8.indefinite IntegrationExercise
8.indefinite IntegrationExercise
8.indefinite IntegrationExercise
27
1 1 x
10. ∫ = dx sec−1 + c (Valid for x > a > 0)
2
x x −a2 a a
eax eax b
ax
(asinbx − bcosbx) + c = sin bx − tan−1 + c
11. ∫ e sinbx=
dx
a2 + b2 a2 + b2 a
ax eax eax b
12. ∫ e cosbx=dx
2
a +b 2
(acosbx + bsinbx) + c =
2
cos bx − tan−1 + c
2 a
a +b
Solved Examples
JEE Main/Boards ∫
dx
n ( sec θ + tan θ )
x2 + a2
= ∫ sec θ dθ = log
x + sinx
Example 1: Evaluate : ∫ 1 + cos xdx
x2 + a2 x
n
= log +
Sol: Here by using the formula a a
x x x
=sinx 2sin cos and
= 1 + cos x 2cos2 x2 +
2 2 2 11 2 2 2 2 a2a2 x x
x2 +
=∴I I x xx x+
∴= + nn
a a + +log + + + + cc
we can solve the given problem.
22 aa a a
x + sinx x + 2sinx/ 2cosx/ 2
∫ 1 + cos xdx = ∫ dx
1 − sinx
2cos2 x / 2 −1
Example 3: Evaluate : ∫ tan 1 + sinx
dx
x 2 x
= ∫ 2 sec x / 2 + tan dx = x tanx / 2 + c
2 Sol: Here first write cos ( ( π / 2) − x ) at the place of sin x
x
2sin2
then by using the formula 1 − cos x =
Example 2: Evaluate : ∫ x2 + a2 dx 2
x
2cos2
And 1 + cos x = we can solve it.
Sol: By applying integration by parts and taking 2
x2 + a2 dx = 2sin2 ( ( π / 4) − (x / 2) )
∫ = ∫ tan
−1
dx
2cos2 ( ( π / 4) − (x / 2) )
(x2 + a2 ) − a2 x2
= x2 + a2 x − ∫ dx x2 + a2 x − ∫ dx
−1 π x π x π x2
x2 + a2 x2 + a2 = ∫ tan tan 4 − 2 dx = ∫ 4 2
− dx = x − +C
4 4
dx
x x2 + a2 + a2 ∫
= x2 + a2 Example 4: Evaluate : ∫ log(2 + x2 )dx
2
Put x = atan θ Sol: Here integrating by parts by taking log(2 + x2 ) as
the first function we can solve the given problem.
2 2 . 2 8 | Indefinite Integration
dt 1 1 1
Sol: Simply put e2x + e−2x = t ⇒ (e2x − e−2x )dx = I= ∫ 1 + 2t2 dt = ∫ dt
( )
2
and then solving we will get the result. 2 2 t2 + 1 / 2
2x −2x
e −e
I= ∫ e2x + e−2x dx =
1 1
−1
tan
t
+ c
21 / 2 1 / 2
dt
t ⇒ (e2x − e−2x )dx =
Put e2x + e−2x =
2 x 2
1 1
1 dt 1 1 = tan−1 ( 2 tan θ) + c = tan−1 +c
2x −2x 2 2
∴I = =∫
2 t 2
log t + C = log e + e
2
+ Cc 1−x
2
x3 − 1 xdx
Example 6: Evaluate : ∫ x3 + xdx Example 8: Evaluate : ∫ (x− 1)(x2 + 4)
Sol: By splitting the given integration as Sol: By partial fractions, we can reduce the given
3 fraction as a sum of two fractions which will be easier
x 1
∫ x(x2 + 1)dx − ∫ x(x2 + 1)dx to integrate.
x A Bx + C
= +
We can solve the given problem. 2
(x − 1)(x + 4) x − 1 x2 + 4
x3 − 1 x3 1 x = 1 ⇒ A = 1/5
∫ x3 + xdx = ∫ x(x2 + 1)dx − ∫ x(x2 + 1)dx x = 2i ⇒ B = –1/5, C = 4/5
x2 1 x
1 4 − 2x
= ∫ x2 + 1dx − ∫ x − (x2 + 1)dx =∴ I ∫ + dx
5 ( x − 1)
5 x2 + 4 ( )
1 1 x
= ∫ 1 − dx − ∫ dx + ∫ 2 dx 11 11 2x
2x 88
2
x +1 x x +1 =
= n(x
logn(x−−1)1)−− 2 −−
55 10
10 xx2++44 xx2 2++44
= x − tan−1 x − logx + log x2 + 1 + c
=
1
n(x(x− −
log 1)1)− −
(
x2 + 24 + 4)2 2 x x
ln log(x
+ −tan
+ tan
)
1 −1
5 10 10 5 5 2 2
M a them a ti cs | 22.29
sinx x2 −1 1 x2 x2 1 (1 − x2 ) − 1
Example 9: Evaluate: ∫ sin 4xdx =
2
sin x − ∫ . dx = sin−1 x + ∫ dx
1 − x2 2 2 2 1 − x2
Sol: By using the formula sin2x = 2sinx.cos x , we can
reduce the given fraction and then by putting sin x = t x2 1 1 1
= sin−1 x + ∫ 1 − x2 dx − ∫ dx
we can solve it. 2 2 2 1 − x2
sinx sinx dx dx
∫ sin 4xdx = ∫ 2cos2x sin2x = ∫ 4 cos x cos2x x2 x 1 − x2 1 −1
= sin−1 x + − sin x + Cc
2 4 4
cos xdx
= ∫ 4(1 − sin2 x)(1 − 2sin2 x) 2x2 − 1 −1 x
= sin x + 1 − x2 + C
c
4 4
Put sin x = t
1 dt ex (2 − x2 )dx
⇒I= ∫
4 (1 − t )(1 − 2t2 )
2 Example 2: Evaluate : ∫
(1 − x) 1 − x2
1 1 2
= ∫ 2 − dt Sol: We can split the given fraction as
4 (t − 1) (2t − 1)
2
1 + x 1
x
11 t −1 1 2t − 1 ∫e
2
+ dx and this integration is
= log − log +c 1 − x (1 − x) 1 − x2
42 t +1 2 2t + 1
in the form of ex (f(x) + f ' (x))
1 sinx − 1 1 2 sinx + 1
= log + log +c ex (2 − x2 )dx x (1 − x2 ) + 1
8 sinx + 1 4 2 2 sinx − 1 I= ∫ = ∫e dx
(1 − x) 1 − x2 (1 − x) 1 − x2
dx 1 + x 1
Example 10: Evaluate : ∫ (x)(x 4 − 1) = ∫e
x
+ dx
2
1 − x (1 − x) 1 − x2
Sol: Here we can write the given integration as
d 1+x 1 x
x −5 −4 But = + (1 + x)
∫ (1 − x −4 ) dx and then by putting 1 − x t
= dx 1 − x2
1−x 2
(1 − x2 )
3
2
(1 − t2 )(1 + 1 − t2 ) (1 − t2 )(2 − t2 )
I= ∫ dt = ∫ t2 (1 + t2 ) dt
t2 (1 + t2 )
=
−=
11
n n
x x+ (1 (( 2 2 2 2
))
/ x/ x) +)3+ 3− −5 5
+ (1
+ C+ C
c
− log
=
2 6
∫ 1 + t2 − 1 + t2 dt = t−
2
t
− 6 tan−1 t + C
c
2 25 5 x2 x+2 (1(( ))
/ x/2 x) 2+)3+ 3+ +5 5
+ (1
Example 6: Evaluate :
= sinx − 2cosec x − 6 tan−1 (sinx) + Cc
x
4e + 6e −x ∫ cosec
22
∫ cosecx.x.log
n
n (cosx++ cos2x
(cosx ) dx for sin x > 0
cos2x) dx
Example 4: Evaluate : ∫ 9ex − 4e−x dx
Sol: By substituting cos2 x − sin2 x in place of . cos2x .
we can reduce the given integration as the sum of two
Sol: By partial fractions we can reduce the given fraction
integrations and then by integrating them separately
as a sum of two fractions and then by integrating them
we will obtain the result.
we will get the result.
x −x ∫∫cosec
2
nn((cosx
cosec2x.x.log cosx++ cos2x ))
cos2x dx
dx
4e + 6e
I= ∫ 9ex − 4e−x dx
2
nncos sin2xxdx
2 2
== ∫∫cosec
cosec2xxlog cosxx++ coscos2xx−−sin dx
x −x x −x
Let 4e + 6e = A 9e − 4e + B ( d
dx
x
9e − 4e )
−x
(
)
nn sinx(cot cot2xx−−1)1) dx
2 2
=
= ∫ ∫ cosecx2 xlog
cosec
sinx(cotxx++ cot
dx
By comparing the coefficients of ex and e− x , we get
== ∫∫cosec cosec22x.x.log
nsinx
nsinxdx dx++ ∫ ∫cosec 22
cosecx. x.log
n[cot
2
cot2xx−−1]dx
n[cotxx++ cot 1]dx
−19 35
A= ,B=
36 36 = I1 + I2
Put t2 + 5 =z 2 ⇒ 2t dt = 2z dz sinx
Example 7: Evaluate : ∫ sin3 x + cos3 xdx
M a them a ti cs | 22.31
t 1 dt 1 t +1 Let x + 1 + x2 = t, then
= ∫1+ t 3
dt = −
3 ∫ 1 + t 3 ∫ t2 − t + 1
+ dt
x t
1 + dx =
dt ⇒ dx = dt
1 1 (2t − 1) + 3 1 + x2 1 + x 2
= − log t + 1 + ∫ dt
3 6 t2 − t + 1
As 1 + x2 + x =t
1 1
= − log t + 1 + log t2 − t + 1 +
3 6 1 1 1 + x2 − x
= =
1 dt t 1 + x2 + x 1
2 t − (1 / 2) 2 + (3 / 4)
( ) 1 t2 + 1
⇒ 2 1 + x2 =t + =
1 1 t t
= − log t + 1 + log t2 − t + 1 +
3 6 t2 + 1 dt
Thus I = ∫ tn
1 2 t − (1 / 2) 2t t
tan−1 + C
2 3 3 /2 1 n− 2 2 1 n n− 2 1 tn+1 tn−1
2∫ 2∫
= t (t + 1)dt = (t +t )dt = + +C
2 n + 1 n − 1
1 1 − tanx + tan2 x 1 2 tanx − 1
= log + tan−1 +C
6
(1 + tanx )
2
3 3 Where t =x + 1 + x2
Example 8: If Im,n = ∫ cosm x.cosnx dx, show that Example 10: Evaluate:
(m + n)I
= m,n cosm x sinnx + mIm−1,n−1 2sin3 ( x / 2 ) dx
I= ∫ for cos x > 0
Sol: By using integration by parts and by taking cosm x ( cos(x/ 2)) cos3 x + 3cos2 x + cos x
as the first term we can prove the given equation.
Integrating by parts, Sol: Here we can reduce the given fraction by using the
x x
sinnx formula sinx = 2sin cos and then by putting
Im,n = cosm x + 2 2
n cos x = t we can solve it.
m
n∫
cosm−1 x sinx sinnx dx
… (i) ( 2sin ( x / 2) cos ( x / 2) ) ( 2sin2 ( x / 2) ) dx
I= ∫
But cos(n-1)x=cos(nx-x) (2cos (x / 2))
2
cos3 x + 3cos2 x + cos x
= cosnx cos nx + sin nx sin x (1 − cos x)sinx dx
⇒ sinx sin nx = cos(n-1)x – cosnx cosx ... (ii)
= ∫
(1 + cos x) cos3 x + 3cos2 x + cos x
From (i) and (ii):
[Put cos x = t]
1
Im,n = cosm x sinnx +
n ( t − 1) dt
m
⇒I= ∫
cosm−1 x[cos(n − 1)x − cosnx cos x]dx (1 + t) t3 + 3t2 + t
n∫
2 2 . 3 2 | Indefinite Integration
(t 2
)
− 1 dt Example 12: Evaluate : I = ∫
dx
= ∫ (1 + t)2 t t + 3 + (1 / t)
1 + x2 + x + 2
1
Then 1 − dt =
2zdz
t2 1 7 π π
Put =
x+ tan θ : − < θ < ; then
2 2 2 2
2zdz dz z −1
⇒ I= ∫ (z2 − 1)z = 2∫
2
z −1
= log
z +1
+ Cc
=dx
7
sec2 θ d θ
2
dθ 7
Sol: Here by putting sin x + cos x = t we can integrate = log sec θ + tan θ − ∫ ; a=
the given fraction using the appropriate formula. a + cos θ 2
dt t3dt dt
= −∫ = −2∫ = 2∫
t ((t2 − 1)(t2 + 1)) / 4 t4 t4 − 1 a(1 + t ) + 1 − t2
2
Put t 4 - 1 = z 2 : z > 0 2 dt
= ∫
a − 1 ( (a + 1) / (a − 1) ) + t2
1 2z dz dz
⇒ I = −2∫ = −∫
4 (z 2 + 1)z 1 + z2 a−1
2 a−1
= tan−1 t + Cc
a−1 a+1 a+1
= − tan−1 z + Cc = − tan−1 t 4 − 1 + Cc
2 a−1 θ
= − tan−1 (1 + sin2 x)2 − 1 + C
c = tan−1 tan + Cc.…(ii)
a+1 2
a2 − 1
= − tan−1 sin2 2 x + 2sin2 x + Cc From (i) and (ii), we get I.
