1. This document provides examples of integration techniques for various functions. 2. It reviews common integration formulas and shows step-by-step workings for integrals involving trigonometric, exponential, logarithmic and other functions. 3. The examples demonstrate various substitution techniques used to evaluate definite and indefinite integrals, such as letting u = functions of x and finding du/dx.

1. 1
6. sec 2t tan 2t dt = sec 2t + c
2
7. (x2 + 4)2 dx = (x4 + 8x2 + 16)dx
x5 8
= + x3 + 16x + c
5 3
Chapter 6
8. x(x2 + 4)2 dx = (x5 + 8x3 + 16x)dx
x6
= + 2x4 + 8x2 + c
Integration 3
6
3 x
9. dx = tan−1 + c
Techniques 16 + x2
2
4
1
4
10. 2
dx = tan−1 x + c
4 + 4x 2
1
11. √ dx
3 − 2x − x2
6.1 Review of Formulas 1 x+1
= dx = arcsin +c
and Techniques 4 − (x + 1)2 2
1 ax x+1
1. eax dx = e + c, for a = 0. 12. √ dx
a 3 − 2x − x2
1 −2(x + 1)
=− dx
1 2 4 − (x + 1)2
2. cos(ax)dx = sin(ax) + c, for a = 0.
a 1
= − · 2[4 − (x + 1)2 ]1/2 + C
2
1 1 1
3. √ dx = dx = − 4 − (x + 1)2 + c
a2 − x2 x 2 a
1− a 4
x 1 13. dx
Let u = , du = dx. 5 + 2x + x2
a a 1 x+1
1 =4 dx = 2 tan−1 +c
= √ du = sin−1 (u) + c 4 + (x + 1)2 2
1 − u2
−1 x
= sin + c, a > 0. 4x + 4
a 14. dx
5 + 2x + x2
b 2(x + 1)
4. √ dx =2 dx = 2 ln | 4 + (x + 1)2 | + c
|x| x2 − a2 4 + (x + 1)2
b 1 4t
= dx 15. dt
x 2 a 5 + 2t + t2
|x| a −1
4t + 4 4
x 1 = dt − dt
Let u = , du = dx and |au| = |x| . 5 + 2t + t2 5 + 2t + t2
a a t+1
b = 2 ln 4 + (t + 1)
2
− 2tan−1 +c
= √ du 2
|au| u2 − 1
b 1 t+1 2 (t + 1)
= √ du 16. dt = dt
|a| |u| u2 − 1 t2 + 2t + 4 2
(t + 1) + 3
b 1
= sec−1 (u) + c 2
= ln (t + 1) + 3 + c
|a| 2
b x
= sec−1 + c, a > 0. 17.
1
e3−2x dx = − e3−2x + c
|a| a 2
1 3
5. sin(6t)dt = − cos(6t) + c 18. 3e−6x dx = − e−6x + c
6 6
360

2. 6.1. REVIEW OF FORMULAS AND TECHNIQUES 361
2 −1/3 30. Let u = ex , du = ex dx
19. Let u = 1 + x2/3 , du = x dx
3 ex 1
4 3 √ dx = √ du
dx = 4 u−1 du 1 − e2x 1 − u2
x1/3 (1 + x2/3 ) 2
= sin−1 u + C = sin−1 ex + c
2/3
= 6 ln |u| + C = 6 ln |1 + x |+c
31. Let u = x2 , du = 2xdx
3 x 1 1
20. Let u = 1 + x3/4 , du = x−1/4 dx √ dx = √ du
4 1 − x4 2 1 − u2
2 2
dx = dx 1 1
x1/4 + x x1/4 (1 + x3/4 ) = sin−1 u + C = sin−1 x2 + c
4 8 2 2
=2 u−1 du = ln |u| + C
3 3 32. Let u = 1 − x4 , du = −4x3 dx
8 2x3 1
= ln |1 + x3/4 | + c √ dx = − u−1/2 du
3 1−x 4 2
√ 1 = −u1/2 + C = −(1 − x4 )1/2 + c
21. Let u = x, du = √ dx
√ 2 x
sin x 1+x
√ dx = 2 sin udu 33. dx
x 1 + x2
√ 1 1 2x
= −2 cos u + C = −2 cos x + c = dx + dx
1 + x2 2 1 + x2
1 1 1
22. Let u = , du = − 2 dx = tan−1 x + ln |1 + x2 | + c
x x 2
cos(1/x)
dx = − cos udu 1
x2 34. √ dx
1 x+x
= − sin u + C = − sin + c
x 1
= x−1/2 · dx
23. Let u = sin x, du = cos xdx 1 + x1/2
π 0
= 2 ln | 1 + x1/2 | + c
cos xesin x dx = eu du = 0
0 0
ln x2 1
2 35. dx = 2 ln x dx
24. Let u = tan x, du = sec xdx x x
π/2 1 1
sec2 xetan x dx = eu du Let u = ln x, du = dx.
