Chapter 08

January 27, 2005 11:45 L24-CH08 Sheet number 1 Page number 337 black

CHAPTER 8
Principles of Integral Valuation

EXERCISE SET 8.1
1 1 1
1. u = 4 − 2x, du = −2dx, − u3 du = − u4 + C = − (4 − 2x)4 + C
2 8 8

3 √
2. u = 4 + 2x, du = 2dx, u du = u3/2 + C = (4 + 2x)3/2 + C
2

1 1 1
3. u = x2 , du = 2xdx, sec2 u du = tan u + C = tan(x2 ) + C
2 2 2

4. u = x2 , du = 2xdx, 2 tan u du = −2 ln | cos u | + C = −2 ln | cos(x2 )| + C

1 du 1 1
5. u = 2 + cos 3x, du = −3 sin 3xdx, − = − ln |u| + C = − ln(2 + cos 3x) + C
3 u 3 3

2 2 1 du 1 1 2
6. u = x, du = dx, = tan−1 u + C = tan−1 x + C
3 3 6 1 + u2 6 6 3

7. u = ex , du = ex dx, sinh u du = cosh u + C = cosh ex + C

1
8. u = ln x, du = dx, sec u tan u du = sec u + C = sec(ln x) + C
x

9. u = tan x, du = sec2 xdx, eu du = eu + C = etan x + C

1 du 1 1
10. u = x2 , du = 2xdx, √ = sin−1 u + C = sin−1 (x2 ) + C
2 1−u 2 2 2

1 1 6 1
11. u = cos 5x, du = −5 sin 5xdx, − u5 du = − u + C = − cos6 5x + C
5 30 30
√
du 1 + 1 + u2 1 + 1 + sin2 x
12. u = sin x, du = cos x dx, √ = − ln + C = − ln +C
u u2 + 1 u sin x

du
13. u = ex , du = ex dx, √ = ln u + u2 + 4 + C = ln ex + e2x + 4 + C
4 + u2

1 −1
14. u = tan−1 x, du = dx, eu du = eu + C = etan x
+C
1 + x2

√ 1 √
15. u = x − 1, du = √ dx, 2 eu du = 2eu + C = 2e x−1
+C
2 x−1

1 1 1
16. u = x2 + 2x, du = (2x + 2)dx, cot u du = ln | sin u| + C = ln sin |x2 + 2x| + C
2 2 2

√ 1 √
17. u = x, du = √ dx, 2 cosh u du = 2 sinh u + C = 2 sinh x + C
2 x

337


338 Chapter 8

dx du 1 1
18. u = ln x, du = , 2
=− +C =− +C
x u u ln x

√ 1 2 du 2 −u ln 3 2 −√x
19. u = x, du = √ dx, =2 e−u ln 3 du = − e +C =− 3 +C
2 x 3u ln 3 ln 3

20. u = sin θ, du = cos θdθ, sec u tan u du = sec u + C = sec(sin θ) + C

2 2 1 1 1 2
21. u = , du = − 2 dx, − csch2 u du = coth u + C = coth + C
x x 2 2 2 x

dx
22. √ = ln x + x2 − 4 + C
x2 − 4

du 1 2+u 1 2 + e−x
23. u = e−x , du = −e−x dx, − = − ln + C = − ln +C
4−u 2 4 2−u 4 2 − e−x

1
24. u = ln x, du = dx, cos u du = sin u + C = sin(ln x) + C
x

ex dx du
25. u = ex , du = ex dx, √ = √ = sin−1 u + C = sin−1 ex + C
1 − e2x 1 − u2

1
26. u = x−1/2 , du = − dx, − 2 sinh u du = −2 cosh u + C = −2 cosh(x−1/2 ) + C
2x3/2

1 du 1 1 1
27. u = x2 , du = 2xdx, = sin u du = − cos u + C = − cos(x2 ) + C
2 csc u 2 2 2

2du
28. 2u = ex , 2du = ex dx, √ = sin−1 u + C = sin−1 (ex /2) + C
4 − 4u2

29. 4−x = e−x
2 2
ln 4
, u = −x2 ln 4, du = −2x ln 4 dx = −x ln 16 dx,
1 1 u 1 −x2 ln 4 1 −x2
− eu du = − e +C =− e +C =− 4 +C
ln 16 ln 16 ln 16 ln 16

1 πx ln 2 1
30. 2πx = eπx ln 2 , 2πx dx = e +C = 2πx + C
π ln 2 π ln 2

1 2 1
31. (a) u = sin x, du = cos x dx, u du = u + C = sin2 x + C
2 2
1 1 1
(b) sin x cos x dx = sin 2x dx = − cos 2x + C = − (cos2 x − sin2 x) + C
2 4 4
1 1 1 1
(c) − (cos2 x − sin2 x) + C = − (1 − sin2 x − sin2 x) + C = − + sin2 x + C,
4 4 4 2
and this is the same as the answer in part (a) except for the constants.

