -
2
o (xi +4) (x + 9)
' is
18UO .
• "
I
• •

d
{ I

m
(A) 4

o
I

.c
I I
I
J

du
\ In l
2

oe
f 1 slnr&
0
ch
J r.
00
te

ta)
2
4
T ,U
m

I I tJ""
fro

15
(, I I
d

I I
,,f l
de

(nl
oa

"
r
nl

Th, u1lut-•
w

(ll l l l la) 4
do

I
t
f l• ◄
I cir
11
!l
" 2
r i ,
i,l•COI t
(l,I l 'I
: .2
4
la 0
" l
J d:i. l j 1 :i. di • J:ti 4r ... J m.t+cos.t
~II\ :r ti,

" (b
2.Z Deneal
a i:

j C{)S•
• .a lb d
J di SUI

1:.• a
4
 1
dx is equal to
~in 2~

(b) log,.{'?- 1
(~) .2 Ing r I)
(d, log( 1

1- J. l d) is equal to

(b) 2 (c) 0 (dj 4

x3
deri,·ativc off (x) -= f t
- - dt , (x > 0), 1s
loge I

m
x2

o
i xr 1 x (x -1) (d) 3 -2:r

.c
(b) l - (c) (In ln
3lnx 2lnx

du
r./2
lio = f x10 sin x dx, then the value of I10 + 9018 is
oe
0
)9 s
ch
9 ?
r (d) 9 .!:
2
te

x54dx =
m

f1 - X)
fro

15 3 (d) _ 16
(c) - - 3
16
d

~: -sin 2x dx is equal to
de

(d) 2 (-.:!-!
oa

(b) 2(.fi+1) (c) 2

(1'CERTEXE l1'1.'1l
nl

2
f 11 - x2 I dx .
15
w

Tr,e Hlu.e of the integral

-2
do

(c) ... 2 (d) 0

(bJ 2

1
, dx i!> eqm,J to
l+crJt~ ,
(d) 4
(b) l
'1/1
J , &in x dx equali; to
(I 1>11, :i: .. co~ x
(d) II ii
(c} 11/ 3
'") It (b) 1t/2
1
Iv ux.,d ,fSein 1 [ 2x
2
J1 dx !ti equ,d to
1 1+x J
(c) n/2
19.122

n/ 2 I to
x. Sin x dx is equa (d) 1
34.
I
0
(l,) n/2
(a) 1t/ 4
,r/2 . dr i5 equal to
sin2x Jog tan x ••
35,
J0 (c) 0
(d) 2 n:
(a) 1t
(b) 1t/2

nJ _!___ dxis
3b The value of 5 + 3 cos x
0
(c) n/2
(d) 0
(b) 1t/8

m
(a) rr./4

l 1) 1,
1

o
3''. "°f log x+- --2 dx=

.c
• X X
0 1t

du
(c) 0 (d) - - ln2
(a)1tln2 (b)-1tln2 2

oe
2a
38. J f (x) dxis equal to ch
0
a
te

(a) 2 JJ (x) dx (b) 0

0
m

a a a 2a
r f (x) dx + I f (2a - x) dx Jf (x) dx + f
fro

(c) (d) f (2a - x) dx

0 0 Q 0
b
J x f (x) dx is equal to
d

3o If f (a Tb - x) =f (x), then
de

a
b
(a) -11.,.b J f (b -
oa

- x) dx a +b b
2
a (b)
2 - f f (b + x) dx.
nl

II
b-a
2 -
bJ
{c) f (x) dx a+ b b
w

a (d)
2 J f ( x) dx
do

n
40. The value of J tan - 1
(-~ - l -) dx, is
() ] 4• X - x2
(a) 1 (b) 0
n/2 (c) -1
(cl) n/ 4
41 The value of J log [ 4 1
3 i-.in XJ dx i
0 4 ~ 3 f<lS X S
(al 2 Cb) 'l
4 (c) 0
(u) 2