x
0 0
2
u
1
=2 u du = u2 + c = (ln x) + c
=e =e−1
0
3 3 3
0 x3 26
25. sec x tan xdx 36. e2 ln x dx = x2 dx = =
−π/4 1 1 3 1 3
0 √ 4
= sec x =1− 2 √
−π/4 37. x x − 3dx
3
π/2 4 √
π/2
26. csc2 xdx = − cot x =1 = (x − 3 + 3) x − 3dx
π/4 π/4 3
4 4
3 2 = (x − 3)3/2 dx + 3 (x − 3)1/2 dx
27. Let u = x , du = 3x dx 3 3
x2 1 1 2
4
2
4
12
dx = du = (x − 3)5/2 + 3 · (x − 3)3/2 =
1 + x6 3 1 + u2 5 3 5
3 3
1 1
= tan−1 u + C = tan−1 x3 + c 1
3 3 38. x(x − 3)2 dx
0
x5 1 1
28. dx = ln(1 + x6 ) + c = (x3 − 6x2 + 9x)dx
1 + x6 6
0
1
1 x x4 9 11
29. √ dx = sin−1 + c = − 2x3 + x2 =
4−x 2 2 4 2 0 4

3. 362 CHAPTER 6. INTEGRATION TECHNIQUES
4 2
x2 + 1
39. √ dx 47. f (x)dx
1 x 0
4 4 1 2
x x2
= x3/2 dx + x−1/2 dx = 2+1
dx + dx
1 1 0 x 1 x2 + 1
1 2
4 1 1
2 5/2 4 72 = ln |x2 + 1| + 1− 2 dx
= x + 2x1/2 = 2 0 1 x +1
5 1 1 5 1 2
= ln 2 + (x − arctan x)
0 0
2 1
2 1 −x2 e−4 − 1 ln 2 π
40. xe−x dx = − e = = + 1 + − arctan 2
−2 2 −2 2 2 4
4x + 1
5 5 x 48. dx
41. dx = √ arctan √ + c 2x2 + 4x + 10
3 + x2 3 3
5 4x + 4 3
dx: N/A = 2 + 4x + 10
dx − 2 + 4x + 10
dx
3 + x3 2x 2x
3 1
= ln |2x2 + 4x + 10| − dx
1 2 (x + 1)2 + 4
42. sin(3x)dx = sin(3x)3dx
3 3 x+1
Let u = 3x, du = 3dx. = ln |2x2 + 4x + 10| − tan−1 +c
1 1 4 2
= (sin u)du = − cos u + c
3 3 1
1 49. dx = tan−1 (x) + c.
= − cos(3x) + c. (1 + x2 )
3 x 1 2x
2)
dx = dx
(1 + x 2 (1 + x2 )
sin3 xdx = (sin2 x) sin xdx 1
= ln 1 + x2 + c.
2
= (1 − cos2 x)sin xdx x2 x2 + 1 − 1
dx = dx
Let u = cos x, du = − sin xdx. (1 + x2 ) (1 + x2 )
2
= 1 − u2 (−du) = u2 du − du x +1 1
= 2 + 1)
dx − dx
(x (1 + x2 )
u3 cos3 x 1
= −u= − cos x. = dx − dx
3 3 (1 + x2 )
= x − tan−1 (x) + c.
43. ln xdx: N/A x3 1 x2
2)
dx = 2xdx
Substituting u = ln x, (1 + x 2 (1 + x2 )
ln x 1
dx = ln2 x + c Let u = x2 , du = 2xdx.
2x 4 1 u 1 u+1−1
= du = du
2 1+u 2 1+u
44. Substituting u = x4 1 u+1 1
= du − du
x3 1 2 1+u 1+u
dx = arctan x4 + c
1 + x8 4 1 1
= du − du
x4 2 1+u
dx: N/A 1
1 + x8 = (u − ln (1 + u)) + c
2
1 1
45. e−x dx: N/A
2
= x2 − ln 1 + x2 + c.
2 2
Substituting u = −x2 Hence we can generalize this as follows,
2 1 2 xn
xe−x dx = − e−x + c 1 + x2
dx
2
1 xn−2
= xn−1 − dx
46. sec xdx: N/A n−1 1 + x2
x 1 1
sec2 xdx = tan x + c 50. dx = 2xdx
1 + x4 2 1 + x4

4. 6.2. INTEGRATION BY PARTS 363
Let u = x2 , du = 2xdx. 1 2x 1 2x
xe2x dx =xe − e dx
1 1 1 2 2
= du = tan−1 (u) + c 1 1
2 1 + u2 2 = xe2x − e2x + c.
1 2 4
= tan−1 x2 + c.
2 4. Let u = ln x, dv = x dx
x3 1 1 1 x2
4
dx = 4x3 dx du = dx and v = .
1+x 4 1 + x4 x 2
Let u = 1 + x4 , du = 4x3 . 1 1
x ln x dx = x2 ln x − x dx
1 1 1 2 2
= du = ln (u) + c 1 1
4 u 4 = x2 ln x − x2 + c.
1 2 4
= ln 1 + x4 + c.
4 5. Let u = ln x, dv = x2 dx
x5 1 1
dx du = dx, v = x3 .