M a them a ti cs | 22.33
JEE Main/Boards
∫ (x )( )
1/m
3m
1 2 1 + x2m + xm 2x2m + 3xm + 6 dx, x〉 0
Q.2 ∫ 1 + − +5 + a dx
x
1+x 2 2
1 − x x x2 − 1 x3 + 3x + 2
−1 sin2x π π
Q.20 ∫ (x2 + 1)2 ( x + 1 )dx
Q.3 ∫ tan 1 + cos2x dx : x ∈ − 2 , 2
dx
Q.4 ∫
1 + tanx
dx
Q.21 ∫ sinx + sec x
x + logsec x
2x − 1 x2 + 1(log(x2 + 1) − 2logx
Q.6 ∫ dx
x2 − x − 1
Q.23 ∫ x4
dx
dx sinx
Q.7 ∫ 1 − 3x − 5 − 3x
Q.24 ∫ sinx − cos x dx
2 x3 3
Q.8 ∫x e cos(ex )dx 1 2a − x
−1
Q.25 ∫ x sin
2 a
dx
sec2 (2 tan−1 x)
Q.9 ∫ 1 + x2
dx
4
Q.26 ∫ sec x cosec2 xdx
dx
Q.10 ∫ (2sinx + 3cos x)2
dθ
Q.27 ∫ :a>b>0
( a + b cos θ )
3/5 3 2
Q.11 ∫ cos x sin xdx
logx dx
Q.12 ∫
2
dx Q.28 Evaluate ∫
x x x4 − 1
−1 x
Q.13 ∫ sin a+ x
dx : a > 0
Q.29 ∫
dx
sin3 x sin ( x + α )
x 2 + sin2x
Q.14 ∫e 1 + cos2x
dx dx
Q.30 ∫
Q.15 ∫
dx (1 + x ) x − x2
x 6 (logx ) + 7logx + 2
2
cos8x − cos7x
Q.31 ∫ 1 + 2cos5x
dx
x2 + 1
Q.16 ∫ (x + 3)(x − 1)2 dx
x3 + 1
Q.32 ∫ dx
1 x ( x − 1)
3
Q.17 ∫ dx
1 − tanx
2 2 . 3 4 | Indefinite Integration
dx x x
Q.33 ∫3 11
2n sec + tan + Cc
(C) 2log
2 2
sin x.cos x
ex n 1 + sinx + Cc
(D) log
Q.34 Evaluate ∫ dx
e2x − 4
Sin2x
Q.5 ∫ dx is equal to
log x sin x + cos4 x
4
Q.35 Evaluate ∫ (1 + x)3 dx
(A) cot −1 (cot2 x) + Cc (B) − cot −1 (tan2 x) + Cc
f(x)
Q.36 Evaluate ∫ x3 − 1 dx , where f(x) is polynomial of (C) tan−1 (tan2 x) + Cc (D) − tan−1 (cos2x) + Cc
the second degree in x such that f(0) = f(1) = 3f(2) = −3 dθ
Q.6 The value of integral ∫ cos3 θ sin2θ
can be
(A)
2 x5
+ Cc (B)
2 x
+ Cc
(C)
5
2
(
tan2 θ + 5 ) tan θ + c
5 1 + x5 5 1 + x5
2
( )
tan θ
2 1 (D) tan2 θ + 5 +c
(C) + Cc (D) None of these 5
5 1 + x5
dx
8 8
Q.7 ∫ a + bx2 a,b≠0 and a/b > 0
cos x − sin x
Q.2 dx equals-
1 − 2sin2 x cos2 x 1 b b b
(A) tan−1 x + c (B) tan−1 x +c
ab a a a
sin2x sin2x
(A) − +c (B) +c a a b
2 2 (C) tan−1 x +c (D) ab tan−1 x +c
b b a
cos2x cos2x
(C) + c (D) − +c
3 1 1 − x7
Q.8 ∫x dx equals
Q.3 Identify the correct expression ( 1 + x7 )
= =xx2 2log
(A) x ∫ logxdx
nxdx x22 +
nx −−x + Cc nx
(A) logx
22 7
(( ))
nn 11++xx7 ++cc (B) logx
nx++ log
77
nx nx
22
nn
− − log
44
(( ))
1 −1 x−7 x7+ c+ c
(B) xx∫∫log
nnxx=
dx xexx ++CCc
dx xe
=
x
nx
(C) logx
22
nx−− log
77
7
((
))
nn 11++xx7 ++cc (D) logx
nx
22
nx++ log
44
7
nn11−−xx7 ++cc (( ))
(C) x ∫ e =dx xex + Cx
cx
dx 1 x
(D) ∫= tan−1 + C
c log | x |
2
a +x 2 a a Q.9 ∫x 1 + log | x |
dx equal
1
(C) 1 + log| x |(log | x | − 2) + c cot x − tanx
3 Q.15 ∫ dx equals
2(cos x + sinx)
(D) 2 1 + log| x |(3log | x | − 2) + c
(A) sec−1 (sin x + cos x ) +cC
(A) cos
cos Xx −
1 1 (x2 − 1)dx
cos2 x + cos 3x + c Q.1 The value of ∫ (2006)
2 3
x3 2x 4 − 2x2 + 1
1 1
(B) cos
cos Xx − cos2 x − cos 3x + c
2 3 2 1 2 1
(A) 2 2 − + + c (B) 2 2 + + + Cc
2 4 2
1 1 x x x x4
(C)cos
cosXx + cos2 x + cos 3x + c
2 3
1 2 1
1 1 (C) 2− + + Cc (D) None of these
(D)cos
cosXx + cos2 x − cos 3x + c 2 x 2
x4
2 3
Q.12 ∫ sin x. cos x. cos 2x. cos 4x. cos 8x. cos16x dx 4ex + 6e− x
equals Q.2 If ∫ dx = Ax + B log (9e2x − 4) + Cc, then
9ex − 4e− x
sin16x cos32x A=……….., B=……… and C=……… (1989)
(A) + c (B) − +c
1024 1024
cos32x cos32x 1 sinx
(C) + c (D) − +c Q.3 Integrate or (1978)
1096 1096 1 − cot x sinx − cos x
x
F(x) , then ∫ x3 f(x2 )dx is equal to
Q.13 If ∫ f(x) dx = Q.4 Integrate the curve (1978)
1 + x4
(A)
1
2
( )
(F(x))2 − ∫ (f(x))2 dx (B)
1 2 2
2
(
x F(x ) − ∫ (f(x2 ) d(x2 ) ) Q.5 Integrate
sinx⋅ sin2 x⋅ sin3 x + sec2 x ⋅ cos2 2x + sin4 x ⋅ cos4 x.
1 1 2 (1979)
(C) F(x) − ∫ (F(x ) dx (D) None of these
x2
2 2
Q.6 Integrate (1979)
(a + bx)2
4ex + 6e− x
Q.14 If ∫ Ax + B log (9e2x − 4) + C then
dx =
9ex − 4e− x
Q.7 Evaluate ∫( tanx + cot x)dx. (1988)
A, B and C are
(x + 1)
3 36 3 Q.8 Evaluate ∫ dx. (1996)
(A)=
A =,B =,C log 3 + cons tant x(1 + xex )2
2 35 2
3 35 −3 Q.9 Integrate the following (1997)
(B)=
A =,B =,C log 3 + cons tant
2 36 2 1/2
1− x dx
3 35 3 ∫
(C) A =
− , B =, C = 1log 3 + cons tant 1+ x x
2 36 2
(D) None of these
2 2 . 3 6 | Indefinite Integration
sinxdx x+
1
Q.10 The value of 2 is (2008) e x
π 1
sin x − Q.14 The integral 1 + x − dx is equal to (2014)
4 x
1 1
x+
(A) ( x + 1 )
π π x+
(A) x + log cos x − + c (B) x − log sin x − + c x + c (B) −xe x +c
4 4 1
1 x+
(C) ( x − 1 )
x+
x + c (D) xe x +c
π π
(C) x + log sin x − + c (D) x − log sicos x − + c
4 4
dx
Q.15 The integral ∫ equals (2015)
(x )
3/ 4
2 4
dy x +1
Q.11 If = y + 3 > 0 and y ( 0 ) = 2 , then y(log2) is
equal to dx
1/ 4
(2011) x4 + 1
( )
1/ 4
(A) + c (B) x 4 + 1 +c
(A) 5 (B) 13 (C) -2 (D) 7 x4
1/ 4
x4 + 1
( )
1/ 4
Q.12 If the integral (C) − x + 1 4
+ c (D) − 4 +c
x
55tanx
tanx
∫∫tanx
tanx−−22
dx xx++aalog
dx =
= nnsinx
sinx−−2cos
2cosxx ++kk
2x12 + 5x9
then a is equal to (2012) Q.16 The integral ∫ dx is equal to (2016)
(x )
3
5 3
(A) -1 (B) -2 (C) 1 (D) 2 + x +1
x10 x5
( )
Q.13 If ∫ f ( x ) dx = ψ ( x ) , then ∫ x5 f x3 dx is equal to (A)
( 5
2 x + x +1 3
)
2
+ c (B)
( 5
2 x + x +1 3
)
2
+c
(2013)
1 3
3
3 2
( ) 3
(A) x ψ x − ∫ x ψ x dx + c
( ) (C)
−x10
+ c (D)
−x5
+c
( ) ( )
2 2
2 x5 + x3 + 1 2 x5 + x3 + 1
(B)
1 3
3
( )
x ψ x3 − ∫ x2 ψ x3 dx + c( )
(C)
1 3
3
( )
x ψ x3 − ∫ x2 ψ x3 dx + c ( )
(D)
1 3
3
( ) ( )
x ψ x3 − ∫ x2 ψ x3 dx + c
JEE Advanced/Boards
Exercise 1
x x e x
dx Q.4 + logxdx
∫ nx dx
Q.1 (i) ∫ ; e x
cot(x/ 2) ⋅ cot(9 x/ 3) ⋅ cot(x/ 6)
cos(x − a)
tan(log x) tan(log(x / 2))tan(log2) Q.5 ∫ dx
(ii) ∫ dx sin(x + a)
x
dx
Q.2 ∫ x5 + 3x 4 − x3 + 8x2 − x + 8
(x − α ) (x − α )(x − β) Q.6 ∫ dx
x 2 +1
(
log log ( (1 + x) / (1 − x) ) ) dx Q.7 ∫
( x + 1)dx
Q.3 ∫ 1 − x2 x( 3 x + 1)
M a them a ti cs | 22.37
−1 x cos2 x
Q.8 ∫ sin dx Q.23 ∫ dx
a+ x 1 + tanx
−1
xInx Q.24 ∫ log x.sin x dx
Q.9 ∫ dx
(x2 − 1)3/2
(x2 + 1)ex
Q.25 Evaluate ∫ (x + 1)2
dx
log6 6 ((sinx)6cos x ) cos x
Q.10 ∫ dx
sinx
esin z 3
Q.26 Evaluate ∫ cos2 x (x cos x − sin x)dx
x2 + 1 log(x2 + 1) − 2logx
Q.11
∫ x 4 dx dx
Q.27 Evaluate ∫
1 − 2x − x2
sinx
Q.12 If f(x) = the antiderivative of
sin2 x + 4 cos2 x dx
Q.28 ∫ dx
sec x + cosecx
1 1
g ( x ) + c is then g (x) is equal to
−1
tan
3 3 2x2 + 3x + 4dx
Q.29 Evaluate ∫
cot x dx dx
Q.13 ∫ Q.30 Evaluate ∫ x(xn + 1)
(1 − sinx)(sec x + 1)
cos x − sinx
3x2 + 1 Q.31 ∫ dx
Q.14 ∫ dx 7 − 9 sin2x
(x2 − 1)3
cot x − tanx
(ax2 − b)dx Q.32 ∫ dx
Q.15 ∫ 1 + 3sin2x
x c2 x2 − (ax2 + b)2
4x5 − 7x 4 + 8x3 − 2x2 + 4x − 7
Q.33 ∫ dx
m tan−1x x2 (x2 + 1)2
e
Q.16 Evaluate ∫ (1 + x2 )3/2 dx Q.34 ∫
dx
cos3 x − sin3 x
−1 x x2
Q.17 Evaluate ∫ sin a+ x
dx Q.35 ∫
(x cos x − sinx)(x sinx + cos x)
dx
(x2 + 1) + x 4 + 1
ecos x (x sin3 x + cos x) (A) ∫
x4 − 1
dx (p) log
In ++Cc
Q.20 ∫ dx x
sin2 x x2 x 4 + x2 + 1
5x 4 + 4x5 2
(B) ∫ x − 1 dx 1 1 x 4 x+41+−1 −2x2 x
Q.21 ∫ dx (q) C C− − In
log 2
2 2 In
(x −21)
(x5 + x + 1)2 x 1 + x4 (x − 1)
a2 sin2 x + b2 cos2 x 1 + x2
Q.22 ∫ dx (C) ∫ dx (r) C − tan−1
1+
1
−1
4 2 4 2
a sin x + b cos x (1 − x2 ) 1 + x 4 4
x
2 2 . 3 8 | Indefinite Integration
α x cos(x/ 2)
x2 − 4 2n cos − cos + c
(A) 2log (B) 2cos−1 +c
Q.2 ∫ dx equals 2 2 cos(α / 2)
x 4 + 24x2 + 16
αα xx cos(x/ 2)
x2 + 4 x2 + 4 (C) 22 22log
nncos cos ++cc (D) −2sin1
cos −−cos +c
1 1 22 22 cos(α / 2)
(A) tan−1 + Cc (B) − cot −1 + Cc
4 4x 4 x
3x 4 − 1
1 4x + 4 1 2 x +4 2 Q.8 Primitive of w.r.t. x is :
(C) − cot −1 + Cc (D) cot −1 + Cc (x 4 + x + 1)2
4 x 4 x
x x
(A) + c (B) − +c
4 4
x + x +1 x + x +1
( x − 1)
2
Q.3 ∫ dx equals
x + 2x2 + 1
4 x +1 x +1
(C) + c (D) − +c
4 4
3 5 3 x + x +1 x + x +1
x x x +x +x+3
(A) +x+ + Cc (B) + Cc
3 2
x +1 3 x2 + 1 ( ) 3x 3x
Q.9 If ∫ e cos 4x dx= e (A sin 4x + B cos 4x) + c
x5 + 4x3 + 3x + 3 Then
(C) +C
c (D) None of these
(
3 x +1 2
) (A) 4A=3b (B) 2A=3B
(C) 3A=4B (D) 4B+3A=1
x4 − 4
Q.4 ∫ dx equals
x2 4 + x2 + x 4 pxp + 2q−1 − qxq−1
Q.10 The evaluation of ∫ dx is
x2p + 2q + 2xp + q + 1
4 + x2 + x 4
(A) c
+C (B) 4 + x2 + x 4 +Cc xp xq
x (A) − + Cc (B) + Cc
xp + q + 1 xp + q + 1
4 + x2 + x 4 4 + x2 + x 4
(C) + Cc (D) + Cc xq xp
2 2x (C) − + Cc (D) +C
c
xp + q + 1 xp + q + 1
M a them a ti cs | 22.39
Q.13 If ∫ eu . Sin 2x dx can be found in terms of known Q.5 Evaluate the following integrals (1980)
functions of x then u can be :
(A) x (B) sin x (C) cos x (D) cos 2x
1 x2
(i) ∫ 1 + sin x dx (ii) ∫ dx
2 1−x
1
1 1− Q.12 Find the indefinite integral
(A) (1 + nxn ) n + c
n(n − 1)
1 log(1 + 6 x )
1 ∫ 3 x + 4 x
+
3
x + x
dx (1992)
1 1−
(B) (1 + nxn ) n + c
n−1
1
1 1+
(C) (1 + nxn ) n + c
n(n + 1)
1
1 1−
(D) (1 + nxn ) n + c
n+1
2 2 . 4 0 | Indefinite Integration
x3 + 3x + 2 sec2 x
Q.13 Integrate ∫ (x2 + 1)2 (x + 1) dx (1999) Q.17 The integral ∫ 9
dx equals
( sec x + tanx ) 2
−1
2x + 2 (for some arbitrary constant K) (2012)
Q.14 Evaluate ∫ sin
2
dx.