1 1
32. (a) sech 2x = = (now multiply top and bottom by sech2 x)
cosh 2x cosh2 x + sinh2 x
sech2 x
=
1 + tanh2 x


Exercise Set 8.2 339

sech2 x
(b) sech2x dx = dx = tan−1 (tanh x) + C, or, replacing 2x with x,
1 + tanh2 x

sechx dx = tan−1 (tanh(x/2)) + C

1 2 2ex
(c) sech x = = x = 2x
cosh x e + e−x e +1
ex
(d) sech x dx = 2 dx = 2 tan−1 (ex ) + C
e2x+1

sec2 x 1 1
33. (a) = =
tan x cos2 x tan x cos x sin x
1 1 1 sec2 x 1
(b) csc 2x = = = , so csc 2x dx = ln tan x + C
sin 2x 2 sin x cos x 2 tan x 2
1 1
(c) sec x = = = csc(π/2 − x), so
cos x sin(π/2 − x)
1
sec x dx = csc(π/2 − x) dx = − ln tan(π/2 − x) + C
2

EXERCISE SET 8.2
1. u = x, dv = e−2x dx, du = dx, v = − 1 e−2x ;
2
1 1 −2x 1 1
xe−2x dx = − xe−2x + e dx = − xe−2x − e−2x + C
2 2 2 4

1 3x 1 3x 1 1 3x 1 3x
2. u = x, dv = e3x dx, du = dx, v = e ; xe3x dx = xe − e3x dx = xe − e + C
3 3 3 3 9

3. u = x2 , dv = ex dx, du = 2x dx, v = ex ; x2 ex dx = x2 ex − 2 xex dx.

For xex dx use u = x, dv = ex dx, du = dx, v = ex to get

xex dx = xex − ex + C1 so x2 ex dx = x2 ex − 2xex + 2ex + C

1 1
4. u = x2 , dv = e−2x dx, du = 2x dx, v = − e−2x ; x2 e−2x dx = − x2 e−2x + xe−2x dx
2 2

For xe−2x dx use u = x, dv = e−2x dx to get

1 1 1 1
xe−2x dx = − xe−2x + e−2x dx = − xe−2x − e−2x + C
2 2 2 4
1 1 1
so x2 e−2x dx = − x2 e−2x − xe−2x − e−2x + C
2 2 4

1
5. u = x, dv = sin 3x dx, du = dx, v = − cos 3x;
3
1 1 1 1
x sin 3x dx = − x cos 3x + cos 3x dx = − x cos 3x + sin 3x + C
3 3 3 9


340 Chapter 8

1
6. u = x, dv = cos 2x dx, du = dx, v = sin 2x;
2
1 1 1 1
x cos 2x dx = x sin 2x − sin 2x dx = x sin 2x + cos 2x + C
2 2 2 4

7. u = x2 , dv = cos x dx, du = 2x dx, v = sin x; x2 cos x dx = x2 sin x − 2 x sin x dx

For x sin x dx use u = x, dv = sin x dx to get

x sin x dx = −x cos x + sin x + C1 so x2 cos x dx = x2 sin x + 2x cos x − 2 sin x + C

8. u = x2 , dv = sin x dx, du = 2x dx, v = − cos x;

x2 sin x dx = −x2 cos x + 2 x cos x dx; for x cos x dx use u = x, dv = cos x dx to get

x cos x dx = x sin x + cos x + C1 so x2 sin x dx = −x2 cos x + 2x sin x + 2 cos x + C

1 1 1 2 1 1 2 1
9. u = ln x, dv = x dx, du = dx, v = x2 ; x ln x dx = x ln x − x dx = x ln x − x2 + C
x 2 2 2 2 4

√ 1 2
10. u = ln x, dv = dx, v = x3/2 ;
x dx, du =
x 3
√ 2 3/2 2 2 4
x ln x dx = x ln x − x1/2 dx = x3/2 ln x − x3/2 + C
3 3 3 9

ln x
11. u = (ln x)2 , dv = dx, du = 2 dx, v = x; (ln x)2 dx = x(ln x)2 − 2 ln x dx.
x

Use u = ln x, dv = dx to get ln x dx = x ln x − dx = x ln x − x + C1 so

(ln x)2 dx = x(ln x)2 − 2x ln x + 2x + C

1 1 √ ln x √ 1 √ √
12. u = ln x, dv = √ dx, du = dx, v = 2 x; √ dx = 2 x ln x−2 √ dx = 2 x ln x−4 x+C
x x x x

3 3x
13. u = ln(3x − 2), dv = dx, du = dx, v = x; ln(3x − 2)dx = x ln(3x − 2) − dx
3x − 2 3x − 2
3x 2 2
but dx = 1+ dx = x + ln(3x − 2) + C1 so
3x − 2 3x − 2 3
2
ln(3x − 2)dx = x ln(3x − 2) − x − ln(3x − 2) + C
3

2x x2
14. u = ln(x2 + 4), dv = dx, du = 2+4
dx, v = x; ln(x2 + 4)dx = x ln(x2 + 4) − 2 dx
x x2 + 4
x2 4 x
but dx = 1− dx = x − 2 tan−1 + C1 so
x2 +4 x2 + 4 2
x
ln(x2 + 4)dx = x ln(x2 + 4) − 2x + 4 tan−1 +C
2



√
15. u = sin−1 x, dv = dx, du = 1/ 1 − x2 dx, v = x;

sin−1 x dx = x sin−1 x − x/ 1 − x2 dx = x sin−1 x + 1 − x2 + C

2
16. u = cos−1 (2x), dv = dx, du = − √ dx, v = x;
1 − 4x2
2x 1
cos−1 (2x)dx = x cos−1 (2x) + √ dx = x cos−1 (2x) − 1 − 4x2 + C
1 − 4x2 2