1 + x4 x 3
1 x4 2 1 3 1 3 1
= 2xdx x ln xdx = x ln x − x · dx
2 1 + x4 3 3 x
Let u = x2 , du = 2xdx. 1 3 1
= x ln x − x2 dx
1 u2 1 u2 + 1 − 1 3 3
= du = du 1 1
2 1+u 2 2 1 + u2 = x3 ln x − x3 + c.
1 u2 + 1 1 3 9
= du − du
2 1 + u2 1 + u2 1
6. Let u = ln x, du = dx.
1 1 x
= du − du ln x u2 1
2 1 + u2 dx = udu = + c = (ln x)2 + c.
1 x 2 2
= u − tan−1 (u) + c
2
1 2 7. Let u = x2 , dv = e−3x dx
= x − tan−1 x2 + c. 1
2 du = 2xdx, v = − e−3x
Hence we can generalize this as follows, 3
x4n+1 1 x2n−2 x4(n−1)+1 I = x2 e−3x dx
dx = − dx
1 + x4 2 n−1 1 + x4 1 1
and = − x2 e−3x − − e−3x · 2xdx
3 3
x4n+3 1 x2n x4(n−1)+3 1 2
dx = − dx = − x2 e−3x + xe−3x dx
1+x 4 4 n 1 + x4 3 3
Let u = x, dv = e−3x dx
1
6.2 Integration by Parts du = dx, v = − e−3x
3
1
I = − x2 e−3x
1. Let u = x, dv = cos xdx 3
du = dx, v = sin x. 2 1 1
+ − xe−3x − − e−3x dx
x cos xdx = x sin x − sin xdx 3 3 3
1 2 2
= x sin x + cos x + c = − x2 e−3x − xe−3x + e−3x dx
3 9 9
1 2 2 −3x
2. Let u = x, dv = sin 4xdx = − x2 e−3x − xe−3x − e +c
1 3 9 27
du = dx, v = − cos 4x
4 8. Let u = x3 , du = 3x2 dx.
3 1 1
x sin 4x dx x2 ex dx = eu dx = eu + c
3 3
1 1 1 x3
= − x cos 4x − − cos 4x dx = e + c.
4 4 3
1 1
= − x cos 4x + sin 4x + c.
4 16 9. Let I = ex sin 4xdx
3. Let u = x, dv = e2x dx u = ex , dv = sin 4xdx
1 1
du = dx, v = e2x . du = ex dx, v = − cos 4x
2 4

5. 364 CHAPTER 6. INTEGRATION TECHNIQUES
1 1 1
I = − ex cos 4x − − cos 4x ex dx du = cos xdx v = − cos 2x
4 4 2
1 x 1 1 1 1
= − e cos 4x + x
e cos 4xdx I = cos x sin 2x + − cos 2x sin x
4 4 2 2 2
1
Use integration by parts again, this time let − − cos 2x cos xdx
u = ex , dv = cos 4xdx 2
1 1 1 1
du = ex dx, v = sin 4x = cos x sin 2x − cos 2x sin x + Idx
4 2 4 4
1 x
I = − e cos 4x So,
4 3 1 1
1 1 x 1 I = cos x sin 2x − cos 2x sin x + c1
+ e sin 4x − (sin 4x)ex dx 4 2 4
4 4 4
1 1 1 2 1
I = − ex cos 4x + ex sin 4x − I I = cos x sin 2x − cos 2x sin x + c
4 16 16 3 3
So,
17 1 1 12. Here we use the trigonometric identity:
I = − ex cos 4x + ex sin 4x + c1 sin 2x = 2 sin x cos x.
16 4 16
4 1 We then make the substitution
I = − ex cos 4x + ex sin 4x + c
17 17 u = sin x, du = cos x dx.
10. Let, u = e2x , dv = cos x dx so that, sin x sin 2x dx = 2 sin2 x cos x dx
du = 2e2x dx and v = sin x.
e2x cos x dx 2 3 2
= 2u2 du = u + c = sin3 x + c
3 3
= e2x sin x − 2 e2x sin x dx This integral can also be done by parts, twice.
