IN1.4: INTEGRATION BY PARTS
The rule for integration by parts is the rule corresponding to the product rule for differentiation.
Formula for integration by parts.
∫ udv=
uv − ∫ vdu
The aim when using integration by parts is to form a new integral, (on the right), that is simpler than the original
(on the left). The correct choice of terms for u and dv/dx is critical.
Choose u such that when it is differentiated it is removed, or simplifies the integral. (provided dv/dx can be easily
integrated for v.)
If the integral involves a natural logarithm or inverse trig function this should be chosen as u
Examples
1. Find
∫ x cos xdx
let u = x ∴ du = dx (x differentiates to 1)
and dv = cosxdx ∴ v = sinx
∫ x cos xdx = ∫ udv=
(cosx is easily integrated)
uv − ∫ vdu
= x sinx −
∫ sin xdx
= xsinx + cosx + c
2. Find
∫ xe
2x
∫ xe
2x
dx
let u = x
∴ du = dx and
1 2x
e
dv = e dx ∴ v = 2
2x
dx = ∫ udv
= uv − ∫ vdu
1 2x
1
e ) − ∫ e 2 x dx
2
2
1 2x 1 2x
=
xe − e + c
2
4
=x(
IN 1.4 – integration by Parts
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June 2012
∫x
3.Find
∫x
3
3
let u =
log e xdx
1
dx (need to let u = log function)
x
x4
3
dv = x dx ∴ v =
4
log e x ∴ du =
= uv − ∫ vdu
log e xdx = ∫ udv
=
x4 1
∫ 4 × x dx
x3
log e x − ∫ dx
4
4
x
+
log e x −
16
log e x (
x4
=
4
x4
=
4
x4
)−
4
∫ arcsin xdx
4. Find
Consider
∫ arcsin xdx as ∫ 1× arcsin xdx
let u =
arcsin x ∴ du =
dv = 1dx
∫ arcsin xdx = ∫ udv=
uv − ∫ vdu
1
∫x
= arcsin x (x)−
1 − x2
1 1
= x arcsin x − − ∫
dt
2
t
= x arcsin x +
dx …….
1
1 − x2
dx
∴v=x
use a substitution t = 1 − x
1
dt =
−2 xdx ∴ xdx =
− dt
2
2
1 − x2 + c
See Exercise 1
Repeated integration by parts
Sometimes after integrating by parts the new integral cannot be integrated except by another use of integration
by parts.
Examples
1. Find
∫ x e dx
let u = x2 ∴ du =2x dx
2 x
and dv =
∫ x e dx = ∫ udv=
2 x
e x dx ∴ v = e x
uv − ∫ vdu
∫ e 2 xdx
− 2 ∫ xe dx
= x2
ex −
= x2
ex
x
x
Integrate this new integral by parts.
u=x
∴ du = dx
x
x
dv = e dx ∴ v = e
IN 1.4 – integration by Parts
= x2
e x − 2(x e x − ∫ e x dx )
= x2
e x − 2x e x +2 e x + c
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June 2012
2. Find
∫e
x
u= e
cos xdx
∴ du = e dx
x
x
dv = cos xdx ∴ v = sin x
∫e
cos xdx = ∫ udv
= uv − ∫ vdu
x
=
e x sin x − ∫ sin xe x dx Integrate this new integral by parts .
u= e
∴ du = e dx
dv = sin xdx ∴ v = − cos x
x
=
=
x
e x sin x − ( e x (− cos x) − ∫ − cos xe x dx )
e x sin x + e x cos x − ∫ e x cos xdx )
The new integral on the right side of the equation is the same as the original integral, on the left, multiplied by
(−1). Shift this to the left side of the equation to give:
2
∫e
∫e
∴
x
x
cos xdx = e x sin x + e x cos x
cos xdx =
1 x
e ( sin x + cos x )
2
When the integral is a product of trig functions, or exponential and trig functions it is often necessary to repeat
integration by parts until the original integral appears on the right-hand side of the equation.
See Exercise 2
Exercises
Find
1. (a)
∫ x sin xdx
(d)
∫ x sin(π − 2 x)dx
(g)
∫x
(b)
∫ xe dx
x
(e)
∫ log
cos xdx
(b)
∫x e
x log e xdx
(e)
∫x e
1
2
e
∫ x cos 4 xdx
(c)
xdx
(f)
∫x
4
log e xdx
log e xdx
2.
(a)
∫x
(d)
∫
(f)
∫x e
2
5 x3
dx (hint: write as
IN 1.4 – integration by Parts
2 3x
dx
2 −2 x
∫x x e
3 2 x3
(c)
∫e
x
sin xdx
dx
dx )
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June 2012
Answers
1.(a) − x cos x + sin x + c
(b)
xe x − e x + c
(c)
1
1
x sin 4 x + cos 4 x + c
4
16
1
x
cos (π − 2 x ) + sin (π − 2 x ) + c (e) xlog e x − x + c
2
4
5
5
x
x
1
1
(f)
(g) − log e x − + c
log e x − + c
5
25
x
x
(d)
2. (a)
x 2 sin x + 2 x cos x − 2sin x + c
1 x
1
e sin x − e x cos x + c
2
2
1 2 −2 x 1 −2 x 1 −2 x
(e) − x e − xe − e + c
2
2
4
(c)
IN 1.4 – integration by Parts
1 2 3x 2 3x 2 3x
x e − xe + e + c
3
9
27
3
3
2
4
(d) x 2 log e x − x 2 + c
3
9
1 3 x3 1 x3
(g) x e − e + c
3
3
(b)
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June 2012