3
17. u = tan−1 (3x), dv = dx, du = dx, v = x;
1 + 9x2
3x 1
tan−1 (3x)dx = x tan−1 (3x) − dx = x tan−1 (3x) − ln(1 + 9x2 ) + C
1 + 9x2 6

1 1 1 1 x2
18. u = tan−1 x, dv = x dx, du = dx, v = x2 ; x tan−1 x dx = x2 tan−1 x − dx
1 + x2 2 2 2 1 + x2
x2 1
but dx = 1− dx = x − tan−1 x + C1 so
1 + x2 1 + x2
1 1 1
x tan−1 x dx = x2 tan−1 x − x + tan−1 x + C
2 2 2

19. u = ex , dv = sin x dx, du = ex dx, v = − cos x; ex sin x dx = −ex cos x + ex cos x dx.

For ex cos x dx use u = ex , dv = cos x dx to get ex cos x = ex sin x − ex sin x dx so

ex sin x dx = −ex cos x + ex sin x − ex sin x dx,

1 x
2 ex sin x dx = ex (sin x − cos x) + C1 , ex sin x dx = e (sin x − cos x) + C
2

1
20. u = e3x , dv = cos 2x dx, du = 3e3x dx, v = sin 2x;
2
1 3x 3
e3x cos 2x dx = e sin 2x − e3x sin 2x dx. Use u = e3x , dv = sin 2x dx to get
2 2
1 3
e3x sin 2x dx = − e3x cos 2x + e3x cos 2x dx, so
2 2
1 3 9
e3x cos 2x dx = e3x sin 2x + e3x cos 2x − e3x cos 2x dx,
2 4 4
13 1 1 3x
e3x cos 2x dx = e3x (2 sin 2x + 3 cos 2x) + C1 , e3x cos 2x dx = e (2 sin 2x + 3 cos 2x) + C
4 4 13

1
21. u = eax , dv = sin bx dx, du = aeax dx, v = − cos bx (b = 0);
b
1 ax a
eax sin bx dx = − e cos bx + eax cos bx dx. Use u = eax , dv = cos bx dx to get
b b
1 a
eax cos bx dx = eax sin bx − eax sin bx dx so
b b
1 a a2
eax sin bx dx = − eax cos bx + 2 eax sin bx − 2 eax sin bx dx,
b b b
eax
eax sin bx dx = 2 (a sin bx − b cos bx) + C
a + b2


342 Chapter 8

e−3θ
22. From Exercise 21 with a = −3, b = 5, x = θ, answer = √ (−3 sin 5θ − 5 cos 5θ) + C
34

cos(ln x)
23. u = sin(ln x), dv = dx, du = dx, v = x;
x

sin(ln x)dx = x sin(ln x) − cos(ln x)dx. Use u = cos(ln x), dv = dx to get

cos(ln x)dx = x cos(ln x) + sin(ln x)dx so

sin(ln x)dx = x sin(ln x) − x cos(ln x) − sin(ln x)dx,

1
sin(ln x)dx = x[sin(ln x) − cos(ln x)] + C
2

1
24. u = cos(ln x), dv = dx, du = − sin(ln x)dx, v = x;
x

cos(ln x)dx = x cos(ln x) + sin(ln x)dx. Use u = sin(ln x), dv = dx to get

sin(ln x)dx = x sin(ln x) − cos(ln x)dx so

cos(ln x)dx = x cos(ln x) + x sin(ln x) − cos(ln x)dx,

1
cos(ln x)dx = x[cos(ln x) + sin(ln x)] + C
2

25. u = x, dv = sec2 x dx, du = dx, v = tan x;
sin x
x sec2 x dx = x tan x − tan x dx = x tan x − dx = x tan x + ln | cos x| + C
cos x

26. u = x, dv = tan2 x dx = (sec2 x − 1)dx, du = dx, v = tan x − x;

x tan2 x dx = x tan x − x2 − (tan x − x)dx

1 1
= x tan x − x2 + ln | cos x| + x2 + C = x tan x − x2 + ln | cos x| + C
2 2

2 1 x2
27. u = x2 , dv = xex dx, du = 2x dx, v = e ;
2
2 1 2 x2 2 1 2 x2 1 x2
x3 ex dx = x e − xex dx = x e − e +C
2 2 2

1 1
28. u = xex , dv = dx, du = (x + 1)ex dx, v = − ;
(x + 1)2 x+1
xex xex xex ex
dx = − + ex dx = − + ex + C = +C
(x + 1)2 x+1 x+1 x+1

1 2x
29. u = x, dv = e2x dx, du = dx, v = e ;
2
2 2 2 2
1 2x 1 1 1
xe2x dx = xe − e2x dx = e4 − e2x = e4 − (e4 − 1) = (3e4 + 1)/4
0 2 0 2 0 4 0 4



1
30. u = x, dv = e−5x dx, du = dx, v = − e−5x ;
5
1 1 1
1 1
xe−5x dx = − xe−5x + e−5x dx
0 5 0 5 0
1
1 1 1 1
= − e−5 − e−5x = − e−5 − (e−5 − 1) = (1 − 6e−5 )/25
5 25 0 5 25