If this is done, an equivalent answer is ob-
Let, u = e2x , dv = sin x dx so that, tained:
du = 2e2x dx and v = − cos x. 1 2
cos x sin 2x − cos 2x sin x + c
e2x sin x dx 3 3
13. Let u = x, dv = sec2 xdx
2x 2x
= −e cos x + 2 e cos x dx du = dx, v = tan x
e2x cos x dx x sec2 xdx = x tan x − tan xdx
sin x
= e2x sin x + 2e2x cos x − 4 e2x cos x dx = x tan x − dx
cos x
Now we notice that the integral on both of Let u = cos x, du = − sin xdx
1
these is the same, so we bring them to one side x sec2 xdx = x tan x + du
of the equation. u
= x tan x + ln |u| + c
5 e2x cos x dx = x tan x + ln |cos x| + c
= e2x sin x + 2e2x cos x + c1 14. Let u = (ln x)2 , dv = dx
ln x
e2x cos x dx du = 2 dx, v = x
x
1 2x 2
= e sin x + e2x cos x + c I = (ln x)2 dx
5 5
ln x
= x(ln x)2 − x·2 dx
11. Let I = cos x cos 2xdx x
and u = cos x, dv = cos 2xdx = x(ln x)2 − 2 ln xdx
1
du = sin xdx, v = sin 2x Integration by parts again,
2 1
1 1 u = ln x, dv = dxdu = dx, v = x
I = cos x sin 2x − sin 2x(− sin x)dx x
2 2 1
2
1 1 I = x(ln x) − 2 x ln x − x · dx
= cos x sin 2x + sin x sin 2xdx x
2 2
Let,u = sin x, dv = sin 2xdx = x(ln x)2 − 2x ln x + 2 dx

6. 6.2. INTEGRATION BY PARTS 365
= x(ln x)2 − 2x ln x + 2x + c 20. Let u = 2x, dv = cos x dx
duπ= 2 dxandv = sin x. π
2
15. Let u = x2 , dv = xex dx so that, du = 2x dx 2x cos xdx = 2x sin x|0 − 2
π
sin xdx
1 2 0 0
and v = ex (v is obtained using substitu- = (2x sin x +
π
2 cos x)|0 = −4.
2
tion).
2 1 2 2 1
x3 ex dx = x2 ex − xex dx 21. x2 cos πxdx
2
0
1 2 1 2
= x2 ex − ex + c Let u = x2 , dv = cos πxdx,
2 2 sin πx
du = 2xdx, v = .
π
x 1 1 1
16. Let u = x2 , dv = dx sin πx sin πx
3/2 x2 cos πxdx = x2 − 2xdx
(4 + x2 ) 0 π 0 0 π
1 2 1
du = 2xdx, v = − √ = (0 − 0) − x sin (πx) dx
4 + x2 π 0
3 1
x x 2
3/2
dx = x2 3/2
dx =− x sin (πx) dx
(4 + x2 ) (4 + x2 ) π 0
x2 1 Let u = x, dv = sin(πx)dx,
= −√ + √ 2xdx cos(πx)
4+x 2 4 + x2 du = dx, v = − .
x2 π
1
=− + 2 (4 + x2 ) + c. 2
(4 + x2 ) − xsin(πx)dx
π 0
1 1
17. Let u = ln(sin x), dv = cos xdx 2 x cos(πx) cos(πx)
=− − − − dx
1 π π 0 0 π
du = · cos xdx, v = sin x 1
sin x 2 cos π 1 sin(πx)
=− (− − 0) +
I = cos x ln(sin x)dx π π π π 0
= sin x ln(sin x) 2 1 1 2
1 =− + (0 − 0) = − 2
− sin x · · cos xdx π π π π
sin x
1
= sin x ln(sin x) − cos xdx
22. x2 e3x dx
= sin x ln(sin x) − sin x + c 0
Let u = x2 , dv = e3x dx,
e3x
18. This is a substitution u = x2 . du = 2xdx, v = .
1 3
x sin x2 dx = sin udu 1
x2 e3x
1 1 3x
e
2 x2 e3x dx = − 2xdx
1 1 3 0 3
= − cos u + c = − cos x2 + c. 0 0
2 2 1 3 2 1 3x
= e −0 − xe dx.
19. Let u = x, dv = sin 2xdx 3 3 0
1 Let u = x, dv = e3x dx,
du = dx, v = − cos 2x e3x
2 dv = dx, v = .
1
3
x sin 2xdx e3 2 1 3x
0 − xe dx
1
1
1
1 3 3 0
= − x cos 2x − − cos 2x dx e3 2 e3x
1 1 3x
e
2 0 0 2 = − x − dx
1 1 1 3 3 3 0 0 3
= − (1 cos 2 − 0 cos 0) + cos 2xdx 1
2 2 0 e3 2 e3 e3x
1 = − − dx
1 1 1 3 3 3 0 3
= − cos 2 + sin 2x 3 1
2 2 2 0 e 2 e3 e3x
1 1 = − −
= − cos 2 + (sin 2 − sin 0) 3 3 3 9 0
2 4 e 3
2 e3 1 3
1 1 = − − e −1
= − cos 2 + sin 2 3 3 3 9
2 4

7. 366 CHAPTER 6. INTEGRATION TECHNIQUES
e3 2e3 2 3 x cos (ax) sin (ax)
= − + e −1 =− + + c, a = 0.
3 9 27 a a2
e3 2e3 2e3 2 5e3 2
= − + − = −
3 9 27 27 27 27 27. (xn ) (ln x) dx = (ln x) (xn ) dx
10
Let u = ln x, dv = xn dx,
23. ln 2xdx 1 xn+1
1 du = dx, v = .
Let u = ln 2x, dv = dx x (n + 1)
1
du = dx, v = x. (ln x)(xn ) dx
x
10 10
10 1 xn+1 xn+1 dx
ln (2x)dx = x ln (2x)|1 − x dx = (ln x) −
1 1 x (n + 1) (n + 1) x
10
= (10 ln(20) − ln 2) − dx xn+1 (ln x) xn
= − dx
1
10
(n + 1) (n + 1)
= (10 ln(20) − ln 2) − [x]1 x n+1
(ln x) x n+1
= (10 ln(20) − ln 2) − (10 − 1) = − 2 + c, n = −1.
