5.

3 Integration by parts
Exercise 5.18: Find the exact value of the definite integral.
1 R1 1
xe−2x dx (1 − x)e− 2 x dx
R 2
i) 0 ii) 0

u=x dv = e−2x dx u=1−x

1
dv = e− 2 x dx
du = dx v = − 12 e−2x 1
du = −dx v = −2e− 2 x

xe−2x dx
R
1
(1 − x)e− 2 x dx
R

= − 12 xe−2x − − 12 e−2x
R
dx 1 1
= − 2(1 − x)e− 2 x − 2e− 2 x dx
R
1 1 −2x
= − 2
xe−2x − 4
e +c 1 1
= − 2(1 − x)e− 2 x + 4e− 2 x + c
1
xe−2x dx
R 2 R1 1
0
0 (1 − x)e− 2 x dx
h i1
2
− 12 xe−2x − 14 e−2x
i1
= 1 1
h
0 = − 2(1 − x)e− 2 x + 4e− 2 x
0
= − 41 e−1 − 14 e−1 + 1
1
4 = 4e− 2 − (−2 + 4)
1 1
= 4
− 2e 4
= √
e
−2

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5.3 Integration by parts
Rπ R4
iii) 0 x cos 12 x dx iv) 2 4x ln x dx

u=x dv = cos 12 x dx u = ln x dv = 4x dx
1
du = dx v = 2 sin 1
2
x du = x
dx v = 2x2

R
x cos 12 x dx
R
4x ln x dx
= 2x2 ln x −
R
= 2x sin 12 x − 2 sin 21 x dx 2x dx
R

= 2x sin 1
x + 4 cos 1
x +c = 2x2 ln x − x2 + c
2 2

Rπ R4
0 x cos 12 x dx 2 4x ln x dx
h iπ h i4
= 2x sin 21 x + 4 cos 12 x = 2x2 ln x − x2
0 2
= 2π − 4 = (32 ln 4 − 16) − (8 ln 2 − 4)
= 64 ln 2 − 8 ln 2 − 12
= 56 ln 2 − 12

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5.3 Integration by parts
R2 ln x
R √3
v) 1 x3
dx vi) 0 x tan−1 x dx

u = ln x dv = 1
dx u = tan−1 x dv = x dx
x3
1 1 1 2
du = x
dx v= − 2x12 du = 1+x2
dx v= 2
x

ln x
x tan−1 x dx
R R
x3
dx
ln x 1 x2
R 1 2 1
tan−1 x −
R
= − 2x 2 + 2x3
dx = x dx
2 2 1+x2
ln x 1 1 2 1 1
= − 2x2 − +c tan−1
R
4x2 = 2
x x− 2
1− 1+x2
dx
1 2 1 1
= 2
x tan−1 x− 2
x + 2
tan−1 x+c
R2 ln x
1 x3
dx
h
ln x 1
i2 R √3
= − 2x2
− 4x2 1 0 x tan−1 x dx
ln 2 1
i√3
+ 14
h
= − 8
− 16 = 1 2
x tan−1 x − 12 x + 1
tan−1 x
2 2 0
3 1 √
= 16
− 8 ln 2 1 3 1
= 2
π − 2
+ 6
π
√
2 3
= 3
π − 2

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5.3 Integration by parts
Exercise 5.19: Find the exact area of the shaded region.
1
i) y = (x + 1)e− 3 x
1
u=x+1 dv = e− 3 x dx
1
du = dx v = −3e− 3 x

1
(x + 1)e− 3 x dx
R

1 1
= − 3(x + 1)e− 3 x + 3e− 3 x dx
R
−1 x
(x + 1)e 3 =0 1 1
= − 3(x + 1)e− 3 x − 9e− 3 x + c
x+1=0
x = −1 R0 1
−1 (x + 1)e− 3 x dx
−1
R0 x
Area is −1 (x + 1)e 3 dx h 1 1
i0
= − 3(x + 1)e− 3 x − 9e− 3 x
−1
1
= (−3 − 9) − (−9e ) 3

1
= 9e 3 − 12

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5.3 Integration by parts
ii) y = x3 ln x
u = ln x dv = x3 dx
1 1 4
du = x
dx v= 4
x

x3 ln x dx
R

1 4 1 3
R
= 4
x ln x − 4
x dx
1 4 1 4
= 4
x ln x − 16
x +c

x3 ln x = 0 R2
x3 ln x dx
1
x = 0, ln x = 0 h i2
1 4 1 4
x=1 = 4
x ln x − 16
x
1
1
R2 = 4 ln 2 − 1 + 16
Area is 1 x3 ln x dx
15
= 4 ln 2 − 16

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5.3 Integration by parts
√ 3
Exercise 5.20: The diagram shows the curve y = x cos x, for 0 ≤ x ≤ 2
π. The
shaded region between the curve and the x-axis is denoted by R.
Find the volume of the solid obtained when the region R is rotated completely
about the x-axis. Give your answer in terms of π.