(2000)
4x + 8x + 13
1 1 1 2
(A) − − ( sec x + tanx ) + K
Q.15 For any natural number m, evaluate (2002) ( sec x + tanx )
11
2 11 7
1 ex − e− x − 1 1 e4x + e2x + 1
(C) log +c (D) log +c
2 ex + e− x + 1
2 e4x − ex + 1
Exercise 1 Exercise 1
Q.14 Q.19 Q.23 Q.25 Q.3 Q.11 Q.18 Q.20
Exercise 2 Exercise 2
Q.4 Q.10 Q.13 Q.15 Q.1 Q.5 Q.7 Q.10
Q.12
Previous Years’ Questions
Q.5 Q.7 Q.9 Previous Years’ Questions
Q.1 Q.2 Q.4 Q.10
Q.12
M a them a ti cs | 22.41
Answer Key
1 1
JEE Main/Boards Q.17
2
x − log cos x − sinx + A
2
Exercise 1 x2 1
Q.18 ƒ(x)= + −1
Q.1 tan x – sec x + A 2 x
ax 1
Q.2 x + tan−1 x − 2sin−1 x + 5 sec−1 x + +A Q.19 (2x3m + 3x2m + 6xm )m+1/m
loga 6(m + 1)
x2 1 1 3 x
Q.3 +A Q.20 log(x2 + 1) − log(x + 1) + tan−1 x +
2 4 2 2 2
x +1
1 3
1 1
Q.8 sin(ex ) + A Q.24 x + log sinx − cos x + Cc
3 2 2
Q.9
x
+ Cc x2 −1 1 2a − x a2 1 x x x2
2 Q.25 sin + sin − 1− +c
1+ x 2 2 a 2 2a 2a 4a2
1
Q.10 − +A
2(2 tanx + 3) 1
Q.26 tan3 x + 2 tanx − cot x + Cc
3
5 5
Q.11 − cos8/5 x + cos18/5 x + A
8 18 2 a−b
=Q.27 tan−1 tan(θ / 2) + c1
a2 − b2 a + b
− logx 1
Q.12 − +A
x x 1 1 1
=
Q.28 ∫ 2 dθ= θ + c= sec−1 (x2 ) + c
2 2
x x
Q.13 tan−1 (a + x) − a + Cc −2
a a = cos α + cot x sin α + c
Q.29
sin α
x
Q.14 e tanx + A
x 1
Q.30= 2 − +c
1− x 1 − x
Q.15 log 2logx + 1 − log 3logx + 2 + A
1 1 1
2∫
Q.31= (cos 3x − cos 2x)dx = sin 3x − sin 2x + c
5 3 1 3 2
Q.16 log x + 3 + log x − 1 − +A
8 8 2(x − 1) 1 1
Q.32=− log | x | +2 log | x − 1 | + − +c
x − 1 (x − 1)2
2 2 . 4 2 | Indefinite Integration
3 3 1
Q.33 2logt − log(2t − 1) − c, where t = x + x2 − x + 1
+C
2 2 (2t − 1)
1 log x 1 x 1
Q.35 I =
− + log + +c
2 (1 + x)2 2 1 + x 2(1 + x)
x2 + x + 1 2 2x + 1
Q.36 log + tan−1 +c
x −1 3 3
Exercise 2
Single Correct Choice Type
1 a2 tanx − cot x
Q.6 a + bx − 2alog(a+ bx) − + c Q.7 2 tan−1 +c
b3 a + bx
2
dt xex x
xe 11
Q.8 ∫ 2 = ln
logn ++ ++cc Q.9 2[cos−1 x − log| 1 + 1 − x | − 1 log| x |] + c
xx xx
t (t − 1) 11++xexe 11++xe xe 2
Q.16 A
JEE Advanced/Boards
Exercise 1
x x x
n sec xx x
3ln sec x x
6ln sec x sec(logx) − x tan log | 2 | + c
22n2nsec
Q.1 (i) 2log sec
− −
−
3ln
3log
3ln sec
sec
2 3 − −
−
6ln
6log
6ln sec
sec
3 6 + +
C
6
+
C C
+ c (ii) log
2 2 3 6 x
sec log
2
M a them a ti cs | 22.43
x x
−2 x −β 11++xx 111+++xxx 111+++xxx 1 + x x e
Q.2 ⋅ ++Cc nn
Q.3 log nnn
log ––log
nnn
–+nc Q.4 − ++Cc
α −β x −α 11−−xx 111−−−xxx 111−−−xxx 1 − x e x
cos x x
cos
Q.5 cos
cosa.arc
a.arccos
cos − sina
− sina .nnsinx
. log sinx 2 2
+ + sinsinx −x sin
2 2
− sin + Cc
a a+ C
cosa
cosa
x 4x 4 3 3 2 2 11 2 2 −1
tt44 tt22 11 2 −1
Q.6 + x+ x− x− x+ 5x n(x
n(x+ 1)
+ + log
+ 5x + 1) + 3tan−x1 +x +
+ 3tan C
+Cc Q.7 66 −− ++tt++ log n(1++tt2))−−tan
n(1 tan−1tt ++CC where
where tt =
=xx1/6
1/6
44 22
44 22 22
+ c where t =x1/6
x logx
Q.8 (a + x)tan−1 − ax + c Q.9 sec−1 x − +c
a x2 − 1
3/2
1 log2 (sinx) x 1 1 1 2
Q.10 + log tan + cos x + c Q.11 – 1 + 2 log 1 + 2 – + c
6 log36 2 3 x x 3
dt 1 sec x 1 1 x x 1 1 2 x2 x x x
∫ f(x)dx
Q.12 = ∫= tan−1 +c Q.13 ln tan
ln tan+ +tantan
log + tan
+ tan+ C+ Cc
t2 + 3 3 3 2 2 2 24 4 2 2 2 2
x ax2 + b
Q.14
C− + C Q.15 sin−1 +k
(x2 − 1)2 cx
−1 x −1
em tan 1 x em tan x (m + x)
Q.16 1
= m ⋅ + = +c +c
m2 + 1 x2 + 1
x2 + 1 (m2 + 1) x2 + 1
−1 x a + x x
=Q.17 a tan
a a
− +c
a (
Q.182 2 cos (e− x
)(
) sin(e x ) + cos (e x
) c
) +C
x +1 1
Q.19 −log
− ln + 1 − + Cc Q.20C − ecos x (x + cosec x) + c
x
e x + 1 + 1
x
e
x +1 x5 1 a2 tanx
Q.21 C − or C
c+ Q.22 x + tan−1 + Cc
5
x + x +1 x
x + x +1 a + b
2 2 b2
1 1 x x1 1 1 − 1 − x2
Q.23 log
ln(cos
ln(cos
x +xsinx)
+ sinx)
+ ++ +(sin2x
(sin2x
+ cos2x)
+ cos2x) c
+ C+ C Q.24 (xlogx − x)Sin−1 x + 1 − x2 logx − log + Cc
4 4 2 28 8 x
x −1 1
= ex f(x)=
Q.25 + c ex +c = ez f(z) +=
Q.26 c ez sin−1 z − c esin x (x − sec x) + c
+=
x +1
1 − z2
−1 1 + x 1 1 11
x x π π
=Q.27 sin + c Q.28 sinx
sinx− −
cos
cos
x−x − log
nntan
tan
+ + + +C C c
2 2 2 22 2 2 8 8
2x2 + 3x + 4
4x + 3 23 4x + 3
Q.29 I
= 2x2 + 3x + 4 + log + +c
8 16 2 4 2
2 2 . 4 4 | Indefinite Integration
1 z −1 1 xn 11 (4(4++3sinx
3sinx++3cos
3cosx)x)
=Q.30 log = +c log + c Q.31 n
n
log ++CCc
n z n n
x +1 24
24 44−−3sinx
3sinx−−3cos
3cosx)x)
−1
2 sin2x 7 6x
Q.32 tan + Cc Q.33 4logx
4nx + + 6 tan−1 (x) + + Cc
sinx + cos x x 1 + x2
22 −1−1 11 22++sinx
sinx++cos
cosxx x sinx + cos x
Q.34 tan (sinx
tan (sinx++cos
cosx)x)++ nn
log ++CC
c Q.35 log
n +c
33 33 22 22−−sinx
sinx−−cos
cosxx x cos x − sinx
1 11 cos
cos θ+ + sin
θ sin θ θ 1 11 2
Q.36 (sin2
(sin2 ))nn cos θ + sin θ− −−n(sec
θ)θlog
n n(sec
2θ2)2θ+θ) )C
++CcC = +c Q.37 A → s; B → p; C → q; D → r
2 22
(sin2 θ
cos
cosθ θ
− − sin
sin θ θ 2 2 logn(sec
cos2θ
cos θ − sin θ 2
Exercise 2
Q.1 D Q.2 A Q.3 D Q.4 A Q.5 A Q.6 C Q.7 B
x x 2 1
Q.5 (i) 4 sin − 4 cos + c (ii) −2 1 − x − (1 − x)3/2 + (1 − x)5/2 + c
4 4 2 5
cos2x ex (x 4 + 1)1/ 4
Q.6 x sinx + cos x − +c Q.7 +c Q.8 − +c
4 (x + 1)2 x
2
Q.9 −2 1 − x + cos−1 x + x(1 − x) + c Q.10 [ x − x2 − (1 − 2x)sin−1 x ] − x + c
π
1 2 + 1 − tan2 x
Q.11 − log| cot x + cot2 x − 1 | + log +c
2 2 − 1 − tan2x
Q.12 3 x2/3 − 12 x7/12 − 4 x1/2 − 12 x5/12 + 1 x1/3 − 4x1/ 4 − 7x1/6 − 12x1/12 + (2x1/2 − 3x1/3
