Chapter 08 by Faheem Ajmal

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CHAPTER 8

Principles of Integral Valuation EXERCISE SET 8.1 1. u = 4 − 2x, du = −2dx, − 2. u = 4 + 2x, du = 2dx,

3. u = x , du = 2xdx,

3 2 1 2

1 2

√

1 1 u3 du = − u4 + C = − (4 − 2x)4 + C 8 8

u du = u3/2 + C = (4 + 2x)3/2 + C

sec2 u du =

1 1 tan u + C = tan(x2 ) + C 2 2

4. u = x , du = 2xdx,

tan u du = −2 ln | cos u | + C = −2 ln | cos(x2 )| + C

1 − 3

5. u = 2 + cos 3x, du = −3 sin 3xdx,

6. u =

2 2 1 x, du = dx, 3 3 6

du 1 1 = − ln |u| + C = − ln(2 + cos 3x) + C u 3 3

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

sinh u du = cosh u + C = cosh ex + C

7. u = ex , du = ex dx,

8. u = ln x, du =

1 dx, x

sec u tan u du = sec u + C = sec(ln x) + C eu du = eu + C = etan x + C

9. u = tan x, du = sec xdx,

du 1 1 = sin−1 u + C = sin−1 (x2 ) + C 2 2 2 1−u 1 1 1 u5 du = − u6 + C = − cos6 5x + C 11. u = cos 5x, du = −5 sin 5xdx, − 5 30 30 1 + 1 + sin2 x 1 + √1 + u 2 du √ 12. u = sin x, du = cos x dx, = − ln + C = − ln +C u sin x u u2 + 1 2

10. u = x , du = 2xdx,

1 2

√

14. u = tan−1 x, du = √

du = ln u + u2 + 4 + C = ln ex + e2x + 4 + C 4 + u2 −1 dx, eu du = eu + C = etan x + C

√

13. u = ex , du = ex dx,

1 1 + x2

1 x − 1, du = √ dx, 2 x−1

16. u = x2 + 2x, du = (2x + 2)dx,

1 2

15. u =

17. u =

√

1 x, du = √ dx, 2 x

√ x−1

eu du = 2eu + C = 2e cot u du =

1 1 ln | sin u| + C = ln sin |x2 + 2x| + C 2 2

√ 2 cosh u du = 2 sinh u + C = 2 sinh x + C

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Chapter 8

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

dx , 18. u = ln x, du = x

20. u = sin θ, du = cos θdθ,

21. u =

2 2 , du = − 2 dx, x x

22.

√

sec u tan u du = sec u + C = sec(sin θ) + C 1 2

−

csch2 u du =

1 2 1 coth u + C = coth + C 2 2 x

dx = ln x + x2 − 4 + C x2 − 4

23. u = e−x , du = −e−x dx,

24. u = ln x, du =

1 dx, x

1 2 + e−x du 1 2 + u + C = − +C = − ln ln 4 − u2 4 2 − u 4 2 − e−x

cos u du = sin u + C = sin(ln x) + C

25. u = ex , du = ex dx,

−

ex dx du √ = = sin−1 u + C = sin−1 ex + C 1 − e2x 1 − u2 dx, − 2 sinh u du = −2 cosh u + C = −2 cosh(x−1/2 ) + C

√

1 2x3/2 du 1 1 1 1 = sin u du = − cos u + C = − cos(x2 ) + C 27. u = x2 , du = 2xdx, 2 csc u 2 2 2 2du √ = sin−1 u + C = sin−1 (ex /2) + C 28. 2u = ex , 2du = ex dx, 4 − 4u2 26. u = x−1/2 , du = −

29. 4−x = e−x ln 4 , u = −x2 ln 4, du = −2x ln 4 dx = −x ln 16 dx, 1 u 1 −x2 ln 4 1 −x2 1 eu du = − e +C =− e 4 +C =− +C − ln 16 ln 16 ln 16 ln 16 1 πx ln 2 1 e 2πx + C 2πx dx = +C = 30. 2πx = eπx ln 2 , π ln 2 π ln 2 1 1 31. (a) u = sin x, du = cos x dx, u du = u2 + 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 2

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. 32. (a) sech 2x = =

1 1 = cosh 2x cosh2 x + sinh2 x sech2 x 1 + tanh2 x

(now multiply top and bottom by sech2 x)

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Exercise Set 8.2

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(b)

sech2x dx =

sech2 x dx = tan−1 (tanh x) + C, 1 + tanh2 x

or, replacing 2x with x,

sechx dx = tan−1 (tanh(x/2)) + C

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

sec2 x 1 1 = = cos2 x tan x cos x sin x tan x

33. (a)