(n + 1) (n + 1)
= (10 ln(20) − ln 2) − 9.
24. Let, u = ln x, dv = x dx
28. (sin ax) (cos bx) dx
1 x2
du = dx, v = .
x 2 Let u = sin ax, dv = (cos bx) dx
2 2 2 sin bx
1 1 du = a (cos ax) dx, v = .
x ln xdx = x2 ln x − xdx b
1 2 1 1 2
2
1 2 1 3 sin ax cos bx dx
= x ln x − x2 = 2 ln 2 − .
2 4 1 4 sin bx sin bx
= (sin ax) − a (cos ax) dx
b b
25. x2 eax dx (sin ax) (sin bx) a
= − (cos ax) (sin bx) dx
b b
Let u = x2 , dv = eax dx, Let u = cos ax, dv = sin bxdx,
eax cos bx
du = 2xdx, v = . du = −a (sin ax) dx, v = − .
aax b
e eax
x2 eax dx = x2 − 2xdx sin ax sin bx a
a a − cos ax sin bx dx
b b
x2 eax 2 sin ax sin bx a − cos bx
= − xeax dx. = − cos ax
a a b b b
Let u = x, dv = eax dx,
eax − cos bx
dv = dx, v = . − (− sin ax) adx
a b
2 ax
x e 2 sin ax sin bx a − cos ax cos bx
− xeax dx = −
a a b b b
x2 eax 2 eax eax a
= − x − dx − cos bx sin ax dx
a a a a b
x2 eax 2 xeax eax sin ax sin bx a cos ax cos bx
= − − 2 +c = +
a a a a b b2
a 2
x2 eax 2xeax 2eax + sin ax cos bx dx
= − + 3 + c, a = 0. b
a a2 a
sin ax cos bx dx
26. x sin (ax) dx sin ax sin bx a cos ax cos bx
= +
Let u = x, dv = sin axdx, b b2
cos ax a 2
du = dx, v = − . + sin ax cos bx dx
a b
a 2
xsin (ax) dx sin ax cos bx dx − sin ax cos bx dx
b
− cos (ax) cos (ax) sin ax sin bx a cos ax cos bx
=x − − dx = +
a a b b2

8. 6.2. INTEGRATION BY PARTS 367
a2
1− sin ax cos bx dx + (n − 1) sinn xdx
b2
sin ax sin bx a cos ax cos bx n sinn xdx
= +
b b2
= − sinn−1 x cos x
sin ax cos bx dx
− (n − 1) sinn−2 xdx
b2 sin ax sin bx a cos ax cos bx
= + 1
b2 − a2 b b2 sinn xdx = − sinn−1 x cos x
n
sin ax cos bx dx n−1
− sinn−2 xdx
1 n
= 2 − a2
(b sin ax sin bx + a cos ax cos bx) ,
b
a = 0 b = 0. 31. x3 ex dx = ex (x3 − 3x2 + 6x − 6) + c
29. Letu = cosn−1 x, dv = cos xdx
du = (n − 1)(cosn−2 x)(− sin x)dx, v = sin x 32. cos5 xdx
cosn xdx 1 4
= cos4 sin x + cos3 xdx
= sin x cos n−1
x 5 5
1
− (sin x)(n − 1)(cosn−2 x)(− sin x)dx = cos4 sin x
5
= sin x cosn−1 x 4 1 2
+ cos2 x sin x + cos xdx
5 3 3
+ (n − 1)(cosn−2 x)(sin2 x)dx 1 4
= cos4 sin x + cos2 x sin x
= sin x cosn−1 x 5 15
8
+ (n − 1)(cosn−2 x)(1 − cos2 x)dx + sin x + c
15
= sin x cosn−1 x
+ (n − 1)(cosn−2 x − cosn x)dx 33. cos3 xdx
1 2
Thus, cosn xdx = cos2 x sin x + cos xdx
3 3
1 2
= sin x cosn−1 x + (n − 1) cosn−2 xdx = cos2 x sin x + sin x + c
3 3
− (n − 1) cosn xdx. 34. sin4 xdx
n cosn xdx = sin x cosn−1 x 1
= − sin3 x cos x +
3
sin2 xdx
4 4
+ (n − 1) cosn−2 xdx 1 3 1 1
= − sin3 x cos x + x − sin 2x
4 4 2 4
cosn xdx
1
1 n−1 35. x4 ex dx
= sin x cosn−1 x + cos n−2
xdx 0
n n 1
= ex (x4 − 4x3 + 12x2 − 24x + 24) 0
30. Let u = sinn−1 x, dv = sin x dx = 9e − 24
du = (n − 1) sinn−2 x cos x, v = − cos x.
sinn xdx 36. Using the work done in Exercise 34,
π/2
= − sinn−1 x cos x sin4 xdx
0
+ (n − 1) cos2 x sinn−2 xdx 1 3 3
π/2
= − sin3 x cos x + x − sin 2x
= − sinn−1 x cos x 4 8 16 0
3π
+ (n − 1) (1 − sin2 x) sinn−2 xdx =
16
= − sinn−1 x cos x π/2
− (n − 1) sinn−2 xdx 37. sin5 xdx
0

9. 368 CHAPTER 6. INTEGRATION TECHNIQUES
π/2 π/2
1 4 π/2 3
= − sin4 x cos x + sin xdx cosm xdx
5 0 5 0 0
1
π/2 (n − 1)(n − 3)(n − 5) · · · 2
= − sin4 x cos x = .