1 1
31. u = ln x, dv = x2 dx, du = dx, v = x3 ;
x 3
e e e e
1 3 1 1 3 1 3 1 3 1 3
x2 ln x dx = x ln x − x2 dx = e − x = e − (e − 1) = (2e3 + 1)/9
1 3 1 3 1 3 9 1 3 9

1 1 1
32. u = ln x, dv = 2
dx, du = dx, v = − ;
x x x
e e e
ln x 1 1
√ 2
dx = − ln x √
+ √
dx
e x x e e x2
√
1 1 √ 1
e
1 1 1 1 3 e−4
= − + √ ln e − √
=− + √ − +√ =
e e x e e 2 e e e 2e

1
33. u = ln(x + 2), dv = dx, du = dx, v = x;
x+2
1 1 1 1
x 2
ln(x + 2)dx = x ln(x + 2) − dx = ln 3 + ln 1 − 1− dx
−1 −1 −1 x+2 −1 x+2
1
= ln 3 − [x − 2 ln(x + 2)] = ln 3 − (1 − 2 ln 3) + (−1 − 2 ln 1) = 3 ln 3 − 2
−1

1
34. u = sin−1 x, dv = dx, du = √ dx, v = x;
1 − x2
√
3/2
√
3/2
√
3/2
√ √ √
3/2
−1 −1 x 3 3
sin x dx = x sin x − √ dx = sin−1 + 1 − x2
0 0 0 1 − x2 2 2 0
√ √
3 π 1 π 3 1
= + −1= −
2 3 2 6 2
√ 1
35. u = sec−1 θ, dv = dθ, du = √ dθ, v = θ;
2θ θ − 1
4 √ √1
4
1 √ 4 √ 4
sec−1 θdθ = θ sec−1 dθ = 4 sec−1 2 − 2 sec−1 2 − θ − 1
θ − √
2 2 2 θ−1 2 2

π π √ 5π √
=4 −2 − 3+1= − 3+1
3 4 6
1 1
36. u = sec−1 x, dv = x dx, du = √ dx, v = x2 ;
x x2−1 2
2 2 2
1 2 1 x
x sec−1 x dx = x sec−1 x − √ dx
1 2 1 2 1 x2 − 1
1 1
2 √
= [(4)(π/3) − (1)(0)] − x2 − 1 = 2π/3 − 3/2
2 2 1


344 Chapter 8

1
37. u = x, dv = sin 2x dx, du = dx, v = − cos 2x;
2
π π π π
1 1 1
x sin 2x dx = − x cos 2x + cos 2x dx = −π/2 + sin 2x = −π/2
0 2 0 2 0 4 0

π π π π
1 2 π2
38. (x + x cos x)dx = x + x cos x dx = + x cos x dx;
0 2 0 0 2 0

u = x, dv = cos x dx, du = dx, v = sin x
π π π π π
x cos x dx = x sin x − sin x dx = cos x = −2 so (x + x cos x)dx = π 2 /2 − 2
0 0 0 0 0

√ √ 1 2
39. u = tan−1 x, dv = xdx, du = √ dx, v = x3/2 ;
2 x(1 + x) 3
3 √ √ 2 3/2 √ 3
1 3
x
x tan−1 xdx = x tan−1 x − dx
1 3 1 3 1 1+x
2 3/2 √ 3
1 3
1
= x tan−1 x − 1− dx
3 1 3 1 1+x
2 3/2 √ 1 1
3 √
= x tan−1 x − x + ln |1 + x| = (2 3π − π/2 − 2 + ln 2)/3
3 3 3 1

2x
40. u = ln(x2 + 1), dv = dx, du = dx, v = x;
x2 + 1
2 2 2 2
2x2 1
ln(x2 + 1)dx = x ln(x2 + 1) − dx = 2 ln 5 − 2 1− dx
0 0 0 x2+1 0 x2 + 1
2
= 2 ln 5 − 2(x − tan−1 x) = 2 ln 5 − 4 + 2 tan−1 2
0
√
41. t = x, t2 = x, dx = 2t dt
√
x
(a) e dx = 2 tet dt; u = t, dv = et dt, du = dt, v = et ,

√ √ √
e x
dx = 2tet − 2 et dt = 2(t − 1)et + C = 2( x − 1)e x + C

√
(b) cos x dx = 2 t cos t dt; u = t, dv = cos tdt, du = dt, v = sin t,

√ √ √ √
cos x dx = 2t sin t − 2 sin tdt = 2t sin t + 2 cos t + C = 2 x sin x + 2 cos x + C



42. Let q1 (x), q2 (x), q3 (x) denote successive antiderivatives of q(x),
so that q3 (x) = q2 (x), q2 (x) = q1 (x), q1 (x) = q(x). Let p(x) = ax2 + bx + c.