2 7 3 5 2
−6x + 11)ln(1 + x ) + 12ln(1 + x1/2 ) − 3[ln(1 + x1/6 )]2 + c
1/6 1/6
−6x1/6 +11)log(1 + x1/6) + 12log(1 + x1/2)−3[log(1 + 1/6)]2 + c
1 1 3 x 2x + 2 3
Q.13 − log| x + 1 | + log| x2 + 1 | + tan−1 x + + c Q.14 (x + 1)tan−1 2
− log(4x + 8x + 13) + c
2 4 2 2
x +1 3 4
1
Q.15 ⋅ (2x3m + 3x2m + 6xm )(m+1)/m + c Q.16 C Q.17 C
6(m + 1)
M a them a ti cs | 22.45
Solutions
1 1
1 + 1 – 2 1 x = – ∫ 1 − 3xdx – ∫ 5 − 3xdx
Sol 2: ∫ 1 + x2 2
+ 5
2
+ a
dx
4 4
1 − x | x | x − 1
1 2
( ) 1 2
( )
3/2 3/2
dx 2dx dx = × 1 – 3x + × 5 − 3x
+ ∫ ax dx 12 3 12 3
= ∫ dx + ∫ 1 + x2 – ∫ 2
+ 5∫
2
1−x | x | x −1
1
( 1 − 3x ) + (5 − 3x ) + c
3/2 3/2
=
ax 18
= x + tan–1x – 2sin–1x + 5sec–1x + +c
loga
2 x3 3
Sol 8: ∫x e cos(ex )dx
sin2x
Sol 3: = tanx 3
ex = t ⇒
3
3x2ex dx = dt
1 + cos2x
1 1 1 3
sin2x
−1 \ ∫ cos tdt = sint + c = sinex + c
∫ tan 1 + cos2x dx =
−1
∫ tan tanxdx 3 3 3
x2 sec2 (2 tan−1 x)
= ∫ xdx = 2
+c Sol 9: ∫ dx
1 + x2
1 + tanx 2
2tan–1x = t ⇒ dx = dt
Sol 4: ∫ x + logsec x dx 1 + x2
Put x + logsecx = t 1 1 1
∫ sec2 tdt = tant + c = tan(2 tan−1 x) + c
2 2 2
1
⇒ 1 + sec x tanx dx =
dt
sec x 1 2 tan(tan−1 x) x
= +c = +c
2 1 + tan (tan x)
2 −1
1 + x2
dt
∴∫ = logt + c = log(x + logsecx) + c
t
dx sec2 x
Sol 5: Put 3cosx + 2sinx = t
Sol 10: ∫ (2sinx + 3cos x)2 = ∫ (2 tanx + 3)dx
dt Put 2tanx + 3 = t ⇒ 2sec2x dx = dt
–3sinx + 2cosx =
dx
1 dt 1 1 1
2 ∫ t2
dt ∴ = – +c = – × +c
∴I = ∫ t = logt = log(3cosx + 2sinx) 2t 2 (2 tanx + 3)
3/5
2x − 1 Sol 11: ∫ cos x sin3 xdx
Sol 6: ∫ dx
x2 − x − 1 3/5
x(1 – cos2 x)sinxdx
= ∫ cos
Put x2 – x – 1= t
2 2 . 4 6 | Indefinite Integration
∫ ( cos )
3/5
= x – cos13/5 x sinxdx x x
= tan−1 (a + x) − a +c
a a
Put cosx = t; –sinxdx = dt
∫ (t )
13/5 5 18/5 5 8/5
= – t3/5 dt = t – t +c 2 + sin2x
x
18 8 Sol 14: ∫e dx
1 + cos2x
5 5
= (cos x)18/5 – (cos x)8/5 + c 1 + cos2x = 2cos2x
18 8
2 + sin2x = 2 + 2sinx cosx
logx
Sol 12: ∫ x2
dx 2 + 2sinx cos x
⇒ ∫ ex x
(sec2 x + tanx)dx
2cos2 x
=
∫e
1 dlogx 1
logx ∫ dx – ∫ ∫ dx dx
x2
dx x 2
tanx = t ⇒ sec2xdx = dt
x
1 1 1
= (logx) – – ∫ × – dx + c
∴This is of form ∫e (f(x) + f '(x)dx = exf(x)
x x x ∴I = extanx + c
1 1 1
= – logx + ∫ dx + c = – (logx + 1) + c dx
x 2 x
x Sol 15: ∫ x[6(logx)2 + 7logx + 2]
−1 x
Sol 13: ∫ sin a+ x
dx , a > 0 1
logx = t; dx = dt
x
x = atan2θ ⇒ dx = 2atanqsec2qdq dt dt
∫ (6t2 + 7t + 2) = ∫ (6t2 + 3t + 4t + 2)
−1 atan2 θ
∫ sin asec2 θ
2a tanθ sec2θ dq dt 3 2
= ∫ (3t + 2)(2t + 1) =– ∫ (3t + 2) – (2t + 1) dt
= 2a∫ (sin−1 sin θ)tan θ sec2 θ dθ
1 1
tanθ = t ⇒ sec2qdθ = dt = –3 ∫ dt + 2∫ dt
3t + 2 2t + 1
= 2a∫ t tan−1 tdt
1 22
= – 3 log(3t + 2) + log
n(2t+ +
n(2t 1)1)+ +c c
1 1 3 22
= 2a tan−1 t ∫ tdt – ∫ × t2dt
2 1+t 2
2 2t + 1 2 2logx + 1
= + cc = log
n(2t + 1) n(2t + 1) + c + c
2 3t + 2 2 3logx + 2
t2 1
= 2a (tan–1 t) – t − tan−1 t + c
2 2 = log | 2log x + 1| – log | 3 log x + 2 | + c
1 1
= at2tan–1t + tan−1 t – t 2a + c
2 2 x2 + 1
Sol 16: ∫ (x + 3)(x – 1)2 dx
x
\t = tanθ = 5 1 3 1 1
a ∫ 8 × (x + 3) dx + ∫ 8(x − 1) dx + 2 ∫ (x – 1)2 dx
I=
x x 1 x 1 x 5 55 3 33 1 11
∴I = a tan−1 + tan−1 – 2a + c = n(x
log
n(x +3)
+ +3)
n(x n(x
+ ++ n(x
3) log – –1)
n(x –1)
– ––
1) + +c+cc
a a 2 a 2 a 8
8 8 8
8 8 2(x – –1)
2(x2(x –1)
1)
1
−1 x 1 x 1 x Sol 17: ∫ (1 − tanx ) dx
= x tan + tan−1 – 2a + c
a 2 a 2 a
Put tanx = t ⇒ sec2xdx = dt
x x x
= xtan–1 a + atan
–1
a –a a +c dt dt
or dx = =
2
1 + tan x 1 + t2
M a them a ti cs | 22.47
1 2
2∫
= (cos2θ + 1)dθ + tan–1x
∴I = ∫ (1 – t)(1 + t2 ) dt
1 1 1 t +1 1 11 1 −1 −1
= tan + x+2x)2–) tan
tan (1(1
−1
– tan−x1 −x −log
n(x + 1)
n(x + c+ c
+ 1)
= ∫
2 (1 – t)
dt + ∫
2 t2 + 1
dt 2 22 2
1 11 1 11 1 11 2 2 2 x xx
= nn
log n
+ ++nlog
n
1 1+1+tan
n +tan
tanx x+x+++ +c+cc
2 22 1 1−1−tanx
tanx 2 22 2 22 1
− tanx Sol 21: I = ∫ sin x + sec x dx
1 1 sec secxx xx
nn
= log + + + +c c
2 2 1 1− −tanx
tanx 2 2 2 cos x (cos x + sin x) + (cos x − sin x)
⇒ ∫ 2 + 2 sin x cos x dx =
∫ 2 + sin 2x
dx
x x 11
= – – log
nn
| cos
| cos
x –x –
sinx
sinx| +| c+c
22 22 cos x + sin x cos x − sin x
⇒ ∫ 2 + 2 sin x
dx + ∫
2 + 2 sin 2x
dx
1
Sol 18: f’(x) = x –
x2 1
⇒ ∫ 2 + [1 − (sin x + cos x)2 ] × d(sin x − cos x)
1 x2 1
f(x) = ∫ f '(x)dx
= ∫ x2
x − dx = + +c
2 x 1
1 1
+ ∫ 2 + [(sin x + cos x)2 − 1] × d(sin x + cos x)
f(1) = + 1 + c = ⇒ c = –1
2 2
1
x2 1 ⇒ ∫( × d(sin x − cos x)
∴f(x) = + –1 3) − (sin x − cos x)2
2
2 x
1
Sol 19: ∫ (x3m + x2m + xm )(2x2m + 3xm + 6)1/mdx + ∫ (1)2 + (sin x + cos x)2 × d(sin x + cos x)
Put xm = t and integrate.
1 1
x3 + 3x + 2 ⇒ ∫( 2
3) − p 2
dp + ∫
1 + q2
2
dq
Sol 20: ∫ (x2 + 1)2 (x + 1) dx
x3 + 3x + 2 = x3 + x + 2x + 2 1 3 +p
⇒ log + tan−1 (q) + c
= x(x2 + 1) + 2(x + 1) 2 3 3 −p
On putting
1 dx 1
= 2∫ dx – ∫ +∫ dx cot x= sec θ & − cosec2 dx= sec θ tan θ dθ
(1 + x2 )2 (x2 + 1)(x + 1) (1 + x2 )
(cosx + sinx)dx = dt
sin2 θ 1 − cos2 θ
= −∫ dθ = − ∫ dθ
cos θ + cos3 θ cos θ(1 + cos2 θ) 1 1
⇒ x + log(sinx − cos x) + c
2 2
(1 + cos2 θ) − 2 cos2 θ cos θ ∴ 1 1
= ∫ cos θ(1 + cos θ) 2
= − ∫ sec θdθ + 2∫
2
1 + cos θ
dθ
∫ t ∫=
dt =
t
dtlogt
nt=
= nt=n(sinx
n(sinx
log – cos
– cos
x) x)
d(sin θ) −1
1 2a − x
⇒ − ∫ sec θdθ + 2∫
2
1 + cos θ
Sol 25: ∫ x sin .
