MATH 137
Integration By Parts
Evaluate the following integrals using integration by parts:
1.
∫
x sinh(6x) dx
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2.
∫
€ dx
x cos(4 x)
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3.
∫
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x sin(2x) dx
u= x
v ′ = sinh(6x) dx
↓ diff
↓ int
1
v = cosh(6x)
6
u′ = 1
€
x
1
∫ x sinh(6x) dx = uv − ∫ u′ v = 6 cosh(6x) − ∫ 6 cosh(6x) dx
x
1
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= cosh(6x) − sinh(6x)+ C
6
36
u= x
v ′ = cos(4 x) dx
↓ diff
↓ int
1
v = sin(4 x)
4
u′ = 1
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x
1
∫ x cos(4 x) dx = uv − ∫ u′ v = 4 sin(4 x) − ∫ 4 sin(4 x) dx
x
1
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= sin(4 x) + cos(4 x)+ C
4
16
u= x
↓ diff
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4.
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∫
x e−10 x dx
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v ′ = sin(2x) dx
↓ int
1
v = − cos(2x)
2
u′ = 1
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x
1
∫ x sin(2x) dx = uv − ∫ u′ v = − 2 cos(2x) + ∫ 2 cos(2x) dx
x
1
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= − cos(2x) + sin(2x)+ C
2
4
u= x
↓ diff
v ′ = e−10 x dx
↓ int
1
v = − e−10x
10
u′ = 1
€
x
1
∫ x e−10x dx = uv − ∫ u′ v = − 10 e−10x + ∫ 10 e−10x dx
x
1 −10x
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= − e−10x −
e
+C
10
100
Solutions
5. ∫
ln x
dx
x6
v ′ = x −6 dx
u = ln x
↓ diff
1
u′ =
x
↓ int
1
v = − x −5
5
ln x
ln x
1 −6
∫ x 6 dx = u v − ∫ u′ v = − 5x 5 − ∫ − 5 x dx
ln x 1 −6
+ ∫ x dx
5x 5 5
ln x
1
=− 5 −
+C
5x
25x 5
=−
u = x2
↓ diff
6. ∫ x 2e −3x dx
u′ = 2 x
v ′ = e−3x dx
↓ int
1
v = − e −3x
3
1 2 −3x
2 −3x
2 −3x
∫ x e dx = u v − ∫ u′ v = − 3 x e − ∫ − 3 x e dx
1
2
= − x 2e −3x + ∫ x e −3x dx
3
3
Use integration by parts again on the last integral:
2
x
3
2
u′ =
3
u=
2
3
v ′ = e−3x dx
1
v = − e −3x
3
2
9
2
9
2
9
2
9
∫ x e −3x dx = u v − ∫ u′ v = − x e−3x − ∫ − e −3x dx = − x e −3x + ∫ e−3x dx
=−
Thus,
2 −3x
2 −3x
xe
−
e
+C
9
27
1
2
2
2 −3x
2 −3x
−3x
−3x
∫ x e dx = − 3 x e − 9 x e − 27 e + C
7. ∫ x 2 cos(4 x) dx
u = x2
↓ diff
v ′ = cos(4 x) dx
↓ int
1
v = sin(4x )
4
u′ = 2 x
1
1
2
2
∫ x cos(4 x) dx = u v − ∫ u′ v = 4 x sin(4 x) − ∫ 2 x sin(4 x) dx
Use integration by parts again on the last integral:
1
u= x
v ′ = sin(4 x) dx
2
1
1
u′ =
v = − cos(4x )
2
4
1
1
1
1
1
∫ 2 x sin(4 x) dx = u v − ∫ u′ v = − 8 x cos(4 x) − ∫ − 8 cos(4 x) dx = − 8 x cos(4 x) + 8 ∫ cos(4x ) dx
1
1
= − x cos(4 x) +
sin(4x ) + C
8
32
Thus,
1
 1
1

2
2
∫ x cos(4 x) dx = 4 x sin(4x ) −  − 8 x cos(4 x) + 32 sin(4x ) + C
1 2
1
1
x sin(4x ) + x cos(4 x) −
sin(4x ) + C
4
8
32
=
8. ∫ ln x dx
u = ln x
v ′ = dx
↓ diff
1
u′ =
x
↓ int
v=x
∫ ln x dx = u v − ∫ u′ v = x ln x − ∫ 1 dx
= x ln x − x + C
9. ∫ x sec2 x dx
u=x
↓ diff
u′ = 1
v ′ = sec 2 x dx
↓ int
v = tan x
2
∫ x sec x dx = u v − ∫ u′ v = x tan x − ∫ tan x dx
= x tan x − ( − ln cos x ) + C
= x tan x + ln cos x + C
10. ∫
x
dx
(4 − x )6
u=x
v ′ = (4 − x )−6 dx
↓ diff
↓ int
(4 − x) −5
v=
5
u′ = 1
x
x
(4 − x) −5
dx
=
u
v
−
u
v
=
−
dx
′
∫
∫
∫
5
(4 − x )6
5(4 − x )5
=
(4 − x )−4
−
+C
5(4 − x)5 5(−4)(−1)
=
x
1
−
+C
5(4 − x)5 20(4 − x)4
x
Use the tabular method of integration by parts to evaluate the following integrals:
11. ∫ x 4 sin(3x ) dx Take derivatives down the first column, integrate down the second
column, then multiply across the rows:
u = x4
4 x3
12 x 2
24 x
24
0
4
∫ x sin(3x ) dx = −
v ′ = sin(3x)
1
− cos(3x )
3
1
− sin(3x)
9
1
cos(3x)
27
1
sin(3x)
81
1
−
cos(3x )
243
+
–
+
–
+
x 4 cos(3x) 4x 3 sin(3x) 12x 2 cos(3x ) 24 x sin(3x ) 24 cos(3x )
+
+
−
−
+C
3
9
27
81
243
12. ∫ x 3e −2 x dx
v ′ = e−2 x
1
− e −2x
2
1 −2x
e
4
1
− e −2x
8
1 −2x
e
16
u = x3
3x 2
6x
6
0
∫
x 3e −2x
+
–
+
–
x 3e−2 x 3x 2 e−2 x 6 x e −2 x 6e−2 x
dx = −
−
−
−
+C
2
4
8
16
=−
e −2x
2
 3 3x 2 3x 3 
 x +
+
+  +C
2
2 4 