Repeated Repeated
Diﬀerentiation Antidiﬀerentiation
ax2 + bx + c q(x)
+
2ax + b q1 (x)
−
2a q2 (x)
+
0 q3 (x)

Then p(x)q(x) dx = (ax2 + bx + c)q1 (x) − (2ax + b)q2 (x) + 2aq3 (x) + C. Check:

d
[(ax2 +bx + c)q1 (x) − (2ax + b)q2 (x) + 2aq3 (x)]
dx
= (2ax + b)q1 (x) + (ax2 + bx + c)q(x) − 2aq2 (x) − (2ax + b)q1 (x) + 2aq2 (x) = p(x)q(x)

43. Repeated Repeated
3x2 − x + 2 e−x
+
6x − 1 −e−x
−
6 e−x
+
0 −e−x

(3x2 − x + 2)e−x = −(3x2 − x + 2)e−x − (6x − 1)e−x − 6e−x + C = −e−x [3x2 + 5x + 7] + C

x2 + x + 1 sin x
+
2x + 1 − cos x
−
2 − sin x
+
0 cos x

(x2 + x + 1) sin x dx = −(x2 + x + 1) cos x + (2x + 1) sin x + 2 cos x + C
= −(x2 + x − 1) cos x + (2x + 1) sin x + C


346 Chapter 8

4x4 sin 2x
+
1
16x3 − cos 2x
2
−
1
48x 2
− sin 2x
+ 4
1
96x cos 2x
8
−
1
96 sin 2x
16
+
1
0 − cos 2x
32

4x4 sin 2x dx = (−2x4 + 6x2 − 3) cos 2x + −(4x3 + 6x) sin 2x + C

√
x3 2x + 1
+
1
3x 2
(2x + 1)3/2
3
−
1
6x (2x + 1)5/2
15
+
1
6 (2x + 1)7/2
105
−
1
0 (2x + 1)9/2
945
√ 1 1 2 2
x3 2x + 1 dx = x3 (2x + 1)3/2 − x2 (2x + 1)5/2 + x(2x + 1)7/2 − (2x + 1)9/2 + C
3 5 35 315

47. (a) We perform a single integration by parts:
u = cos x, dv = sin x dx, du = − sin x dx, v = − cos x,

sin x cos x dx = − cos2 x − sin x cos x dx. Thus

1
2 sin x cos x dx = − cos2 x + C, sin x cos x dx = − cos2 x + C
2
1 2 1
(b) u = sin x, du = cos x dx, sin x cos x dx = u du = u + C = sin2 x + C
2 2
x
48. (a) u = x2 , dv = √ , du = 2x dx, v = x2 + 1,
x2 +1
1
x3
1 1 √ 2
1
1√ 2
√ dx = x2 x2 + 1 − 2x x2 + 1 dx = 2 − (x2 + 1)3/2 =− 2+
0 x2 + 1 0 0 3 0 3 3



√ √
2 2
x 1 3
(b) u = x2 + 1, du = √ dx, (u − 1) du =
2
u −u
x2 + 1 1 3 1
2√ √ 1 1√ 2
= 2− 2− +1=− 2+ .
3 3 3 3
e e
49. (a) A = ln x dx = (x ln x − x) =1
1 1
e e
(b) V = π (ln x)2 dx = π (x(ln x)2 − 2x ln x + 2x) = π(e − 2)
1 1

π/2 π/2 π/2 π/2
1 2 π2
50. A = (x − x sin x)dx = x − x sin x dx = − (−x cos x + sin x) = π 2 /8 − 1
0 2 0 0 8 0

π π
51. V = 2π x sin x dx = 2π(−x cos x + sin x) = 2π 2
0 0

π/2 π/2
52. V = 2π x cos x dx = 2π(cos x + x sin x) = π(π − 2)
0 0

π
53. distance = t3 sin tdt;
0

Repeated Repeated
t3 sin t
+
3t2 − cos t
−
6t − sin t
+
6 cos t
−
0 sin t

π π
t3 sin t dx = [(−t3 cos t + 3t2 sin t + 6t cos t − 6 sin t)] = π 3 − 6π
0 0

1
54. u = 2t, dv = sin(kωt)dt, du = 2dt, v = − cos(kωt); the integrand is an even function of t so
kω
π/ω π/ω π/ω π/ω
2 1
t sin(kωt) dt = 2 t sin(kωt) dt = − t cos(kωt) +2 cos(kωt) dt
−π/ω 0 kω 0 0 kω
k+1 π/ω k+1
2π(−1) 2 2π(−1)
= + sin(kωt) =
kω 2 k2 ω2 0 kω 2

1 3
55. (a) sin4 x dx = − sin3 x cos x + sin2 x dx
4 4
1 3 1 1
= − sin3 x cos x + − sin x cos x + x + C
4 4 2 2
1 3 3
= − sin3 x cos x − sin x cos x + x + C
4 8 8


348 Chapter 8

π/2 π/2 π/2
1 4
(b) sin5 x dx = − sin4 x cos x + sin3 x dx
0 5 0 5 0
π/2 π/2
4 1 2
= − sin2 x cos x + sin x dx
5 3 0 3 0
π/2
8 8
=− cos x =
15 0 15

1 4 1 4 1 2
56. (a) cos5 x dx = cos4 x sin x + cos3 x dx = cos4 x sin x + cos2 x sin x + sin x + C
5 5 5 5 3 3
1 4 8
= cos4 x sin x + cos2 x sin x + sin x + C
5 15 15
1 5
(b) cos6 x dx = cos5 x sin x + cos4 x dx
6 6
1 5 1 3
= cos5 x sin x + cos3 x sin x + cos2 x dx
6 6 4 4
1 5 5 1 1
= cos5 x sin x + cos3 x sin x + cos x sin x + x + C,
6 24 8 2 2
π/2
1 5 5 5
cos5 x sin x + cos3 x sin x + cos x sin x + x = 5π/32
6 24 16 16 0