2 a
dx
d(sin θ) 1 2a − x
⇒ − ∫ sec θ dθ + 2∫ dsin−1
2 − sin2 θ −1 1 2a − x 2 a
sin . ∫ xdx − ∫ xdx dx
∫
2 a dx
1 2 + sin θ
⇒ − log | sec θ + tan θ | +2 × log +c
2 2 − sin θ
x2 1 2a − x
⇒ sin−1
1 2 + 1 − cos2 θ 2 2 a
⇒ − log | sec θ + sec2 θ − 1 | + log +c
2 2 − 1 − cos2 θ
2
x 1 1 –1
1 2 + 1 − tan2 x – ∫ × × × dx
− log | cot x + cot2 x − 1 | +
⇒
2
log +c
2 2a − x 2a 2 2a − x
1−
2 − 1 − tan2 x
4a2
x2 1 2a − x
Sol 23: ∫
(
x2 + 1 log(x2 + 1) – logx2 )dx ⇒
2
sin−1
2
a
x4
2a x2 1
1 1
+
8a ∫ ×
2a – x
dx
1 + 2 log 1 + 2 4a2 – 2a + x
x x
=∫ dx
x3
1
1+
1 2
= t ⇒ – dx = dt
Sol 26: ∫ cos4x sin2 x dx
2
x x3
1 1 t3/2 1 t3/2 (sin2 x + cos2 x)2
– ∫ t logtdt – (logt)
=
2 2 3/2
–∫ × dt
t 3 / 2
∫ sin2 x cos4 x
dx
sin2 x 1 2
1 2
3/2 3/2
1 1 4 1
= – log 1 + 1 + + 1 + + c = ∫ cos4 x + sin2 x + cos2 x dx
2 3 x2 x2 9 x2
2
1 1
3/2
1 2 = ∫ tan x sec2 xdx + ∫ cosec2 xdx + 2∫ sec2 xdx
= – 1 + log 1 + 2 – + c
3 x2 x 3 tan3 x
= – cotx + 2tanx + c
3
sinx
Sol 24: ∫ sinx − cos x dx 1
Sol 27: I
= ∫ (a + b cos θ)2 dθ
1 sinx − cos x + sinx + cos x 1 1 dt
2∫
= dx = ∫ 1dx + ∫
(sinx − cos x) 2 2 t sin θ
Let P =
(a + b cos θ)
Put sinx – cosx = t
M a them a ti cs | 22.49
a 1 b2 − a2 1 1 1
∫ b ∫ (a + b cos θ) sin α ∫ cos α + cot x sin α
=P dθ + dθ =− d(cos α + cot x sin α )
b a + b cos θ
−(b2 − a2 ) 1 a 1 1 1
⇒
b ∫ (a +=
b cos θ)2
dθ
b ∫ a + b cos θ
dθ − P =−
sin α ∫ t
dt ; where t = cos α + cot x sin α
1 b a 1 −2
⇒ ∫
2 b ∫ a + b cos θ
= dθ dθ =
− P cos α + cot x sin α + c
2 2
(a + b cos θ) (a − b ) sin α
b a sin θ 1
= I1 −
b (a b cos )
+c Sol 30: I = ∫ dx
a − b2
2
+ θ (1 + x ) x − x2
dx dx 5x 5x
Now, ∫ = ∫ …(i) 1
2sin
2
cos 8x − 2 sin cos 7x
2
x x4 − 1 x (x2 )2 − 1 = ∫ dx
2 5x 5x
sin + 2 cos 5x sin
Let x2= sec θ, then 2x dx= sec θ tan θ dθ 2 2
2 2 . 5 0 | Indefinite Integration
21x 11x 19x sin 9x Sol 34: [Here e2x − 4 =− (ex )2 − 22 , which is of the
sin − sin − sin −
1 2 2 2 2
= ∫ form x2 − a2 , hence substitution= ex 2sec θ may be
2 15 x
sin tried]
2
ex ex
15x 15x Now, ∫ dx = ∫ dx …(i)
2sin cos 3x − 2sin cos 2x e2x − 4 (ex )2 − 22
1 2 2
= ∫ dx
2 15 x
sin ex 2 sec θ, then ex=
Let = dx 2sec θ tan θdθ
2
1 1 1 Now from (i),
= ∫ (cos 3x − cos 2x)dx = sin 3x − sin 2x + c
2 3 2 ex 2sec θ tan θ 2sec θ tan θ
∫ 2x dx
= ∫ = dθ ∫ 2 tan θ
dθ
e −4 4 sec2 θ − 4
x3 + 1
Sol 32: I = ∫ x(x − 1)3 dx = ∫ sec θ=
dθ log | sec θ + tan θ | +c …..(ii)
x3 + 1 A B C D ex
=+ + + ex 2 sec θ
= ∴ sec
= θ
x(x − 1)3 x x − 1 (x − 1)2
(x − 1)3 2
put x =
−1 & x =
2 ⇒ B=
2 & c=
1 ex ex e2x − 4
From (ii), ∫ dx =
log
2
+
2
+c
−1 2 1 2 22x − 4
∴=I ∫ dx + ∫ dx + ∫ dx + ∫ dx
x x −1 (x − 1)2 (x − 1)3 ex + e2x − 4
= log +c [ ex + e2x − 4 > 0]
1 1 2
=− log | x | +2 log | x − 1 | + − +c
x − 1 (x − 1)2
= log(ex + e2x − 4 ) − log 2 +=
c log(ex + e2x − 4 ) + c'
1
Sol 33: I = ∫ 2
dx
1 1
x + x − x +1
2 ∫
Sol 35:
= I (log x) ⋅ − − − dx
2
2(1 + x) 2(1 + x) x
put t = x + x2 − x + 1 [Taking u = log x]
t2 − 1 2t2 − 2t + 2 1 log x 1 dx
⇒ x
= = and dx dt =− ⋅ + ∫ ….(i)
2t − 1 (2t − 1)2 2 (1 + x)2 2 x(1 + x)2
t2 − t + 1 1 1+x−x 1 1
⇒ 2∫
I= dt Now, = = −
t(2t − 1)2 x(1 + x)2
x(1 + x)2 x(1 + x) (1 + x)2
t2 − t + 1
A B C
let =+ + 1+x−x 1 1 1 1
t(2t − 1) t 2t − 1 (2t − 1)2
2 = − =− −
x(1 + x) (1 + x)2 x 1 + x (1 + x)2
Solving by partial fraction method, we get
dx 1 1 1
A = 1, C =
3
and B = −
3 ∴ ∫ x(1 + x)2 =∫ x − 1 + x − (1 + x)2 dx
2 2
3 3 1 1 x 1
=I 2 log t − log (2t − 1) − +c = log x − log(1 + x) + = log +
2 2 (2t − 1) 1+x 1+x 1+x
or 4a + 2b = 2 3/2
x x −6
∴f(x) = x – x – 3
2 ∫ 3/2
dx = ∫ 3/2
dx
15/2 1 1
x 1 + 5 1 + 5
x2 − x − 3 x(x − 1) 3 x x
∫ dx = ∫ dx – ∫ dx
x3 – 1 (x − 1)(x2 + x + 1) (x3 − 1) 1
Put 1 + =t
x5
x 3
= ∫ (x2 + x + 1) dx – ∫ (x3 − 1) dx ⇒ –5x–6dx = dt or x–6dx = –
1
dt
5
1 dt 1 1
∴I= – ∫
5 t 3/2
– × (–2)
=
5 t
+c
1 2x + 1 dx 3
∫ dx – ∫ –∫ dx
2 (x + x + 1)
2 2
2 (x 3
− 1)
1 3 2 1 2 x5
x + +
∴I= +c= +c
2 2 5 5 1 + x5
1
1+
1 x5
xx++ 1
11 2 11 2 2 −1 22
= log n(x
n(x2++xx++1)1)–– ×× tan tan−1 (sin8 x − cos8 x)
22 22 33 33 Sol 2: (B) − ∫ dx
22 1 − 2sin2 x cos2 x
1 x+2
– ∫ − dx (sin2 x − cos2 x)(sin2 x + cos2 x)(sin4 x + cos4 x)
x − 1 (x + x + 1)
2 = –∫ dx
1 − 2sin2 x cos2 x
(sin2 x − cos2 x) (sin2 x + cos2 x)2 − 2sin2 x cos2 x)
1 11 2 2 2 1 1 1 −1 −2x 2x1+ 1
+2x +1 = –∫ dx
= n(xn(x
log + x+++x1)
n(x +x 1)
+– 1) tan1−1 – –n(x
– – tantan –n(x – 1)
–n(x
1)
log(x –−1)1)
2 22 3 33 3 33 1 − 2sin2 x cos2 x
1 11 2x 2x
+2x1+ 1
+1 3 33 1
+ +∫+ ∫ ∫ 2 2 + ++ 2 2 dx
dx dx = – ∫ (sin2 x − cos2 x)dx = – ∫ (– cos2x)dx = sin2x + c
2 22x x+xx+++x1+
2
x1+ (x 2
1 (x+(xx+++x1)
+x 1)
+ 1) 2
1 2x + 1 Sol 3: (C)
= log(x2 + x + 1) – tan− – log(x
n(x + 1)
− 1)
3 3 nx)x − x ×11 1dx
(A) xx∫ ∫xlog
∫nxdx
=
= =xx(x((logx)x
nxdx
nxdx nx)x − ∫−−∫x∫×x ×dxdx
(nx)x = x logx – x + cx
2 2
xx x
3 2x + 1
+ tan−1 x ∫x ∫ | x| x|dx
(B)=
=
2
|dx x x2log
2
| –x2++cxcx
nn| x| x| –x
3 3
2×
2 x
(C) x ∫ e= dx x ex + c = xex + cx
2
(x + x + 1) 2 2x + 1
= log
n + tan−1 +c
| x −1 | 3 3 dx
(D) ∫
a2 + x2
x = atanθ, dx = asec2qdq
sec2 θdθ
⇒I= ∫ =
asec θ ∫ sec θdθ
x x2 + a2
= log|secθ + tanq| + c = log + +c
a a
2 2 . 5 2 | Indefinite Integration
x
Sol 4: (B) ∫ 1 + 2 tanx(secx + tanx)dx dx 1 −1
Sol 7: (A) ∫ a 2 =
a
tan a
b + x
= ∫ 1 + 2 tanxsecx + 2 tan2 x dx b× b
b b
I= 2
(tan2 θ + 5) tan θ + c
Sol 11: (B) I = ∫ 2sinx(cos2x + cos x)dx
5 2
= ∫ 2sinx(2cos x − 1 + cos x)dx
M a them a ti cs | 22.53
2∫
= f(t)dt – ∫ 1 dt 1 t 2
2
4∫ t 4 1
∴I
= = · +c
1 2 2 2
= x F(x ) − ∫ f(x2 )d(x2 )
2
1 2 1
= 2− + +c
2 x 2
x4
4e2x + 6
Sol 14: (D) I = ∫ 9e2x − 4dx
4ex + 6e− x
⇒ 4e + 6 = a(9e – 4) + b× 18 × e
2x 2x 2x
Sol 2: Given, ∫ 9ex − 4e−x dx = Ax + B log (9e 2x
– 4) + c
⇒ 9a + 18b = 4 4e2x + 6
–4a = 6
LHS = ∫ 9e2x − 4 dx
3
∴a=– Let 4e2x + 6 = A (9e2x – 4) + B(18e2x)
2
27 35 ⇒ 9A + 18B = 4
18b = 4 + ⇒b=
2 36 and –4A = 6
3 35
3 35 18e2x ⇒A= – and B =
∴I = ∫ 2 36 9e2x − 4 dx
– dx + 2 36
3x 35 A(9e2x − 4) + B(18e2x )
= – – 3x+ + 35log
n(9e2x
n(9e2x− −4)4)+ +c c
∴∫ dx
2 2 3636 9e2x − 4
3 35 1
A= – and B = 9e2x − 4
= A ∫ 1dx + B ∫ dt, where t =
2 36 t
=Ax + B log (9e2x–4) + c
cos x − sinx
Sol 15: (A) I = ∫ 2sinx cos x ( cos x + sinx )
dx
3 35
= – x + log(9e2x − 4) + c
2 36
2 2 . 5 4 | Indefinite Integration
3 35 3 35
∴ A= – , B= ∴ A= – , B=
2 36 2 36
and C = any real number and C = any real number
4
sinx and I3 = ∫ sin x cos4 x dx
Sol 3: Let I = ∫ sinx − cos x dx
1 3x sin 4x sin8x
Again, let sin x = A (cos x + sin x) + B (sin x – cos x),
=
128 ∫ (3 − 4 cos 4x + cos8x)dx = –
128 128
+
1024
then A + B = 1 and A – B = 0
∴I = I1 + I2 + I3
1 1
⇒A= ,B=
2 2 cos 4x cos2x cos6x
= − – + + sin2x + tanx − 2x
1 1 16 8 24
(cos x + sinx) + (sinx − cos x) 3x sin 4x sin8x
∴I = ∫ 2 2 dx + − +
(sinx − cos x) 128 128 1024
1 cos x + sinx 1 x2
= ∫
2 sinx − cos x
dx + ∫ 1dx + c
2
Sol 6: Let I =
(a + bx)2
Put a + bx = t
1 1
= log(sinx − cos x) + x + c