1 1 1 sec2 x (b) csc 2x = = = , so sin 2x 2 sin x cos x 2 tan x

csc 2x dx =

1 ln tan x + C 2

1 1 = = csc(π/2 − x), so (c) sec x = 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 = − 12 e−2x ; 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 1 3x 3x e3x dx = xe3x − e3x + C 2. u = x, dv = e dx, du = dx, v = e ; xe dx = xe − 3 3 3 3 9 3. u = x2 , dv = ex dx, du = 2x dx, v = ex ; x2 ex dx = x2 ex − 2 xex dx. xex dx use u = x, dv = ex dx, du = dx, v = ex to get

For

xex dx = xex − ex + C1 so

x2 ex dx = x2 ex − 2xex + 2ex + C

1 4. u = x2 , dv = e−2x dx, du = 2x dx, v = − e−2x ; 2 For xe−2x dx use u = x, dv = e−2x dx to get

−2x

xe so

1 1 dx = − xe−2x + 2 2

1 x2 e−2x dx = − x2 e−2x + 2

1 1 e−2x dx = − xe−2x − e−2x + C 2 4

1 1 1 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 cos 3x dx = − x cos 3x + sin 3x + C x sin 3x dx = − x cos 3x + 3 3 3 9

xe−2x dx

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Chapter 8

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

9. u = ln x, dv = x dx, du =

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

1 1 dx, v = x2 ; x 2

x ln x dx =

1 2 1 x ln x − 2 2

x dx =

1 2 1 x ln x − x2 + C 2 4

√ 2 1 10. u = ln x, dv = x dx, du = dx, v = x3/2 ; x 3 √ 2 3/2 2 2 4 x1/2 dx = x3/2 ln x − x3/2 + C x ln x dx = x ln x − 3 3 3 9 ln x 2 2 2 11. u = (ln x) , dv = dx, du = 2 dx, v = x; (ln x) dx = x(ln x) − 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 12. u = ln x, dv = √ dx, du = dx, v = 2 x; x x

√ ln x √ dx = 2 x ln x−2 x

√ √ 1 √ dx = 2 x ln x−4 x+C x

3 3x dx, v = x; ln(3x − 2)dx = x ln(3x − 2) − dx 3x − 2 3x − 2 2 2 3x dx = 1+ dx = x + ln(3x − 2) + C1 so but 3x − 2 3x − 2 3 2 ln(3x − 2)dx = x ln(3x − 2) − x − ln(3x − 2) + C 3

13. u = ln(3x − 2), dv = dx, du =

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

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√ 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 dx, v = x; 16. u = cos−1 (2x), dv = dx, du = − √ 1 − 4x2 2x 1 √ cos−1 (2x)dx = x cos−1 (2x) + dx = x cos−1 (2x) − 1 − 4x2 + C 2 1 − 4x2 3 dx, v = x; 17. u = tan−1 (3x), dv = dx, du = 1 + 9x2 3x 1 tan−1 (3x)dx = x tan−1 (3x) − dx = x tan−1 (3x) − ln(1 + 9x2 ) + C 1 + 9x2 6 x2 1 1 2 1 2 1 −1 −1 18. u = tan−1 x, dv = x dx, du = x x dx, v = ; x tan x dx = tan x − dx 2 1+x 2 2 2 1 + x2 x2 1 but dx = x − tan−1 x + C1 so dx = 1 − 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 x x x e sin x dx = −e cos x + e sin x − ex sin x dx, 1 x x 2 e sin x dx = e (sin x − cos x) + C1 , ex sin x dx = ex (sin x − cos x) + C 2 1 20. u = e3x , dv = cos 2x dx, du = 3e3x dx, v = sin 2x; 2 1 3x 3 3x e3x sin 2x dx. Use u = e3x , dv = sin 2x dx to get e cos 2x dx = e sin 2x − 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 a eax cos bx dx. Use u = eax , dv = cos bx dx to get eax sin bx dx = − eax cos bx + b b 1 ax a ax e cos bx dx = e sin bx − eax sin bx dx so b b 1 ax a ax a2 ax e sin bx dx = − e cos bx + 2 e 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

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

sin(ln x)dx,

1 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 =

cos(ln x)dx,

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

1 1 dx, du = (x + 1)ex dx, v = − ; (x + 1)2 x+1 xex xex ex xex x x + e + e +C dx = − dx = − + C = (x + 1)2 x+1 x+1 x+1