(b) 2
(c) it
(d) l
INTEGRALS
19.123
1
- - - dx is equal to
1 -,. co52X
(b) 2 (c) 3
(d) 4 IN( f:R I' I X~Ml'LARI
is equal to

b b b- t
(b) jf (x+ c) dx (c) f,, f (.,·) dx (d) Jf(x) dx
a C

~d g are continuous functions in (0, l] satisf . _ [ N ( fRT T'XEMPLAR I

.f ,,[ dx. ymgf(x)-J(a-x)andg(x)+g(a ~x)=a

m
(x) .,,x) 1s equal to '
l0

o
.c
a a b
(b) f
~ f(x) dx (c) f f(x) dx (d) a J f(x) dx

du
20 0 0
INC~RT EXE"1PLARI

{sm - 1 x) dxequal<;
oe
ch
(c) - 1 (d) 1 {CBSE 2020!
(b) 0
te

_ _ _ __ ___ _ _ _ _ _ _....,..-"--~ - ' 1 - - - - - - ANSWERS

. 6. (b) 7. (a) S. (d) 9. (b)
m

(c) 3. (a) 4. (d) 5. (c)

~ J7.
1s. J a) 1~. (a) 26. (c) (b) 18 (b)
(a) 12. (c) 13. (a) 14. (a) 27. (b)
11
fro

22 (b) 23. (c) 24. (b) ' 25. , (b)

20. (b) 21. (b) 33, (d) 34, (c) 35. (c) 36. (a)
30. (a) 31. (d) 32. (d) 44. (b) iS. (b)
29 (d) 41. (c) 42. (c) 43. (a)
36. (c} 39. (d) 40. (b)
d
de

_ _ _ __ _ _ _ _ _ FILL IN THE BLANKS TYPE QUESTIONS {FBQs )

oa

1 n
dx =-,thena=,................................... .
-4x2 8
nl

;t .,
J sin 3 x cos- x dx is
w

value of ......................... ·
do

-:'t
::/2
" value of f esill x cos x d."< is ................................ ·
0
f _X_ - ................................. .
v?- +x2 dx - .......................

The value of th(: integral

2/tt ~m
J
. (1'ix J ii~
.
iii ·" ........... · .................
1/n X
l -IX ••• u•u• ' , •.. , ••

The value of the integralf 1,3n -;- dx i5 .................... ·

0 J+ X

The value of the integral J/:( 1_ - _!2 Jd:r is .................,· .................. .

l X X
- - - - - - - - - - - - - - - - - - - - - - - - - - - - -- ,m...,......,,..,

19.124
.,., .....
l:' ······•"''"'"'
::_ (1 + logx) d~ = ................... ..

CJ
1 x
11,e value of the u1 c,,. •
. t uril f2z (ax5 +ln··'~ I
/)
1 where a b, c, d a re con.stan
ex t i ':<, I

............ ,........... . 2 {)
011-ly on ............. ,............. ..
10. "f --- L
- d.\ - .......................... ,........... .
· 1 +:-in.,
0

11
'r: gi.n --.-4
xcos~_- d,· = ...................................
'
·
~ l +~m .\

m
it ~
~ J tan"xscc2 xdx = .............................. ·

o
1

.c
0
rr/41 + tanxd .

du
1
Th value of
e f0 -·- - -
15
1-tan.x
X .... ........... ·

oe
,-
3✓X •
The value of J -1-dxis ........................ .
ch
0
..,x
2
n/2 . o.--;. l:
smx a .
i
I, .· .
te

The value o f J0 - - - 2- x 1s ..............,.. :··,··.. ···i •. , , '

1 + COS X I• , •.
m

a a/2 • ,1 • .
J
Iff(a -x) =X and .f(x) dx=k ff(x) dx, thenk ,' .. :......: .................,. ........ .
fro

1
0 (j .,
2,i
J cos7 xsin4 x dx is .................,....................... .
d

The value of the integral

de

0
1
Jx I x I dx is ...................................................... .
oa

The value of the integral

1
nl

• The value of the integral J log ( ~ ::) dx is, ................... ,......,,.....,..... .

w

-1 .
do

1
2 The value of the integral JIi -x I dx i!'i ............................,.................... . ,I -

-1
1t/2
21 The valm> of the integral J l
lJX if. .............. ,. .............. ,"''•·· .. •
() J + ta11 3. X

- • lf or ,,~ X dx. = kn ih1•11 k

"2
......... ...... ,... ,, ........ ...... ............ .. ,........ ,.
-n
Va+ X ' ,
,

23. lf f(x) = f(a - x) and J x f(x) dx

u
f
"
k /Ct) t/\, l'lw11 k
0 0
24 ·Th. e value of the intl!gral 10
f 1-
IU
<Ix is ........... .................... '
() (l(l -X) 10 ~ X 10
 19.125