5 n(n − 2)(n − 4) · · · 3
0
π/2
4 1 2 41. Let u = cos−1 x, dv = dx
+ − sin2 x cos x − cos x 1
5 3 3 0 du = − √ dx, v = x
(Using Exercise 30) 1 − x2
1 π π cos−1 xdx
=− sin4 cos − sin4 0 cos 0 I=
5 2 2
4 1 2 π π 2 π 1
+ − sin cos − cos = x cos−1 x − x −√ dx
5 3 2 2 3 2 1 − x2
8 x
= = x cos−1 x + dx √
15 1 − x2
Substituting u = 1 − x2 , du = −2xdx
38. Here we will again use the work we did in Ex-
1 1
ercise 34. I = x cos−1 x + √ − du
u 2
sin6 xdx 1
= x cos1 x − u−1/2 du
1 5 2
= − sin5 x cos x + sin4 xdx 1
6 6 = x cos−1 x − · 2u1/2 + c
1 2
= − sin5 x cos x = x cos−1 x − 1 − x2 + c
6
5 1 3 3
+ − sin3 x cos x + x − sin 2x + c 42. Let u = tan−1 x, dv = dx
6 4 8 16 1
1 5 du = dx, v = x
= − sin5 x cos x − sin3 x cos x 1 + x2
6 24 x
15
+ x−
15
sin 2x + c I= tan−1 xdx = x tan−1 x − dx
48 96 1 + x2
Substituting u = 1 + x2 ,
We now just have to plug in the endpoints: 1
π/2 I = x tan−1 x − ln(1 + x2 ) + c.
sin6 xdx 2
0 √ 1
1 5 43. Substituting u = x, du = √ dx
= − sin5 x cos x − sin3 x cos x 2 x
6 24 √
π/2 I = sin xdx = 2 u sin udu
15 15
+ x− sin 2x
48 96 0 = 2(−u cos u + sin u) + c
√ √ √
15π = 2(− x cos x + sin x) + c
=
96 √
44. Substituting w = x
39. m even : 1 1
π/2 dw = √ dx = dx
sinm xdx 2 x 2w
√
x
0
(m − 1)(m − 3) . . . 1 π I= e dx = 2wew dx
= · Next, using integration by parts
m(m − 2) . . . 2 2
m odd: u = 2w, dv = ew dw
π/2
du = 2dw, v = ew
sinm xdx
0 I = 2wew − 2 ew dw
(m − 1)(m − 3) . . . 2 √ √ √
=
m(m − 2) . . . 3 = 2wew − 2ew + c = 2 xe x − 2e x + c
40. m even: 45. Let u = sin(ln x), dv = dx
π/2 dx
cosm xdx du = cos(ln x) , v = x
0
x
π(n − 1)(n − 3)(n − 5) · · · 1 I = sin(ln x)dx
=
2n(n − 2)(n − 4) · · · 2
m odd: = x sin(ln x) − cos(ln x)dx

10. 6.2. INTEGRATION BY PARTS 369
Integration by parts again, = 3 u2 eu − 2 ueu du
u = cos(ln x), dv = dx
dx
du = − sin(ln x) , v = x = 3u2 eu − 6 ueu − eu du
x
cos(ln x)dx = 3u2 eu − 6ueu + 6eu + c
8 √ 2
3 x
Hence e dx = 3u2 eu du
= x cos(ln x) + sin(ln x)dx 0 0
2
I = x sin(ln x) − x cos(ln x) − I = 3u e − 6ueu + 6e
2 u u
0
= 6e2 − 6
2I = x sin(ln x) − x cos(ln x) + c1
1 1 50. Let u = tan−1 x, dv = xdx
I = x sin(ln x) − x cos(ln x) + c dx x2
2 2 du = , v=
1 + x2 2
46. Let u = 4 + x2 , du = 2xdx I= x tan−1 xdx
I= x ln(4 + x2 )dx x2 1 x2
= tan−1 x − dx
1 1 2 2 1 + x2
= ln udu = (u ln u − u) + C x2
2 2 = tan−1 x
1 2
= [(4 + x ) ln(4 + x2 ) − 4 − x2 ] + c
2
1 1
2 − 1dx − dx
2 1 + x2
47. Let u = e2x , du = 2e2x dx x2 1
1 = tan−1 x − x − tan−1 x + C
I = e6x sin(e2x )dx = u2 sin udu 2 2
2 −1 x
2
x 1
Let v = u2 , dw = sin udu = tan x − + tan−1 x + c
2 2 2
dv = 2udu, w = − cos u 1
1 Hence x tan−1 xdx
I= −u2 cos u + 2 u cos udu 0
2 x2 x 1
1
π 1
1 = tan−1 x − + tan−1 x = −
= − u2 cos u + u cos udu 2 2 2 0 4 2
2
1 2 51. n times. Each integration reduces the power of
= − u cos u + (u sin u + cos u) + c
2 x by 1.