28. u = xex , dv =

29. u = x, dv = e2x dx, du = dx, v =

1 xe dx = xe2x 2 2x

1 − 2 0

1 2x e ; 2

1 e dx = e − e2x 4 2x

1 = e4 − (e4 − 1) = (3e4 + 1)/4 4

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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 5 5 0 0 0 1 1 1 1 1 = − e−5 − e−5x = − e−5 − (e−5 − 1) = (1 − 6e−5 )/25 25 5 25 5 0 1 1 31. u = ln x, dv = x2 dx, du = dx, v = x3 ; x 3 e e e e 1 1 3 1 3 1 1 1 3 2 2 x dx = e − x = e3 − (e3 − 1) = (2e3 + 1)/9 x ln x dx = x ln x − 3 3 1 3 9 3 9 1 1 1 1 1 1 dx, du = dx, v = − ; 2 x x x e e ln x 1 1 dx = − ln x √ + √ 2 dx x2 x e x e e √ √ 1 1 1 3 e−4 1 1 1 1 = − + √ ln e − =− + √ − +√ = e x √e e 2 e e 2e e e

32. u = ln x, dv =

√ e

1 dx, v = x; x+2 1 1 1 1 2 x dx 1− ln(x + 2)dx = x ln(x + 2) − dx = ln 3 + ln 1 − x+2 −1 −1 x + 2 −1 −1 1 = ln 3 − [x − 2 ln(x + 2)] = ln 3 − (1 − 2 ln 3) + (−1 − 2 ln 1) = 3 ln 3 − 2

33. u = ln(x + 2), dv = dx, du =

−1

1 dx, v = x; 1 − x2 √ √ √3/2 √3/2 √3/2 √3/2 x 3 3 −1 −1 −1 √ sin + 1 − x2 sin x dx = x sin x − dx = 2 2 1 − x2 0 0 0 0 √ √ 1 π 3 1 3 π + −1= − = 2 3 2 6 2

34. u = sin−1 x, dv = dx, du = √

√

1 √ dθ, v = θ; 2θ θ − 1 4 4 √ √ 4 1 4 √ √ 1 √ sec−1 θdθ = θ sec−1 θ − dθ = 4 sec−1 2 − 2 sec−1 2 − θ − 1 2 2 θ−1 2 2 2 π √ π 5π √ −2 − 3+1= − 3+1 =4 3 4 6

35. u = sec−1

θ, dv = dθ, du =

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

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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 2 2 0 4 0 0 0

38.

(x + x cos x)dx = 0

1 2 x 2

x cos x dx = 0

π2 + 2

x cos x dx; 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

√

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

39. u = tan−1

x, dv =

2x dx, v = x; x2 + 1 2 2 2 2 1 2x2 dx = 2 ln 5 − 2 dx 1 − ln(x2 + 1)dx = x ln(x2 + 1) − 2 x2 + 1 0 0 x +1 0 0 2 = 2 ln 5 − 2(x − tan−1 x) = 2 ln 5 − 4 + 2 tan−1 2

40. u = ln(x2 + 1), dv = dx, du =

√

x, t2 = x, dx = 2t dt √ (a) e x dx = 2 tet dt; u = t, dv = et dt, du = dt, v = et ,

41. t =

√ x

e (b)

cos cos

dx = 2tet − 2

√ √

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

x dx = 2

t cos t dt; u = t, dv = cos tdt, du = dt, v = sin t,

x dx = 2t sin t − 2

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

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Exercise Set 8.2

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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 Diﬀerentiation

Repeated Antidiﬀerentiation

ax2 + bx + c

q(x) +

2ax + b −

q1 (x) q2 (x)

2a +

q3 (x)

0 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 Diﬀerentiation

Repeated Antidiﬀerentiation

3x2 − x + 2

e−x +

6x − 1

−

6 + 0

44.

−e−x e−x −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 Repeated Diﬀerentiation

Repeated Antidiﬀerentiation

x2 + x + 1

sin x +

2x + 1 − 2

− cos x − 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

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Chapter 8

45.

Repeated Diﬀerentiation

Repeated Antidiﬀerentiation

4x4

sin 2x +

16x3 − 2

48x

1 − cos 2x 2 1 − sin 2x 4 1 cos 2x 8

96x −

1 sin 2x 16

96 + −

1 cos 2x 32

4x4 sin 2x dx = (−2x4 + 6x2 − 3) cos 2x + −(4x3 + 6x) sin 2x + C 46.