57. u = sinn−1 x, dv = sin x dx, du = (n − 1) sinn−2 x cos x dx, v = − cos x;

sinn x dx = − sinn−1 x cos x + (n − 1) sinn−2 x cos2 x dx

= − sinn−1 x cos x + (n − 1) sinn−2 x (1 − sin2 x)dx

= − sinn−1 x cos x + (n − 1) sinn−2 x dx − (n − 1) sinn x dx,

n sinn x dx = − sinn−1 x cos x + (n − 1) sinn−2 x dx,

1 n−1
sinn x dx = − sinn−1 x cos x + sinn−2 x dx
n n

58. (a) u = secn−2 x, dv = sec2 x dx, du = (n − 2) secn−2 x tan x dx, v = tan x;

secn x dx = secn−2 x tan x − (n − 2) secn−2 x tan2 x dx

= secn−2 x tan x − (n − 2) secn−2 x (sec2 x − 1)dx

= secn−2 x tan x − (n − 2) secn x dx + (n − 2) secn−2 x dx,

(n − 1) secn x dx = secn−2 x tan x + (n − 2) secn−2 x dx,

1 n−2
secn x dx = secn−2 x tan x + secn−2 x dx
n−1 n−1



(b) tann x dx = tann−2 x (sec2 x − 1) dx = tann−1 x sec2 x dx − tann−2 x dx

1
= tann−1 x − tann−2 x dx
n−1

(c) u = xn , dv = ex dx, du = nxn−1 dx, v = ex ; xn ex dx = xn ex − n xn−1 ex dx

1 1 1
59. (a) tan4 x dx = tan3 x − tan2 x dx = tan3 x − tan x + dx = tan3 x − tan x + x + C
3 3 3
1 2 1 2
(b) sec4 x dx = sec2 x tan x + sec2 x dx = sec2 x tan x + tan x + C
3 3 3 3

(c) x3 ex dx = x3 ex − 3 x2 ex dx = x3 ex − 3 x2 ex − 2 xex dx

= x3 ex − 3x2 ex + 6 xex − ex dx = x3 ex − 3x2 ex + 6xex − 6ex + C

60. (a) u = 3x,
1 1 1 2 u 2
x2 e3x dx = u2 eu du = u2 eu − 2 ueu du = u e − ueu − eu du
27 27 27 27
1 2 u 2 2 1 2 2
= u e − ueu + eu + C = x2 e3x − xe3x + e3x + C
27 27 27 3 9 27
√
(b) u = − x,
1 √ −1
xe− x
dx = 2 u3 eu du,
0 0

u3 eu du = u3 eu − 3 u2 eu du = u3 eu − 3 u2 eu − 2 ueu du

= u3 eu − 3u2 eu + 6 ueu − eu du = u3 eu − 3u2 eu + 6ueu − 6eu + C,

−1 −1
2 u3 eu du = 2(u3 − 3u2 + 6u − 6)eu = 12 − 32e−1
0 0

61. u = x, dv = f (x)dx, du = dx, v = f (x);
1 1 1
x f (x)dx = xf (x) − f (x)dx
−1 −1 −1

1
= f (1) + f (−1) − f (x) = f (1) + f (−1) − f (1) + f (−1)
−1

62. (a) u dv = uv − v du = x(sin x + C1 ) + cos x − C1 x + C2 = x sin x + cos x + C2 ;

the constant C1 cancels out and hence plays no role in the answer.

(b) u(v + C1 ) − (v + C1 )du = uv + C1 u − v du − C1 u = uv − v du


350 Chapter 8

dx
63. u = ln(x + 1), dv = dx, du = , v = x + 1;
x+1
ln(x + 1) dx = u dv = uv − v du = (x + 1) ln(x + 1) − dx = (x + 1) ln(x + 1) − x + C

3dx 2
64. u = ln(3x − 2), dv = dx, du = ,v = x − ;
3x − 2 3
2 2 1
ln(3x − 2) dx = u dv = uv − v du = x− ln(3x − 2) − x− dx
3 3 x − 2/3
2 2
= x− ln(3x − 2) − x − +C
3 3

1 1
65. u = tan−1 x, dv = x dx, du = 2
dx, v = (x2 + 1)
1+x 2
1 2 1
x tan−1 x dx = u dv = uv − v du = (x + 1) tan−1 x − dx
2 2
1 2 1
= (x + 1) tan−1 x − x + C
2 2

1
66. u = , dv = x dx, du = − x(ln x)2 dx,
1 1
v = ln x
ln x
1 1
dx = 1 + dx.
x ln x x ln x
This seems to imply that 1 = 0, but recall that both sides represent a function plus an arbitrary
constant; these two arbitrary constants will take care of the 1.