I,..'.-- - -- - ·· 'I· I·

2. 0 ANSWERS
a; I

7
e2 -e
2. 10, 2
12. .:!. l
13. --2 loi>
.,, 2 2 '2 r.
7 14. - - (3" ~ l) 15
log,3 4
17. 0 18. 0 19, () 20, 2
22. a a

m
23. 2 24, 5 25, 2(e-l)

o
;,.-- -- -~ -- - - ~ VERY SHORT ANSWER QUESTIONS (VSAQs)

.c
ch of the following questions in one .wore/ or.011e sentq,ice .or 11s per exa,t requirement of the

du
each of the fallowing integrals1 (1-30)
, n/2
-sin f <ms2 x·dx. 11,/2

oe
2 x dx. t
2, 3. J sir? x dx
0 :t/'l.
ch
2 it/2
cos2 xdx. .Q• f x cos x dx.
2
te

•-n/2
12
I !
m

,. Jl ..::.Cdx.
' X
fro

-2
3
I
12. J -y-dx.
1
d

O X -~ 9
de

2 "
12
(3+ 5co,, X)
g _ _ dx.
~ --,,--
.JI - cos 2.x dx. JS. J 1 0 3+5 sin X
oa

e
10
2 . n 17. 1cos5 .r
nl

dx
f . n
sin x
n
dx , 11 1;; N. 0
w

ip sm x -;- cos x I
J rl~I
do

t./2 ) J'l. d.r

f log ( a s'.n 8 d() ., -d
,,.12 a+ sm 0 ri/4
1
!
:?;. J Mt\~ 111 IC ll'll ZIH II
J f (x) dY: i1. f -1 +~ dx rclJS62tll18l
()
• f (x) + f (a -i. b - X) n X

a1
J -dx ICBSE 2tr12l Z4 j J~
()
0
\'~ 11,,
2 X
I I 2
1
j 2x [C6S'E .20l2l
21,. JI l 11\
--- dx
o 1 + x2 0
,,2
L 1n 1su20111
:t/4 ,II
j sin 2.x dx rcns11: 29141
2!1. J x lo~
f
1

0
 19.126

29.
rr/2
f e·\sinx -cosx) d:x [CBSE2014
l 30. J4 XyX+-l dx
2
C

31. If f (3.i + 2x+k) d.r=O,findthevalueofk.

0
a
32. If J 3.i dx = S, write the value of n.
0

33. If f(x) f
= t sin t dt, then wrlte the value off' (X). INTR
0 atio
a ej
fo4+x-
- _!.__,, dx =~,8 find the value of n.

m
34. l£

ii

o
.c
3
35, vVrite the coefficient a, b, c of which the value of the intes-ral f (ax2 + bx+ c)

du
AR
-3 EO REM
independent.

oe
== f(x), the
3 b
36. Evaluate f 3x dx. f
ch
2
te

lf[-] and 1-/ denote respectively the greatest integer and fraptional partfandions r_espectively,
following integrals: · ,
m

2 15 1 1
37. f [x] dx 38, J [x] dx 39: f (ai dx 40. J i"I dx
fro

D O o·. O
2 l 2 .Ji
f J 2x-[x] dx 'f' log [xJ fix. J [x2J dx
d

4L x[x] dx 42. 43.. 44.

de

8
0
r.J4 D 1 0
J sin {xi dx
oa

45.
0
nl

1
. 2:4
1t
w

2. - 3. 1t
1ll
4 2
4.
5. 0
do

1t 2 Ii 0
7. 1-- 8. 7t
4 4 9, -1 10. 1 ·11. 7!: It
l :!.
1,3. ✓2 14. 0 2 I:?
1s. 0 16.
1C

4 17. 0 l S. (l
19. 0 b-4
20, ~-- 21.
1C
1

25. loge2
2
J
4 22. 2 Iog 2
'.l. log,{ ~ )
26. l
2(e-1) 27, ~
2 30.
31. -2 32. a"' 2
· 16 33. f' (x) 0 x sin :r 3,l. II = 2
36. - -- 3'7. I 1 :l'i. b
' loge 3 ~8. - 19. J. IJO. <' - I 3
2 2 ;I I
42,' _,'__
1 2
_lpgi> 2 . 44. 2