2 2 ⇒ bdx = dt
2
xdx t −a
Sol 4: Let I = ∫ 1 + x4
b dt 1 t2 − 2at + a2
∴I = ∫ · = ∫ dt
t 2 b b3 t2
1 2x
= ∫
2 1 + (x2 )2
dx
1 2a a2 1 a2
Putx2 = u ⇒ 2xdx = du
= ∫
b3
1 − +
t t2
dt =
b3
t − 2a log t − +c
t
1 du 1 1 −1 2
tan−1 (u) + c = tan (x ) + c
2 ∫ 1 + u2 2
=∴ I = 1 a2
2 = a + bx − 2a log (a + bx) − + c
3 a + bx
b
Sol 5: Let I1 = ∫ sinx sin2x sin3x dx
tanx + 1
1
= ∫ (sin 4x + sin2x − sin6x)dx
Sol 7:
= Let I ∫( tanx + cot x )dx = ∫ tanx
dx
4
Put tanx = t2 ⇒ sec2x dx = 2t dt
cos 4x cos2x cos6x
= − − + 2t
16 8 24 ⇒ dx = dt
1 + t4
2 2
I2 = ∫ sec2 x ·cos2 2xdx = ∫ sec x (2cos x − 1)2 dx
t2 + 1 2t t2 + 1
=
= ∫ (4 cos2 x + sec2 x − 4)dx = ∫ (2cos2x + sec2 x − 2)dx
∴I ∫ 2 t4 + 1
= · dt 2 ∫ t 4 + 1 dt
t
1 1
= sin2x + tanx – 2x 1+ 2
1+
= 2∫ t dt = 2∫ t2 dt
1 2
( 2)
2x 2x 2 1 2
A(9e − 4) + B(18e ) t + −2+2
∴∫ dx t2 t − +
2x
9e −4 t
1 1
1 Put t – = u ⇒ 1+ dt = du
9e2x − 4
= A ∫ 1dx + B ∫ dt, where t = t t2
t
du
∴I = 2∫
= Ax + B log (9e2x – 4) + c u + ( 2)2
2
3 35 2 u
=– x+ log (9e2x − 4) + Cc ⇒I= tan−1 +c
2 36
2 2
M a them a ti cs | 22.55
tanx − cot x 1
= 2 tan−1 +c I = 2 cos−1 x − log 1 + 1 − x − log | x | + c
2 2
(x + 1) π π
sin x − + dx
Sol 8: ∫ x(1 + xex )2 dx sinxdx 4 4
Sol 10: (C) 2 = 2∫
π π
ex (x + 1) sin x − sin x −
This can be rewritten as ∫ 2ex (1 + xex )2 dx 4 4
π π π
let 1 + xex =t ⇒ ex (1 + x)dx =dt = 2 ∫ cos + cot x − sin dx
4 4 4
dt
Now integration becomes ∫ t2 (t − 1) π π
∫ dx + ∫ cot x − 4 dx =x + n sin x − 4 + c
=
1 A Bt + C
⇒ =+ (using partial fraction)
t (t − 1) t − 1
2
t2 dy dy
Sol 11: (D) = y +3⇒ = dx
dx y +3
⇒ 1= t2 (A + B) + (C − B)t − C
dy dy
Comparing, we get C = -1, B =- 1 and A = 1 = y +3⇒ = dx
dx y +3
Now our integration becomes log ( y + 3) =x + c
dt 1 t +1 1 1 2
x =0⇒ y =2
∫ t2 (t − 1)= ∫ t − 1 dt − ∫ t 2
dt
= ∫ t − 1 dt − ∫ t dt − ∫ t dt ⇒ log5= 0 + c
−2 +1
t t −1 1 c = log5
= log (t − 1) − log(t) − = + C log + + Cc
−2 + 1 t t
log ( y + 3) =x + log5
In5
x
Putting t= 1 + xe , we get y + 3 = ex − log5 ⇒ y + 3 elog2 + log5
x x y + 3 = 10 ⇒ y = 7
dtdt xexe 11
∫ ∫t2t(t2 (t− −1)=1)= lnlog
ln
x x
1 1+ +xexe
++ + +c c
x x
1 1+ +xexe
Sol 12: (D)
1
1 − x dx 2 5sinx
Sol 9: Let I = ∫ · ∫ sinx − 2cos xdx
1 + x x
2 ( cos x + 2sinx ) + ( sinx − 2cos x )
Put x = cos θ ⇒ dx = –2cos θ sin θ d θ
2
⇒ ∫ dx
sinx − 2cos x
1
1 − cos θ 2 −2cos θ ·sin θ
∴ I ∫ 1 + cos θ · cos2 θ dθ cos x + 2sinx
= ∫ sinx − 2cos
dx + ∫ dx + k
θ θ θ θ
sin 2sin · 2sin · cos = 2log sinx − 2cos x + x + k
2· −2sin θ 2 2 2
= ∫ dθ = − ∫ dθ
θ cos θ θ ∴a=2
cos cos · cos θ
2 2
θ
2sin2
2 dθ = −2 1 − cos θ dθ
Sol 13: (C) ∫ f ( x ) dx = ψ ( x )
= −2∫ ∫ cos θ
cos θ
( )
I = ∫ x5 f x3 dx
JEE Advanced/Boards
=
1 3
3
( )
1
3
( )
x ψ x3 − 3∫ x2 ψ x3 dx + c =x3 ψ x3 dx + c ( )
x + 1 1
Exercise 1
1 x+
Sol 14: (D)
= I ∫ e x + x 1 − e x dx
x2 x x x
1
Sol 1: (i) ∫ tan 2 tan 3 tan 6 dx
x+
= x.e x +c
x x
tan + tan
As ∫ ( xf ' ( x ) + f ( x ) ) dx =xf ( x ) + c x
= ∫ tan 1 −
3 6
dx
2 x
tan
2
dx
Sol 15: (D) ∫ x x x
( )
3/ 4
x2 x 4 + 1 = ∫ tan 2 − tan 3 − tan 6 dx
dx 1 x x x
∫ 3/ 4
⇒ 1+
x 4
t4
= = ∫ tan 2 dx – ∫ tan 3 dx – ∫ tan 6 dx
1
x3 1 + 4
x x x x
= 2logsec – 3logsec – 6logsec + c
1 2 3 6
−4 =dx =4t3dt
5
x x
tan(logx)tan log tan(log2)
dx 2
(ii)
x3
= t3dt ∫ x
dx
1/ 4 1
−t3dt 1 Put logx = t ⇒ dx = dt
∫ t3
=−t + c =− 1 +
x4
+c x
2 dt 2 t –3dt cos(x − a)
( α − β) ∫ t 2 1 + t 2 (α − β) ∫ t –2 + 1 ∫
∴I = = Sol 5: dx
sin(x + a)
∴1 + t–2 = u ⇒ –2t–3dt = du
cos x cosa − sinx sina
1 ∫ sinx cosa + cos x sina
dx
Or t–3dt = – du
2
1 2 –du –1 1 − tanx tana
= ×
2 ( α − β) ∫ =
( α − β)
×2 u = ∫ tanx + tana
dx = ∫ cot(a + x) dx
u
–2 1
= 1+ x5 + 3x 4 – x3 + 8x2 – x + 8
( α − β) t2 Sol 6: ∫ (x2 + 1)
dx
–2 1 –2 sec2 θ
= 1+ = (x3 + 3x2 – 2x + 5)(x2 + 1) x+3
( α − β) tan2 θ (α − β) tan2 θ = ∫ 2
(x + 1)
dx + ∫
x2 + 1
dx
–2 (α − β)sec2 θ –2 (x − β) x4 1
= = = + x3 – x2 + 5x + log(x
n(x22++4)4)++6 tan−1‒1 x + c
3tan x+c
α − β (α − β)tan2 θ α − β (x − α ) 4 2
1 + x ( x + 1) (x)1/2 + 1
Sol 7: ∫ = ∫ x1/2 dx
n n dx
1 − x
x(3 x + 1) (x1/3
+1)
Sol 3: ∫ 1 − x2 Put x1/6 = t ⇒ dx = 6t5dt
1 + x
n
Put log =t (t3 + 1)t5 (t3 + 1)t2
1 − x 6∫ dt = 6 ∫ dt
t3 (t2 + 1) (t2 + 1)
1 − x 1 − x + 1 + x
⇒ × dx =
dt
1 + x (1 − x)2 (t3 – t + 1)(t2 + 1)
= 6∫
t −1
2
+ dt
2 (t + 1) (t2 + 1)
Or dx = dt
2
1 − x t 4 t2
dt = 6 – + t + 3log(1 + t2) – 6tan–1t + c
∴∴I I=
= lnln( t()t ) dt
∫∫log 4 2
22
Where t = x1/6
1
= ∴ I∫ 1. lnln( t( t) )dtdt
=∫log
2 2
−1 x
Integration by parts,
Sol 8: ∫ sin a+ x
dx , a > 0
1 d x = atan2q
=I log(t)∫ 1dt − ∫ (Iog(t) ) ∫ 1dt dt
2 dt dx = 2atanqsec2qdq
1 atan2 θ
= t(logt − 1) + c −1
2 ∫ sin asec2 θ
2a tanθ sec2θ dq
1 1 + x 1 + x
log log log − 1 + c 2a∫ (sin−1 sin θ)tan θ sec2 θ dθ
2 1 − x 1 − x
tanθ = t ⇒ sec2qdθ = dt
x x
x x
e e
x x 2a∫ t tan−1 tdt
nx and d = [−logx]
Sol 4: d = [logx] [– nx]
e
e
x
x 1 1
2a tan−1 t ∫ tdt – ∫ × t2dt
x x x x x x 2 1+t 2
x x e e x e
∴ ∫ ∫ logxdx
nxdx+++∫ ∫ nxdx
nxdx nxdx = – + c
logxdx
e
e x
x e x
2 2 . 5 8 | Indefinite Integration
t2 1
= 2a (tan–1 t) – t − tan−1 t + c (
x2 + 1 log(x2 + 1) – logx2 )dx
2 2
Sol 11: ∫ x 4
1 1
= at2tan–1t + tan−1 t – t 2a + c 1 1
2 2 1 + 2 log 1 + 2
x x
∴ t = tanθ =
x ∫ x3
dx
a
1 2
1+ =t⇒– dx = dt
x x 1 x 1 x x 2
x3
∴ I = a tan−1 + tan−1 – 2a + c
a a 2 a 2 a
1 1 t3/2 1 t3/2
– ∫ t logtdt – (logt)
= –∫ × dt
2 2 3/2 t 3 / 2
x 1 x 1 x
= x tan−1 + tan−1 – 2a + c
a 2 a 2 a
1 2 3/2 2 1/2
= – (logt)t + ∫ t dt
x x 2 3 3
x
= xtan–1 a + atan–1 a – a +c
a
1 2
3/2 3/2
1 1 4 1
= – log 1 + 1 + + 1 + + c
−1 x x 2 3 x2 x2 9 x2
= tan (a + x) − a +c
a a
3/2
1 1 1 2
x x = – 1 + log 1 + 2 – + c
= x x tan−1 − a + atan−1 +c 3 x2 x 3
a a
3x2 + 1 emz 1
= cos z + ∫ emz sin z dz
Sol 14: ∫ (x2 − 1)3 dx
m m
x π 5x 4 + 4x5
dsin e + e
– x
+
4
Sol 21: ∫ (x5 + x + 1)2 dx
2∫
∴I =
π
+dsin e x − e– x + 5x5 + 5x 4 + x + 1 – x5 – x – 1
4 = ∫ (x5 + x + 1)2
dx
π π
= 2 sin e x
+ e– x
+ + sin e
x
– e– x
+ (x5 + x + 1) + (5x 4 + 1)(x + 1)
4 4 = ∫– (x5 + x + 1)2
dx
= 2sin e
x
+
π
cos e
4 ( )
– x
=
x +1 x +1
∫ x5 + x + 1 x5 + x + 1
–d = – + c or
x5
+c
x5 + x + 1
= 2 sin e
( ) + cos (e ) cos (e ) + c
x x – x
a2 tan2 x + b2
Sol 22: ∫ a4 tan2 x + b4 dx
(x2 + x)
Sol 19: ∫ (ex + x + 1)2 dx 1 b2 (a2 + b2 ) + a2 (b2 + a2 )tan2 x
⇒
(a2 + b2 )
∫ a4 tan2 x + b 4
dx
x(ex + x + 1) − xex
= ∫ (ex + x + 1)2 dx
1
2 2 2x
a b (1 + tan + 1 dx
⇒ ∫
a2 + b2 a4 tan2 x + b 4
x xex
= ∫ ex + x + 1 (ex + x + 1)2 dx
–
1 a2b2 + b 4 + (a2b2 + a4 )tan2 x
⇒ ∫
a2 + b2 a4 tan2 x + b 4
dx
1 × x − 1 x
= ∫ x dx
x 2
1 + x + 1 e x +1 e 1 a2 2 1
1 + x 2 ∫ 2
⇒ sec x × + 1 dx
ex e
2
a + b b a 4
2
4 tan x + 1
x +1 –x b
Put +1 =t ⇒ dx = dt
x
e ex 2
a 2
dt 1 1 sec x
1 1 b
∫ – + dt = – logt –
t t2 t
+c ⇒ 2
a + b2
x+
a2 + b2
∫ 2
dx
a2
1 + 2 tanx
x +1 1 b
= –log + 1 – +c
x x + 1
e
x + 1 a2 a2
e Put tanx = t ⇒ sec2 xdx = dt
b2 b2
1 1 dt
ecos x (x sin3 x + cos x) ∴ x+ ∫ 1 + t2