1
= − e4x cos(e2x ) + e2x sin(e2x )
2 52. 1 time. The first integration by parts gets rid
+ cos(e2x ) + c of the ln x and turns the integrand into a sim-
ple integral. See, for example, Problem 4.
√ 1 −2/3
48. Let u = 3
x = x1/3 , du = x dx, 53. (a) As the given problem, x sin x2 dx can
3 be simplified by substituting x2 = u, we
3u2 du = dx
can solve the example using substitution
I = cos x1/3 dx = 3 u2 cos udu method.
Let v = u2 , dw = cos udu (b) As the given integral, x2 sin x dx can not
dv = 2udu, w = sin u be simplified by substitution method and
I = 3 u2 sin u − 2 u sin udu can be solved using method of integration
by parts.
= 3u2 sin u − 6 u sin udu (c) As the integral, x ln x dx can not be sim-
plified by substitution and can be solved
= 3u2 sin u − 6 −u cos u + cos udu using the method of integration by parts.
3 ln x
= 3u sin u + 6u√ u −√ sin u + c √
√ cos 6 (d) As the given problem, dx can be
= 3x sin 3 x + 6 3 x cos 3 x − 6 sin 3 x + c x
simplified by substituting , ln x = u we
√ 1 −2/3 can solve the example by substitution
49. Let u = 3
x = x1/3 , du = x dx, method.
3
3u2 du = dx
√
3 54. (a) As this integral, x3 e4x dx can not be
I = e x dx = 3 u2 eu du simplified by substitution method and can

11. 370 CHAPTER 6. INTEGRATION TECHNIQUES
be solved by using the method of integra- 59.
tion by parts. e2x
4 4 2x
(b) As the given problem, x3 ex dx can be x e /2 +
simplified by substituting x4 = u, we can 4x3 e2x /4 −
solve the example using the substitution 12x2 e2x /8 +
2x
method. 24x e /16 −
4 24 e2x /32 +
(c) As the given problem, x−2 e x dx can be
1
simplified by substituting = u, we can x4 e2x dx
x
solve the example using the substitution x4 3x2 3x 3
= − x3 + − + e2x + c
method. 2 2 2 4
(d) As this integral, x2 e−4x dx can not be 60.
simplified by substitution and can be cos 2x
solved by using the method of integration x 5
sin 2x/2 +
by parts.
5x4 − cos 2x/4 −
55. First column: each row is the derivative of the 20x3 − sin 2x/8 +
previous row; Second column: each row is the 60x2 cos 2x/16 −
antiderivative of the previous row. 120x sin 2x/32 +
120 − cos 2x/64 −
56.
sin x
4 x5 cos 2xdx
x − cos x +
4x3 − sin x − 1 5 5
= x sin 2x + x4 cos 2x
12x2 cos x + 2 4
24x sin x − 20 60
− x3 sin 2x − x2 cos 2x
24 − cos x + 8 16
120 120
x4 sin xdx + x sin 2x + cos 2x + c
32 64
= −x4 cos x + 4x3 sin x + 12x2 cos x 61.
− 24x sin x − 24 cos x + c e−3x
3 −3x
57. x −e /3 +
cos x 3x2 e−3x /9 −
x4 sin x + 6x −e−3x /27 +
4x3 − cos x − 6 e−3x /81 −
12x2 − sin x +
x3 e−3x dx
24x cos x −
24 sin x + x3 x2 2x 2
= − − − − e−3x + c
3 3 9 27
x4 cos xdx
62.
= x4 sin x + 4x3 cos x − 12x2 sin x x2
− 24x cos x + 24 sin x + c ln x x3 /3 +
−1
58. x x4 /12 +
ex −x −2
x5 /60 +
4
x ex + The table will never terminate.
4x3 ex −
12x2 ex + 63. (a) Use the identity
24x ex − cos A cos B
1
24 ex + = [cos(A − B) + cos(A + B)]
2
x4 ex dx This identity gives
π
4 3 2
= (x − 4x + 12x − 24x + 24)e + c x cos(mx) cos(nx)dx
−π

12. 6.2. INTEGRATION BY PARTS 371
π nπ
1 1
= [cos((m − n)x) = cos2 udu
−π 2 n −nπ
+ cos((m + n)x)]dx nπ
1 1 1
1 sin((m − n)x) = u + cos(2u) =π
= n 2 4 −nπ
2 m−n π
π And then sin2 (nx)dx
sin((m + n)x) −π
+ π
m+n −π = (1 − cos2 (nx))dx
=0 −π
π π
It is important that m = n because oth- = dx − cos2 (nx)dx
erwise cos((m − n)x) = cos 0 = 1 −π −π
(b) Use the identity = 2π − π = π
sin A sin B
1 65. The only mistake is the misunderstanding of
= [cos(A − B) − cos(A + B)]
2 antiderivatives. In this problem, ex e−x dx
This identity gives
π is understood as a group of antiderivatives of
sin(mx) sin(nx)dx ex e−x , not a fixed function. So the subtraction
−π
π
1 by ex e−x dx on both sides of
= [cos((m − n)x)
−π 2
ex e−x dx = −1 + ex e−x dx
− cos((m + n)x)]dx
1 sin((m − n)x) does not make sense.