Repeated Diﬀerentiation x3

Repeated Antidiﬀerentiation √ 2x + 1

+ 2

− 6x + 6 − 0

1 (2x + 1)3/2 3 1 (2x + 1)5/2 15 1 (2x + 1)7/2 105 1 (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 2 sin x cos x dx = − cos x + C, sin x cos x dx = − cos2 x + C 2 1 1 (b) u = sin x, du = cos x dx, sin x cos x dx = u du = u2 + C = sin2 x + C 2 2 48. (a) u = x2 , dv = √ 0

√

x3 x2 + 1

x x2

, du = 2x dx, v =

+1 1 dx = x2 x2 + 1 − 0

x2 + 1,

1 √ 2 1√ 2 2x x2 + 1 dx = 2 − (x2 + 1)3/2 = − 2+ 3 3 3 0

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(b) u =

+ 1, du = √

x2 + 1

√ 2

(u − 1) du = 2

dx, 1

1 3 u −u 3

√2 1

√ 1√ 2√ 1 2 = 2− 2− +1=− 2+ . 3 3 3 3 e e 49. (a) A = ln x dx = (x ln x − x) = 1 1

(b) V = π

e (ln x)2 dx = π (x(ln x)2 − 2x ln x + 2x) = π(e − 2) 1

π/2

50. A = 0

1 (x − x sin x)dx = x2 2 π

51. V = 2π

π/2

− 0

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

π x sin x dx = 2π(−x cos x + sin x) = 2π 2

π/2

0 π/2

52. V = 2π

π/2 x cos x dx = 2π(cos x + x sin x) = π(π − 2)

t3 sin tdt;

53. distance = 0

Repeated Diﬀerentiation

Repeated Antidiﬀerentiation

sin t +

3t2 − 6t + 6

− cos t − sin t cos t

−

sin t

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

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

54. u = 2t, dv = sin(kωt)dt, du = 2dt, v = −

55. (a)

1 3 sin4 x dx = − sin3 x cos x + 4 4

sin2 x dx

1 1 1 3 3 − sin x cos x + x + C = − sin x cos x + 4 4 2 2

1 3 3 = − sin3 x cos x − sin x cos x + x + C 4 8 8

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Chapter 8

π/2

π/2 1 4 π/2 3 4 sin x dx = − sin x cos x + sin x dx 5 5 0 0

π/2 1 4 2 π/2 2 − sin x cos x = + sin x dx 5 3 3 0 0 π/2 8 8 = − cos x = 15 15 0 5

(b) 0

1 4 cos x dx = cos4 x sin x + 5 5

56. (a)

1 4 1 2 4 2 cos x dx = cos x sin x + cos x sin x + sin x + C 5 5 3 3 3

4 8 1 cos2 x sin x + sin x + C cos4 x sin x + 15 15 5 1 5 6 5 (b) cos x dx = cos x sin x + cos4 x dx 6 6 1 5 1 3 5 3 2 cos x sin x + cos x dx = cos x sin x + 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 5 5 1 5 = 5π/32 cos5 x sin x + cos3 x sin x + cos x sin x + x 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; n n−1 sin x dx = − sin 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,

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

sinn x dx = −

1 n−1 sinn−1 x cos x + n n

sinn−2 x dx,

sinn−2 x dx

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

= sec

n−2

x tan x − (n − 2)

sec x dx + (n − 2) n

(n − 1)

secn x dx = secn−2 x tan x + (n − 2)

secn x dx =

1 n−2 secn−2 x tan x + n−1 n−1

secn−2 x dx,

secn−2 x dx

secn−2 x dx,

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Exercise Set 8.2

349

tan x dx =

(b)

n−2

tan

x (sec x − 1) dx = 2

1 tann−1 x − n−1

tan

tan4 x dx =

1 tan3 x − 3

x sec x dx − 2

tann−2 x dx

59. (a)

n−1

tan2 x dx =

xn ex dx = xn ex − n

1 tan3 x − tan x + 3

dx =

xn−1 ex dx 1 tan3 x − tan x + x + C 3

1 2 1 2 sec2 x dx = sec2 x tan x + tan x + C sec2 x tan x + 3 3 3 3 (c) x3 ex dx = x3 ex − 3 x2 ex dx = x3 ex − 3 x2 ex − 2 xex dx sec4 x dx =

(b)

x x = x e − 3x e + 6 xe − e dx = x3 ex − 3x2 ex + 6xex − 6ex + C 3 x

2 x

60. (a) u = 3x, 1 1 1 2 u 2 u2 eu du = u2 eu − 2 ueu du = ueu − eu du x2 e3x dx = u e − 27 27 27 27 = √ (b) u = − x,