67. (a) u = f (x), dv = dx, du = f (x), v = x;
b b b b
f (x) dx = xf (x) − xf (x) dx = bf (b) − af (a) − xf (x) dx
a a a a

(b) Substitute y = f (x), dy = f (x) dx, x = a when y = f (a), x = b when y = f (b),
b f (b) f (b)
xf (x) dx = x dy = f −1 (y) dy
a f (a) f (a)
y
(c) From a = f −1 (α) and b = f −1 (β) we get b
bf (b) − af (a) = βf −1 (β) − αf −1 (α); then A1
β β f (b)
f −1 (x) dx = f −1 (y) dy = f −1 (y) dy, a A2
α α f (a) x
which, by Part (b), yields a= f –1(a) b=f –1(b)

β b
f −1 (x) dx = bf (b) − af (a) − f (x) dx
α a
f −1 (β)
−1 −1
= βf (β) − αf (α) − f (x) dx
f −1 (α)
β f −1 (β)
Note from the ﬁgure that A1 = f −1 (x) dx, A2 = f (x) dx, and
α f −1 (α)

A1 + A2 = βf −1 (β) − αf −1 (α), a “picture proof”.



68. (a) Use Exercise 67(c);
1/2 sin−1 (1/2) π/6
1 1 1 1
sin−1 x dx = sin−1 −0·sin−1 0− sin x dx = sin−1 − sin x dx
0 2 2 sin−1 (0) 2 2 0

(b) Use Exercise 67(b);
e2 ln e2 2 2
ln x dx = e2 ln e2 − e ln e − f −1 (y) dy = 2e2 − e − ey dy = 2e2 − e − ex dx
e ln e 1 1

EXERCISE SET 8.3
1
1. u = cos x, − u3 du = − cos4 x + C
4

1 1
2. u = sin 3x, u5 du = sin6 3x + C
3 18

1 1 1
3. sin2 5θ = (1 − cos 10θ) dθ = θ− sin 10θ + C
2 2 20

1 1 1
4. cos2 3x dx = (1 + cos 6x)dx = x+ sin 6x + C
2 2 12

1 1
5. sin3 aθ dθ = sin aθ(1 − cos2 aθ) dθ = − cos aθ − cos3 aθ + C (a = 0)
a 3a

6. cos3 at dt = (1 − sin2 at) cos at dt

1 1
= cos at dt − sin2 at cos at dt = sin at − sin3 at + C (a = 0)
a 3a

1 1
7. u = sin ax, u du = sin2 ax + C, a = 0
a 2a

8. sin3 x cos3 x dx = sin3 x(1 − sin2 x) cos x dx

1 1
= (sin3 x − sin5 x) cos x dx = sin4 x − sin6 x + C
4 6

9. sin2 t cos3 t dt = sin2 t(1 − sin2 t) cos t dt = (sin2 t − sin4 t) cos t dt

1 1
= sin3 t − sin5 t + C
3 5

10. sin3 x cos2 x dx = (1 − cos2 x) cos2 x sin x dx

1 1
= (cos2 x − cos4 x) sin x dx = − cos3 x + cos5 x + C
3 5

1 1 1 1
11. sin2 x cos2 x dx = sin2 2x dx = (1 − cos 4x)dx = x− sin 4x + C
4 8 8 32


352 Chapter 8

1 1
12. sin2 x cos4 x dx = (1 − cos 2x)(1 + cos 2x)2 dx = (1 − cos2 2x)(1 + cos 2x)dx
8 8
1 1 1 1
= sin2 2x dx + sin2 2x cos 2x dx = (1 − cos 4x)dx + sin3 2x
8 8 16 48
1 1 1
= x− sin 4x + sin3 2x + C
16 64 48
1 1 1
13. sin 2x cos 3x dx = (sin 5x − sin x)dx = − cos 5x + cos x + C
2 10 2

1 1 1
14. sin 3θ cos 2θdθ = (sin 5θ + sin θ)dθ = − cos 5θ − cos θ + C
2 10 2

1 1
15. sin x cos(x/2)dx = [sin(3x/2) + sin(x/2)]dx = − cos(3x/2) − cos(x/2) + C
2 3

3
16. u = cos x, − u1/3 du = − cos4/3 x + C
4

π/2 π/2
17. cos3 x dx = (1 − sin2 x) cos x dx
0 0
π/2
1 2
= sin x − sin3 x =
3 0 3

π/2 π/2 π/2
1 1
18. sin2 (x/2) cos2 (x/2)dx = sin2 x dx = (1 − cos 2x)dx
0 4 0 8 0
π/2
1 1
= x− sin 2x = π/16
8 2 0

π/3 π/3 π/3
1 1
19. sin4 3x cos3 3x dx = sin4 3x(1 − sin2 3x) cos 3x dx = sin5 3x − sin7 3x =0
0 0 15 21 0

π π π
1 1 1
20. cos2 5θ dθ = (1 + cos 10θ)dθ = θ+ sin 10θ =π
−π 2 −π 2 10 −π

π/6 π/6 π/6
1 1 1
21. sin 4x cos 2x dx = (sin 2x + sin 6x)dx = − cos 2x − cos 6x
0 2 0 4 12 0
= [(−1/4)(1/2) − (1/12)(−1)] − [−1/4 − 1/12] = 7/24

2π 2π 2π
1 1 1 1
22. sin2 kx dx = (1 − cos 2kx)dx = x− sin 2kx =π− sin 4πk (k = 0)
0 2 0 2 2k 0 4k