Sol 20: ∫ sin2 x
dx 2
a +b 2 2
a +b 2
cos x cos x 1 a2
∫ e sin2 x + x sinx dx Or x + tan−1 tanx + c
2 2 2
a + b b
cos x
∫e (cot x cosecx + x sinx)dx
cos2 x cos3 x
⇒– ∫ e cos x
(1 − cos tx cosecx − x sinx − 1)dx
Sol 23: ∫ 1 + tanxdx = ∫ sinx + cos x dx
= – ∫ ecos x (1 − cosecx cot x) + (x + cosecx)ecos x (– sinx)dx
1 3cos x 1 cos3x
4 ∫ sinx + cos x
= dx + ∫ dx
4 sinx + cos x
= – ∫ decos x (x + cosecx) = –ecosx(x + cosecx) + c
M a them a ti cs | 22.61
dx dx
= − cos θ log sin θ + log | cosec θ − cot θ | + cos=
θ ∫=2
∫ ...(i)
2 − (1 + x) (2) − (1 + x)2
2
1 − 1 − x2
=− 1 − x2 log x + log + 1 − x2 Let z = 1 + x, then dz = dx
x
2 2 . 6 2 | Indefinite Integration
From (i), z 2 a2
∫ z 2 + a2 dz
= z + a2 + log z + z 2 + a2 + c
dz 1 + x 2 2
−1 z
=I ∫ = sin = + c sin−1 +c
( 2)2 − z 2 2 2 3
−x + 2
4 3 23
= x + +
dx sinx cos x 2 4 16
Sol 28: ∫ sec x + cosecx = ∫ sinx + cos x dx
2
23 3 3 23
+ log x + + x + + +c
1 1 + 2sinx cos x − 1 32 4 4 16
2 ∫ sinx + cos x
= dx
4x + 3
1 (sinx + cos x)2 − 1 =I 2x2 + 3x + 4
= ∫ dx 8
2 (sinx + cos x)
2x2 + 3x + 4
23 4x + 3
1 1 1 + log + +c
= ∫ (sinx + cos x)dx – ∫ dx 16 2 4 2
2 2 2 1 (sinx + cos x)
2
1 1 dx z xn + 1 , then dz = nxn−1dx
Sol 30: Let =
=
2
[sinx − cos x] –
2 2
∫ π dx dx
sin x + Now, I
= ∫= ∫ nxn−1 ⋅ x
4
(
x xn + 1 ) (x n
+1 )
x π
sec2 + 1 dz 1 dz
…(i)
n ∫ xn xn + 1 n ∫ (z − 1)z
1 1 2 8 dx = =
= [sinx − cos x] –
2 2 2
∫ x π
2 tan +
( )
2 6 1 A B A(z − 1) + Bz …(ii)
Let =+ =
11 11 xx ππ z(z − 1) z z − 1 z(z − 1)
= [sinx
[sinx−−cos
cosx]x]–– ntan
ntan ++ ++cc
log
22 22 22 22 88
1
∴ A(z − 1) + Bz = …(iii)
11 1 xx ππ
sinx−−cos
= [sinx cosx]x–– ntan
ntan ++ ++cc
log
Putting Z = 0 we get, ‒A = 1
22 2 22 22 88
∴ A =‒1
3 Putting Z = 1, we get B = 1
Sol 29=I ∫ 2x2 + 3x + 4 dx
= ∫ 2 x2 + x + 2 dx
2
1 dz 1 1 1
∴ From =
I ∫ = ∫ − + dz
3 n z(z − 1) n z z − 1
= 2 ∫ x2 + x + 2dx
2
1
= [ − log | z | + log | z − 1 |] + c
2 2 n
3 3 3
= 2 ∫ x2 + 2 ⋅ x ⋅ + − + 2dx
4 4 4 1 z −1 1 xn
= log = +c log +c
n z n xn + 1
2
3 23
= 2∫ x + + dx ...(i)
4 16 cos x − sinx
Sol 31: ∫ 16 − 9 (1 + sin2x ) dx
3 23 3
Let, z =
x + , then dz =
dx. Let = a
=
4 16 4
(cos x − sinx)
Then from
= (i), I 2 2
2 ∫ z + a dz ∫ 16 − 9(sinx + cos x)2 dx
Now, Let 3(sinx + cosx) = t
M a them a ti cs | 22.63
1 dt 11 11 44++t t 11 44++3cos
3cosxx++3sinx
3sinx 7 1 − tan2 θ
∫ = . . log
nn = log
nn = 4logx + +6tan x + 6 ∫
–1 dθ
3 16 − t 2 2.4
2.4 33 44−−t t 24
24 44−−3cos
3cosxx−−3sinx
3sinx x (1 + tan2 θ)
7
= 4logx + + 6tan–1x + 6 ∫ cos2θdθ
co tx − tanx x
Sol 32: ∫ 1 + 3sin2x
dx
7 1
= 4logx + + 6tan–1x + 6 sin2θ
x 2
cos x − sinx
2∫ dx
( )
sin2x (1 + 3sin2x) ∵sin2θ =
2 tan θ
1 + tan θ 2
=
2x
1 + x2
7 6x
(cos2x cos x + sin2 x sinx) + (cos2x sinx − sin2x cos x)
2∫ dx ∴I = 4logx + + 6tan–1x + +c
(
sin2x (1 + 3sin2x) )
x 1 + x2
dx
cos2x(ssinx + cos x) − sin2x(cos x − sinx)
2∫ dx
Sol 34: ∫ cos3 x − sin3 x
2sin2x(1 + 3sin2x)
dx
1 ∫ (cos x − sinx)(1 + cos x sinx)
⇒ ∫ (sinx + cos x)2 + 2sin2x
(sinx + cos x)2 (cos x − sinx)2 + 2sinx cos x
(sinx + cos x)2.cos2x − 2sin2 x(cos x − sinx)
= ∫ (cos x − sinx)(1 + cos x sinx) dx
×∫ dx
(sinx + cos x)2 2sin2x (cos x − sinx) sinx cos x
∫ (1 + cos x sinx) dx + 2∫ (cos x − sinx)(1 + cos x sinx) dx
1 2sin2x dx
⇒ ∫ 2 cos x − sinx 2 (cos x − sinx)
2sin2x sinx + cos
=
x ∫ dx + ∫ dx
1 + 3 1 + (sinx + cos x)2 3 2 − (sinx + cos x)2
(sinx + cos x)2
2 1 2 + sinx + cos x
2 sin2x = tan−1 (sinx + cos x) + log +c
= tan −1
+c 3 3 2 2 − sinx − cos x
sinx + cos x
x2
5 4
4x – 7x + 8x − 2x + 4x − 7 3 2 Sol 35: ∫ (x cos x − sinx)(x sinx + cos x)dx
Sol 33: ∫ x2 (x2 + 1)2
dx
cos
cosθθ++sin
sinθθ 1
Sol 36: ∫∫cos2
cos2θθlog
nn ddθθ Put x + =t
cos
cosθθ−−sin
sinθθ x
cos θ + sin θ ⇒(1 – x–2)dx = dt
⇒ n
⇒log ∫ cos2θdθ
cos θ − sin θ dt
∫
dn cos
cosθθ ++ sin
sinθθ
t2 − 2
dlog cos θ − sin θ
cos θ − sin θ Put t = 2 secq
− ∫∫
−
d θ
∫∫
cos2
cos2θdθddθθ
θd θ
dθ
2 sec θ tan θdθ
∫ 2 tan θ
= log|secθ + tanq|
11 cos
cosθθ++sin
sinθθ 22 sin2
sin2θθ
⇒⇒ sin2
sin2θθlog
n
n ––∫ ∫ ×× ddθθ
22 cos
cosθθ− −sin
sinθθ cos2
cos2θθ 22 x2 + 1 x4 + 1
= log
n + +c
11 cos
cosθθ++sin
sinθθ 2 sin2θ 2x 2x
⇒ sin2
sin2
θθ nn
log ––∫ ∫ tan2θd×θ dθ
22 cos
cosθθ−−sin
sinθθ cos2θ 2
x2 + 1 x4 + 1
111 cos
cos
cos
θθ+θ+sin
+sin
sin
θθθ 1 1 2 sin2θ Or log
n + +c
⇒ sin2
sin2
sin2
θθ θnnn
log ––∫– log
n(sec2
n(sec2
× θθ)d)dθθ dθ x x
222 cos
cos
cos
θθ−θ−sin
−sin
sin
θθθ 22cos2θ 2
(1 + x2 )
1 − tan θ (1 − tan θ × sec2 θ (C) ∫ dx
= − (1 + tan θ)sec2 θ 1 1
1 + tan θ (1 − tan θ) 2 x − x x2 + 2
2
x x
(1 − tan θ) 2sec2 θ 2(1 + tan2 θ) 2 (1 + x –2 )
= × = =
(1 – tan θ) (1 – tan θ)2 2
(1 − tan θ) cos2θ ∫ 2
dx
1 1
− x x − + 2
x x
Sol 37: A → s; B → p; C → q; D → r 1
x– = t ⇒ (1 + x–2)dx = dt
x −14 x
(A) ∫ dx dt
3 2 1 = –∫
x x + +1
x2 t t2 + 2
x − x –3 t= 2 tanθ
∫ 1
dx
x2 + +1 dt = 2 sec2θ dq
x2
1 2 sec2 θdθ 1
Put x2 + +1=t = –∫
x 2
2 tan θ 2 sec θ
–
2
∫ cosecdθ
2 dt 1 dt 11
2x − 3 dx = or x − dx = = nncosec
log cosecθθ−−cot
cotθθ++cc
x 2
3
x 2 22
1 dt 1 x 4 + x2 + 1 11 xx44 ++11−− xx
x2 + ⇒ −− log
nn ++cc
2∫ t
= t +c = + 1+ c = +c
x2 x 22 (x22 −−1)
(x 1)
x2 − 1
(B) ∫ dx
1 + x 4 + x2
x 1 + x4 (D) ∫ dx
2 –2
1 + x4
x −1 (1 − x )
∫ 1
dx ⇒ ∫
2
dx
x 2
x + 2 1 1 1 1 –4
x2 x + − 2 = ∫ × ×
x 1 1 x5
1
1 + 4 2 1 + 4 – 1 2 1+
x x x4
M a them a ti cs | 22.65
1 4
– tan−1 1+ –1+ c ⇒ 1 − dx = dt
x4 x2
dt 1 t
∴∫ = tan−1 + c
1 2
t (4) 2 4 4
dtan−1 1+ 4 –1
∴ x
4
dx 1 x+
−1
∴I = tan x +c
1 1 4 4
= × 1+ –1
1 x4
1+ 1 + 4 – 1
x 1 x2 + 4
Or I = tan−1 +c
4 4x
1 1 1 –4
= × (x − 1)2 x2 − 1 − 2x + 2
x 1 1
1 + 2 1 + 4 − 1 2 1 + 4
x5 Sol 3: (D) ∫ x 4 + 2x2 + 1 dx = ∫ (x2 + 1)2
dx
4 x x
(x2 − 1) –2x 2dx
= ∫ 2
dx + ∫ (x2 + 1)2 dx + ∫ (1 + x2 )2
1
Exercise 2 x2 x +
x
= tan−1−1xx–– log
xxtan n(1++xx2 2) ) –x + c
n(1
ππ ππ (1 − x) 1 2× x
= + + tan−1 x + c
1+x 2 2 1 + x2
x2 − 4 x2 − 4
Sol 2: (A) ∫ x 4 + 24x2 + 16 dx = ∫ 16
dx
=
(1 − x)
+
x
+ tan−1 x + c =
1
+ tan−1 x + c
x2 x2 + 24 + 2 1+x 2
1 + x2 1+x 2
x
4 4 x4 − 4
1 – 2 1 − 2
x
Sol 4: (A) ∫ dx
x x2 x 4 + x2 + 4
= ∫
2 16
dx = ∫ 2
dx
x + + 8 + 16
4 2 x4 − 4
x + + (4)
x2 x ∫ 4
dx
2 2
x ×x x +1+
4 x2
Put x + =t
x
2 2 . 6 6 | Indefinite Integration
4 x
2 sin
x − 3 2
x I= ∫
∫ 4
dx
2cos2
α
− 2cos2
x
x2 + 1 + 2 2
x2
x
2 sin
2
Put x2 +
4 8
+ 1 = t ⇒ 2x − dx =
dt
= ∫ 2
dx
x 2
x3 x
α cos
2 cos 1− 2
1 dt 1 4 2 cos α
x2 + 1 +
2∫ t 2
= = .2 t + c = +c
x2 2
x
x 4 + x2 + 4 – sin
1 2
= +c × 2x
x 2 cos α
2 dx
= ∫ 2
sec x + tanx − 1 x
Sol 5: (A) ∫ dx cos
tanx − sec x + 1 1− 2
cos α
sec x + tanx − (sec2 x − tan2 x)
2
= ∫ (tanx − sec x + 1)
dx
x
cos
(sec x + tanx)[1 − sec x + tanx] 3 =t ∴I= 2
= ∫ [tanx − sec x + 1]
dx Let
α ∫– dt
cos 1 − t2
2
= ∫ (sec x + tanx)dx x
cos
= log|secx + tanx| + logsecx + c 2
= 2cos–1(t) + c = 2cos–1 +c
cos α
s = 1 + 2x + 3x2 + 4x3 + …………. 2
sx = x + 2x2 + 3x3 + ………… 3x 4 – 1
s(1 – x) = 1 + x + x2 + x3 + ……….