= π √ π
2 m−n
π
66. V = π (x sin x)2 dx = π x2 sin xdx
sin((m + n)x) 0 0
−
m+n −π
Using integration by parts twice we get
=0 x2 sin xdx
It is important that m = n because oth-
erwise cos((m − n)x) = cos 0 = 1 = −x2 cos x + 2 x cos xdx
64. (a) Use the identity = −x2 cos x + 2(x sin x − sin xdx)
cos A sin B = −x2 cos x + 2x sin x + 2 cos x + c
1
= [sin(B + A) − sin(B − A)] Hence,
2 π
This identity gives V = (−x2 cos x + 2x sin x + 2 cos x) 0
π = π 2 − 4 ≈ 5.87
cos(mx) sin(nx) dx
−π
π
1 67. Let u = ln x, dv = ex dx
= [sin((n + m)x) dx
−π 2 du = , v = ex
x
− sin((n − m)x)] dx ex
ex ln xdx = ex ln x − dx
1 cos((n + m)x) x
= − ex
2 n+m ex ln xdx + dx = ex ln x + C
π x
cos((n − m)x) Hence,
+
n−m 1
−π ex ln x + dx = ex ln x + c
=0 x
(b) We have seen that 68. We can guess the formula:
1 1
cos2 xdx = x + cos(2x) + c ex (f (x) + f (x))dx = ex f (x) + c
2 4
Hence by letting u = nx: and prove it by taking the derivative:
π
cos2 (nx)dx d x
(e f (x)) = ex f (x) + ex f (x)
−π dx

13. 372 CHAPTER 6. INTEGRATION TECHNIQUES
b
= ex (f (x) + f (x)) + f (x) (b − x) dx
a
1 b
69. Consider, f (x)g (x) dx Consider x sin (b − x) dx
0 0
Choose u = g (x) and dv = f (x)dx, b b
so that du = g (x) dx and , v = f (x) . = (b − x) sin xdx = (sin x) (b − x) dx
0 0
Hence, we have Now, consider
1
g (x)f (x)dx f (x) = x − sin x ⇒ f (x) = 1 − cos x
0 and f (x) = sin x.
1
1
= g (x) f (x)|0 − f (x)g (x) dx Therefore, using
0 f (b) = f (a) + f (a) (b − a)
= (g (1) f (1) − g (0) f (0)) b
1 + f (x) (b − x) dx,
− g (x)f (x) dx a
0 we get
From the given data. b − sin b = 0 − sin 0 + f (0) (b − 0)
1
b
= (0 − 0) − g (x)f (x) dx. + (sin x) (b − x) dx
0
0
Choose, u = g (x) and dv = f (x)dx, b
so that,du = g (x) dx and v = f (x) . ⇒ |sin b − b| = x sin (b − x) dx .
0
Hence, we have
1 Further,
− g (x)f (x) dx b b
0 |sin b − b| = x sin (b − x) dx ≤ xdx ,
1 0 0
1
=− g (x) f (x)|0 − f (x) g (x) dx as sin (b − x) ≤ 1.
0
= − {(g (1) f (1) − g (0) f (0) ) b2
Thus, |sin b − b| ≤ .
1 2
− f (x) g (x) dx Therefore the error in the approximation
0 1
From the given data. sin x ≈ x is at most x2 .
1 2
= − (0 − 0 ) − f (x) g (x) dx
1
0
6.3 Trigonometric
= f (x) g (x) dx.
0
Techniques of
70. Consider, Integration
b b
f (x) (b − x) dx = (b − x) f (x) dx 1. Let u = sin x, du = cos xdx
a a
Choose u = (b − x) and dv = f (x) dx, cos x sin4 xdx = u4 du
so that du = −dx and v = f (x) . 1 5 1
= u + c = sin5 x + c
Hence, we have: 5 5
b
(b − x)f (x) dx 2. Let u = sin x, du = cos xdx
a
b cos3 x sin4 xdx = (1 − u2 )u4 du
b
= (b − x) f (x)|a + f (x) dx
a u5 u7
b = − +c
5 7
= (0 − [(b − a) f (a)]) + f (x) dx sin x sin7 x
5
a
b
= − +c
= − [(b − a) f (a)] + f (x)|a 5 7
= − [(b − a) f (a)] + f (b) − f (a) 3. Let u = sin 2x, du = 2 cos 2xdx.
b π/4
f (x) (b − x) dx cos 2xsin3 2xdx
a 0
= − [(b − a) f (a)] + f (b) − f (a) 1 1
1 1 u4 1
= u3 du = =
f (b) = f (a) + (b − a) f (a) 2 0 2 4 0 8