u=x dv = cos2 x dx
1 1
= 2
cos 2x + 2
dx
1 1
du = dx v= 4
sin 2x + 2
x

x cos2 x dx
R

1
+ 12 x2 − 1
sin 2x + 12 x dx
R
√ = 4
x sin 2x 4
x cos x = 0 1
= 4
x sin 2x + 12 x2 + 1
8
cos 2x − 14 x2 + c
x = 0, cos x = 0
1 1 2 1
= 4
x sin 2x + 4
x + 8
cos 2x + c
1
cos−1 (0) = 2
π
1π
x cos2 x dx
(1 R 2
π + 2kπ π 0
2
x= 3 i1π
π + 2kπ
h
1 2
2 =π 4
x sin 2x + 14 x2 + 1
8
cos 2x
0
y 2 = x cos2 x 1 2
= π( 16 π − 1
− 18 )
8
1π 1 3
Volume is π
R 2
x cos2 x dx = 16
π − 41 π
0
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5.3 Integration by parts
Exercise 5.21: Show that the definite integral is equal to the expression given.
i) 03 x2 e−x dx = 2 − 17
R
e3

u = x2 dv = e−x dx x2 e−x dx = −x2 e−x + 2xe−x dx

R R

du = 2x dx v = −e−x

2xe−x dx:
R

u = 2x dv = e−x dx 2xe−x dx = −2xe−x + 2e−x dx

R R

du = 2 dx v= −e−x = −2xe−x − 2e−x + c

x2 e−x dx = −x2 e−x − 2xe−x − 2e−x + c

R

R 3 2 −x h i3
0 x e dx = − x2 e−x − 2xe−x − 2e−x
0
= (−9e−3 − 6e−3 − 2e−3 ) − (−2)
17
=2− e3

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5.3 Integration by parts
R2 1
ii) 0 (1 + x2 )e− 2 x dx = 18 − 42
e

1 1
dv = e− 2 x dx (1 + x2 )e− 2 x dx
R
u = 1 + x2
1 1 1
v = −2e− 2 x = − 2(1 + x2 )e− 2 x + 4xe− 2 x dx
R
du = 2x dx

1
4xe− 2 x dx:
R

1 1
dv = e− 2 x dx 4xe− 2 x dx
R
u = 4x
1 1 1
v = −2e− 2 x = − 8xe− 2 x + 8e− 2 x dx
R
du = 4 dx
1 1
= − 8xe− 2 x − 16e− 2 x + c

1 1 1 1
(1 + x2 )e− 2 x dx = −2(1 + x2 )e− 2 x − 8xe− 2 x − 16e− 2 x + c
R

i2
2 −1 1 1 1
R2 h
0 (1 + x )e
2
x
dx = − 2(1 + x2 )e− 2 x − 8xe− 2 x − 16e− 2 x
0
= (−10e−1 − 16e−1 − 16e−1 ) − (−2 − 16)
42
= 18 − e

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5.3 Integration by parts
Rπ
iii) 0 x2 sin x dx = π 2 − 4

u = x2 dv = sin x dx x2 sin x dx
R

du = 2x dx v = − cos x = − x2 cos x +
R
2x cos x dx

R
2x cos x dx:

u = 2x dv = cos x dx
R
2x cos x dx
R
du = 2 dx v = sin x = 2x sin x − 2 sin x dx
= 2x sin x + 2 cos x + c

x2 sin x dx = −x2 cos x + 2x sin x + 2 cos x + c

R

Rπ 2 h iπ
2
0 x sin x dx = − x cos x + 2x sin x + 2 cos x 0
= (π 2 − 2) − 2
= π2 − 4

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5.3 Integration by parts
1π
1
x2 cos 2x dx = (π 2
R 4
iv) 0 32
− 8)

u = x2 dv = cos 2x x2 cos 2x dx
R
1
du = 2x dx v= sin 2x = 12 x2 sin 2x − x sin 2x dx
R
2

R
x sin 2x dx:

u=x dv = sin 2x dx
R
x sin 2x dx
du = dx v= − 12 cos 2x = − 12 x cos 2x + 1
R
2
cos 2x dx
= − 12 x cos 2x + 1
4
sin 2x + c

1 2
x2 cos 2x dx = sin 2x + 12 x cos 2x − 1
R
2
x 4
sin 2x + c
1π h i1π
1 2 4
x2 cos 2x dx = sin 2x + 12 x cos 2x − 1
R 4
0 2
x 4
sin 2x
0
1 2
= 32
π − 14
1
= 32
(π 2 − 8)

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5.3 Integration by parts
Exercise 5.22: The diagram shows the curve y = (ln x)2 . The x-coordinate of the
point P is equal to e, and the normal to the curve at P meets the x-axis at Q.

R
i) Find the x-coordinate of Q. ii) Show that ln x dx = x ln x − x + c,
where c is a constant.
At x = e, y = (ln e)2 = 1. P (e, 1)
u = ln x dv = dx
dy 2 ln x
= 1
dx x du = x
dx v=x
dy 2 ln e
At P , dx
= e
= 2e R R
ln x dx = x ln x − 1 dx
Gradient of P Q is − 2e = x ln x − x + c
Equation of P Q is

y − 1 = − 2e (x − e)

Q on x-axis (i.e. y = 0)
0 − 1 = − 2e (x − e) =⇒ x = 2
e
+e
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5.3 Integration by parts
iii) Using integration by parts, or otherwise, find the exact value of the area of the
shaded region between the curve, the x-axis and the normal P Q.
(ln x)2 = 0 =⇒ ln x = 0 =⇒ x = 1

Area is 1e (ln x)2 dx + 12 ( 2e + e − e)(1) = 1e (ln x)2 dx + 1

R R
e

u = ln x dv = ln x dx
1
du = x
dx v = x ln x − x

(ln x)2 dx = x(ln x)2 − x ln x −

R R
ln x − 1 dx
2
= x(ln x) − x ln x − (x ln x − x − x) + c
= x(ln x)2 − 2x ln x + 2x + c
Re h ie
2 2
1 (ln x) dx = x(ln x) − 2x ln x + 2x 1
= (e − 2e + 2e) − 2
=e−2
1
Area is e − 2 + e

12/12