Sol 8: (B) I = ∫ (x 4 + x + 1)2 dx
1 1
s(1 – x) = ∴s= 4x 4 + x – (x 4 + x + 1)
1−x (1 − x)2 = ∫ (x 4 + x + 1)2
dx
1 (1 − x)−1
∴∫ dx = + c = (1 – x)–1 + c
(1 − x)2 –1 × –1 x(4x2 − 1) 1
= ∫ (x 4 + x + 1)2 – (x 4 + x + 1) dx
(x2 − 3) x (x2 − 9 + 6) 1
Sol 6: (C) ∫e
x
(x + 3)2
dx = ∫e (x + 3)2
dx Let ∫ x 4 + x + 1 = I1
x −3 6
= ∫e
x
+ dx (4x3 + 1) 4x3 + 1
2 ∴I = x ∫ – ∫∫ dx – I1
x + 3 (x + 3) (x 4 + x + 1)2 (x 4 + x + 1)2
∫ e ( f(x) + f '(x)) dx
x
= = exf(x) + c
(x 4 + x + 1)−1 (x 4 + x + 1)−1
(x − 3)
=x×
–1
– ∫ –1
dx – I1
= ex +c
(x + 3) –x
= + c + I1 – I1
4
(x + x + 1)
1 − cos x
Sol 7: (B) I = ∫ cos α − cos x
dx
3x
Sol 9: (C) ∫e cos 4xdx
Let 3x =t
M a them a ti cs | 22.67
1 t 4t
1/3
∴I =
3 ∫ e cos dt
3 1/3
x x
1 t
= ∫=
4t 1 t 4t 4 4t I = ∫ 4 dx = ∫ 4
dx
3
e cos dt
3 3 ∫ e cos − sin dt
3 3 3
4
(x − 1) 16 1
x 1 − 4
x
4 t 4π
+
9 ∫e sin dt
3 = ∫
1 1
dx
4/3 5
1 x
1 t 4t 4 t 4t 4 4t 1 − 4
x
= ∫
3
e cos + ∫ e sin + cos dt
3 9 3 3 3
1 4
16 4t Let 1 – = t ∴ dx = dt
– ∫ et cos dt x 4
x5
27 3
25 1 4t 4 4t 1 1 1 t –1/3 3
I = et cos + et sin + c ∴I = ∫ t 4/3 × 4 dt =
4
×
1
+ c = – t–1/3 + c
4
9 3 3 9 3 –
3
et 4t 4t −1/3 1/3
I= 3 + 4 sin +c 3 1 3 x4
25 3 3 =– 1 – 4 +c=– +c
4 x 4 x 4 − 1
∴3A = 4B
u
Sol 13: (A) I = ∫e sin2xdx
Sol 10: (C) When u = x
p + xp + 2q−1 + qxp + 2q−1 − q(xq−1 + xp + 2q−1 ) x x
I= ∫ p+q 2
dx I= ∫ e sin2xdx= ∫ e (sin2x + 2cos2x − 2cos2x)dx
(x + 1)
x
(p + q)xp + q−1 xq − qxq−1 (xp + q ) + 1 = ∫e (sin2x + 2cos2x) − 2∫ ex cos2x
= ∫ dx
(xp + q + 1)2 x
= ∫e (sin2x + 2cos2x)
uv '− vu'
It is of the form –2∫ ex (cos2x − 2sin2x + 2sin2x)dx
2
u
∴ Where u = x p+q
+ 1 and v = –xq ∴ 5 I = ∫ ex (sin2x + 2cos2x)dx
∫ tdx = ∫ tf '(t)dt
= tf(t) – ∫ f(t)dt = tf(t) – g(t)
Previous Years’ Questions
= f–1(x) (x – g)(f–1(x))
cos3 x + cos5 x
1/3
Sol 1: (C) Let I = ∫ sin2 x + sin4 x dx
x
Sol 12: (B) dx
(cos2 x + cos4 )·cos x dx
( )
4
x4 − 1
= ∫ (sin2 x + sin4 x)
Put sinx = t ⇒ cosx dx = dt
2 2 . 6 8 | Indefinite Integration
⇒ A = 2, B = –6 1 dt
⇒ 1 − dt = ∫
du =
u2 2
t −1
y2 − 3 y + 2 2 6
∴ =1 + −
y (y + 1) y y +1 1 t −1 1 u2 − u + 1
= log
= +c log +c
2 t +1 2 u2 + u + 1
∴ Eq. (i) reduces to,
2 6 2 1 e2x − ex + 1
I= ∫ 1 + t2 − 1 + t2 dt = t−
t
− 6 tan−1 (t) + c =
2
log
e2x + ex + 1
+c
2
= sin x – – 6 tan–1(sinx) + c
sinx 2sinx − sin2x
Sol 4: Given, f(x) = ∫ dx
x3
x
Sol 2: (A) Given,f(x) = for n ≥ 2
(1 + xn )1/n On differentiating w.r.t. x, we get
1 x
∴ I=
1 1−
(1 + nxn ) n + c Sol 5: (i) Let I = ∫ 1 + sin dx
2
n(n − 1)
x x x x
ex ⇒I= ∫ cos2 + sin2 + 2sin cos dx
Sol 3: (C) Since, I = ∫ dx 4 4 4 4
e4x + e2x + 1
e3x x x
J =∫ dx ⇒I= ∫ cos 4 + sin 4 dx
1 + e2x + e4x
x x
(e3x − ex ) = 4sin – 4cos + c
∴J – I = ∫ dx 4 4
1 + e2x + e4x
M a them a ti cs | 22.69
1/ 4
x2 1 (x 4 + 1)1/ 4
(ii) Let I = ∫ 1−x
dx = − 1 +
x4
+c= −
x
+c
Put 1 – x = t2 ⇒ –dx = 2t dt
1− x
(1 − t2 )2 · ( −2t)
Sol 9: Let I = ∫ 1+ x
dx
∴I = ∫ dt
t
2t3 t5 Put x = cos2 θ ⇒ dx = – 2sin θ cos θ d θ
2 4
= −2∫ (1 − 2t + t ) dt = −2 t − + +c
3 5 1 − cos θ
=∴I ∫ 1 + cos θ
· ( −2sin θ cos θ)dθ
2 1
= −2 1 − x − (1 − x)3/2 + (1 − x)5/2 θ θ
3 5 = − ∫ 2 tan · sin θ cos θ dθ = −2∫ 2sin2 · cos θ dθ
2 2
Sol 6:=
Let I ∫ (e
log x
+ sinx)cos x dx = −2∫ (1 − cos θ) cos θ dθ = −2∫ (cos θ − cos2 θ) dθ
1 1 4
sin−1 x dx − x + c
π∫
x
dx − 2∫ ex · = ...(i)
= ∫e ·
(x + 1)2 (x + 1)3
dx
4 1 1 1
I= − (1 − 2x)sin
−1
x+ x − x2 − x + c Now, I1 = ∫ dx
3 4
π 2 2 x+ x
2 Put x = t12 ⇒ dx = 12t11dt
= x − x2 − (1 − 2x)sin−1 x − x + c
π
t11 t8 dt
∴ I1 = 12∫ dt = 12∫
cos2x t4 + t 3 t +1
Sol 11: Let I = ∫ sinx
dx
dt
= 12∫ (t7 − t6 + t5 − t 4 + t3 − t2 + t − 1)dt + 12∫
2
cos x − sin x 2 t +1
= ∫ = dx ∫ cot2 x − 1 dx
sin2 x t8 t7 t6 t5 t 4 t3 t2
= 12 − + − + − + − 1 + 12 log
In (t(t++1)1)
Put cot x = sec θ ⇒ –cosec2xdx = sec θ tan θ d θ 8 7 6 5 4 3 2
sec θ · tan θ In (1 + 6 x )
∴I
= ∫ sec2 θ − 1 · dθ And I2 = ∫ dx
2 3
−(1 + sec θ) x + x
sec θ · tan2 θ Put x = u6 ⇒ dx = 6u5 du
= −∫ dθ
2
1 + sec θ
log (1 + u) 5 log (1 + u)
2 =∴ I2 ∫ =6u du ∫ ⋅ 6u5 du
sin θ u2 + u3 u2 (1 + u)
= −∫ dθ
cos θ + cos3 θ
u3 u3 − 1 + 1
6∫ In(1 + u)du= 6 ∫
+ u)du
log(1 In(1
log(1+ u)du
+ u)du
1 − cos2 θ (u + 1) u+1
= −∫ dθ
cos θ(1 + cos2 θ)
1
= 6 ∫ u2 − u + 1 − log(1
In(1 + u)du
+ u)du
(1 + cos2 θ) − 2cos2 θ u+1
= −∫ dθ
cos θ (1 + cos2 θ) In(1 + u)+ u)
log(1
= 6 ∫ (u2 − u + 1) log
In(1(1++u)du 6 ∫6 ∫
u)du− − du du
(u +(u1)+ 1)
cos θ II I|
= − ∫ sec θ dθ + 2∫ dθ
1 + cos2 θ u3 2
= 6 − u + u log(1
In (1 ++u)u)
3 2
cos θ
= − log | sec θ + tan θ | + 2∫ dθ
2 − sin2 θ 2u3 − 3u2 + 6u 1 2
−∫ du − 6 [In(1 + u)]
[log(1 + u)]2
u+1 2
dt
= − log | sec θ + tan θ | + ∫ 2 − t2 , where sin θ = t = (2u3 – 3u2 + 6u) log (1 + u)
11u
= − log | sec θ + tan θ | +2 ·
1
log
2 + sin θ
+c − ∫ 2u2 − 5u + du − 3[In(1
[log(1+ u)]2 2
+ u)]
u+1
2 2 2 − sin θ
= (2u3 – 3u2 + 6u) log (1 + u)
− log cot x + cot2 x − 1
=
2u3
2u3 55 22
=
−− −− uu ++11u
11u−−11
11log
InIn(u
(u++1)
1) [log(1
−−3[In(1 +22u)]2
+ u)]
1 2 + 1 − tan2 x
= 33 22 3[In(1 + u)]
+ log +c
2 2 − 1 − tan2 x
3 2/3 12 7/12 12
∴=I x − x + 2x1/2 − x5/12 + 3x1/3 − 4x1/ 4
2 7 5
1 In(1 + 6 x ) – 6x1/6 – 12x1/12 + 12 log (x1/12+1)
12: Let I
Sol= ∫ 3 x + 4 x + 3
x + x
dx
+ (2x1/2 – 3x1/3 + 6x1/61111/1) log(1+ x1/6)
∴ I = I 1 + I2 ,
2 2 1/2 5 5 1/31/3 1/61/6
− − x1/2
x − − x x 11x 11x − −1111 In(1+ +x1/6
In(1
log x1/6))
1 In(1 + 6 x ) 3 3 2 2
=where I1 ∫=
3 dx , I2 ∫ dx
4 3
x+ x x+ x
M a them a ti cs | 22.71
−1 sin θ 3 2
⇒ x = (Ax + B) (x + 1) + C (x2 + 1) = ∫ sin
cos θ · sec θ
· sec θ dθ
2
Putting x = –1,we get –1 = 2C ⇒ C = –1/2
3
sin−1 (sin θ)·sec2 θ dθ
2∫
Equation coefficient of x2, we get =
0 = A + C ⇒ A = –C = 1/2
3 3
= ∫ θ ·sec2 θ dθ = [θ · tan θ – ∫ 1· tan θ dθ]
Putting x = 0,we obtain 2 2
0 = B + C ⇒ B = –C = 1/2
3
= [θ tanθ − log sec θ ] + c
3
x + 3x + 2 x +1
1 2 2
= +−
2 2
(x + 1) (x + 1) 2(x + 1) 2(x
2 + 1) (x + 1)2
2
2
3 −1 2x + 2 2x + 2 2x + 2
= tan ·
− log 1 + +c
x3 + 3x + 2 2 3 3 3 1
∴I = ∫ dx
(x2 + 1)2 (x + 1) 2x + 2 3 2x + 2
2
= (x + 1)tan−1 − log 1 + + c
1 dx 1 x +1 dx 3 4 3 1
=– ∫
2 x + 1 2 ∫ x2 + 1
+ dx + 2∫
(x2 + 1)2
2x + 2 3
1 1 1 = (x + 1)tan−1 2
− log(4x + 8x + 13) + c
⇒ I = – log |x+1| + log |x +1| + tan x + 2I1 ...(i)
2 –1
3 4
2 4 2
dx 3
where I1 = ∫ let log3 + c1 = c
(x2 + 1)2 2
(
ex e2x − 1 ) (z 2
−1 ) where z = ex
=J−I ∫ e=4x
+ e2x + 1
dx ∫ z 4 + z + 1 dz
1 1
1 −1 − 2 dx dx
1 e e+x + e−ex−−x 1− 1
x
z z2 = 1 In
x
In
= log
J −J −
I= ∫ ∫ z +z +1 1− 1− 1 2 2 e e+x +e−ex−+x 1+ 1
I=
z z
1 exex++e−ex− x−−11
∴ J − I = 1In
log
In + c.
22 exex++e−ex− x++11
sec2 x
Sol 17: (C) I = ∫ 9
dx
( sec x + tanx ) 2
1
t + 1 −9 2 −12
1 t dt
I= ∫
2
=
9 t
∫
2
t + t 2 dt
t 2
−9 +1 −13 +1
1t 2 t 2
=
2 9 13
− + 1 − 2 + 1
2
−7 −11
1 t 2 t 2
= +
2 7 11
− −
2 2
1 −7 1 −11
− t 2 − t 2
=
7 11
1 1 1 1
=
− −
7 2 11 11 2
7
t t