1 1
23. tan(2x − 1) + C 24. − ln | cos 5x| + C
2 5

25. u = e−x , du = −e−x dx; − tan u du = ln | cos u| + C = ln | cos(e−x )| + C

1 1
26. ln | sin 3x| + C 27. ln | sec 4x + tan 4x| + C
3 4



√ 1 √ √
28. u = x, du = √ dx; 2 sec u du = 2 ln | sec u + tan u| + C = 2 ln sec x + tan x + C
2 x

1
29. u = tan x, u2 du = tan3 x + C
3

1 1
30. tan5 x(1 + tan2 x) sec2 x dx = (tan5 x + tan7 x) sec2 x dx = tan6 x + tan8 x + C
6 8

1 1
31. tan 4x(1 + tan2 4x) sec2 4x dx = (tan 4x + tan3 4x) sec2 4x dx = tan2 4x + tan4 4x + C
8 16

1 1
32. tan4 θ(1 + tan2 θ) sec2 θ dθ = tan5 θ + tan7 θ + C
5 7

1 1
33. sec4 x(sec2 x − 1) sec x tan x dx = (sec6 x − sec4 x) sec x tan x dx = sec7 x − sec5 x + C
7 5

1 2
34. (sec2 θ − 1)2 sec θ tan θdθ = (sec4 θ − 2 sec2 θ + 1) sec θ tan θdθ = sec5 θ − sec3 θ + sec θ + C
5 3

35. (sec2 x − 1)2 sec x dx = (sec5 x − 2 sec3 x + sec x)dx = sec5 x dx − 2 sec3 x dx + sec x dx

1 3
= sec3 x tan x + sec3 x dx − 2 sec3 x dx + ln | sec x + tan x|
4 4
1 5 1 1
= sec3 x tan x − sec x tan x + ln | sec x + tan x| + ln | sec x + tan x| + C
4 4 2 2
1 5 3
= sec3 x tan x − sec x tan x + ln | sec x + tan x| + C
4 8 8

36. [sec2 x − 1] sec3 x dx = [sec5 x − sec3 x]dx

1 3
= sec3 x tan x + sec3 x dx − sec3 x dx (equation (20))
4 4
1 1
= sec3 x tan x − sec3 x dx
4 4
1 1 1
= sec3 x tan x − sec x tan x − ln | sec x + tan x| + C (equation (20), (22))
4 8 8
1 1
37. sec2 t(sec t tan t)dt = sec3 t + C 38. sec4 x(sec x tan x)dx = sec5 x + C
3 5

1
39. sec4 x dx = (1 + tan2 x) sec2 x dx = (sec2 x + tan2 x sec2 x)dx = tan x + tan3 x + C
3

40. Using equation (20),
1 3
sec5 x dx = sec3 x tan x + sec3 x dx
4 4
1 3 3
= sec3 x tan x + sec x tan x + ln | sec x + tan x| + C
4 8 8


354 Chapter 8

41. u = 4x, use equation (19) to get
1 1 1 1 1
tan3 u du = tan2 u + ln | cos u| + C = tan2 4x + ln | cos 4x| + C
4 4 2 8 4

1
42. Use equation (19) to get tan4 x dx = tan3 x − tan x + x + C
3
√ 2 2
43. tan x(1 + tan2 x) sec2 x dx = tan3/2 x + tan7/2 x + C
3 7

2
44. sec1/2 x(sec x tan x)dx = sec3/2 x + C
3

π/8 π/8
1
45. (sec2 2x − 1)dx = tan 2x − x = 1/2 − π/8
0 2 0

π/6 π/6
1
46. sec2 2θ(sec 2θ tan 2θ)dθ = sec3 2θ = (1/6)(2)3 − (1/6)(1) = 7/6
0 6 0

47. u = x/2,
π/4
1
π/4 √
2 tan5 u du = tan4 u − tan2 u − 2 ln | cos u| = 1/2 − 1 − 2 ln(1/ 2) = −1/2 + ln 2
0 2 0

1 π/4
1
π/4 √
48. u = πx, sec u tan u du = sec u = ( 2 − 1)/π
π 0 π 0

1 1
49. (csc2 x − 1) csc2 x(csc x cot x)dx = (csc4 x − csc2 x)(csc x cot x)dx = − csc5 x + csc3 x + C
5 3

cos2 3t 1 1
50. · dt = csc 3t cot 3t dt = − csc 3t + C
sin2 3t cos 3t 3

cos x 1
51. (csc2 x − 1) cot x dx = csc x(csc x cot x)dx − dx = − csc2 x − ln | sin x| + C
sin x 2

1
52. (cot2 x + 1) csc2 x dx = − cot3 x − cot x + C
3
2π 2π
1
53. (a) sin mx cos nx dx = [sin(m + n)x + sin(m − n)x]dx
0 2 0
2π
cos(m + n)x cos(m − n)x
= − −
2(m + n) 2(m − n) 0
2π 2π
but cos(m + n)x = 0, cos(m − n)x = 0.
0 0
2π
1 2π
(b) cos mx cos nx dx = [cos(m + n)x + cos(m − n)x]dx;
0 2 0
since m = n, evaluate sin at integer multiples of 2π to get 0.
2π
1 2π
(c) sin mx sin nx dx = [cos(m − n)x − cos(m + n)x] dx;
0 2 0
since m = n, evaluate sin at integer multiples of 2π to get 0.

Chapter 08

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Chapter 08