√ x

xe−

2 2 1 2 2 1 2 u u e − ueu + eu + C = x2 e3x − xe3x + e3x + C 27 27 27 3 9 27

−1

dx = 2

u3 eu du,

u e du = u e − 3 3 u

3 u

2 u u u e du = u e − 3 u e − 2 ue du 2 u

3 u

= u3 eu − 3u2 eu + 6 ueu − eu du = u3 eu − 3u2 eu + 6ueu − 6eu + C, 2

−1

−1 u3 eu du = 2(u3 − 3u2 + 6u − 6)eu

= 12 − 32e−1

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 = 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

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350

Chapter 8

dx , v = x + 1; 63. u = ln(x + 1), dv = dx, du = 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 ,v = x − ; 3x − 2 3 2 1 2 ln(3x − 2) − x− dx ln(3x − 2) dx = u dv = uv − v du = x − 3 3 x − 2/3 2 2 = x− ln(3x − 2) − x − +C 3 3

64. u = ln(3x − 2), dv = dx, du =

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

1 2 1 (x + 1) tan−1 x − x + C 2 2

1 , dv = x1 dx, du = − x(ln1x)2 dx, 66. u = ln x 1 1 dx = 1 + dx. x ln x x ln x

v = 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 f (x) dx = xf (x) − xf (x) dx = bf (b) − af (a) − a

xf (x) dx

(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)

bf (b) − af (a) = βf −1 (β) − αf −1 (α); then β β f (b) −1 −1 f (x) dx = f (y) dy = f −1 (y) dy, α

f (a)

which, by Part (b), yields β f −1 (x) dx = bf (b) − af (a) − α

−1

(β) − αf

−1

f (x) dx

(α) −

Note from the ﬁgure that A1 =

f –1(a)

= βf

f −1 (β)

f (x) dx f −1 (α)

f −1 (x) dx, A2 =

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

f −1 (β)

f (x) dx, and f −1 (α)

b=f

–1(b)

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Exercise Set 8.3

351

68. (a) Use Exercise 67(c); π/6 1/2 sin−1 (1/2) 1 1 1 1 −0·sin−1 0− − sin−1 x dx = sin−1 sin x dx = sin−1 sin x dx 2 2 2 2 sin−1 (0) 0 0 (b) Use Exercise 67(b); e2 ln x dx = e2 ln e2 − e ln e − e

ln e2

f −1 (y) dy = 2e2 − e −

ey dy = 2e2 − e − 1

ln e

1 u3 du = − cos4 x + C 4

1. u = cos x, − 2.

1 1 u5 du = u = sin 3x, sin6 3x + C 3 18 1 1 1 2 sin 5θ = (1 − cos 10θ) dθ = θ − sin 10θ + C 2 2 20 1 1 1 (1 + cos 6x)dx = x + cos2 3x dx = sin 6x + C 2 2 12 1 1 cos3 aθ + C sin3 aθ dθ = sin aθ(1 − cos2 aθ) dθ = − cos aθ − a 3a cos3 at dt = (1 − sin2 at) cos at dt

cos at dt −

= 1 a

7. u = sin ax,

u du =

1 1 sin at − sin3 at + C (a = 0) a 3a

1 sin2 ax + C, a = 0 2a

sin3 x(1 − sin2 x) cos x dx

sin x cos x dx =

(sin3 x − sin5 x) cos x dx =

(a = 0)

sin2 at cos at dt =

sin t cos t dt =

1 1 sin4 x − sin6 x + C 4 6

sin t(1 − sin t) cos t dt = 2

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

ex dx 1

EXERCISE SET 8.3

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352

Chapter 8

12.

1 (1 − cos 2x)(1 + cos 2x) dx = (1 − cos2 2x)(1 + cos 2x)dx 8 1 1 1 1 = sin2 2x dx + sin2 2x cos 2x dx = (1 − cos 4x)dx + sin3 2x 8 8 16 48

1 sin x cos x dx = 8 2

1 1 1 x− sin 4x + sin3 2x + C 16 64 48 1 1 1 (sin 5x − sin x)dx = − cos 5x + cos x + C sin 2x cos 3x dx = 2 10 2 1 1 1 (sin 5θ + sin θ)dθ = − cos 5θ − cos θ + C sin 3θ cos 2θdθ = 2 10 2 1 1 sin x cos(x/2)dx = [sin(3x/2) + sin(x/2)]dx = − cos(3x/2) − cos(x/2) + C 2 3 3 u = cos x, − u1/3 du = − cos4/3 x + C 4 =

13.

14.

15.

16.

π/2

(1 − sin2 x) cos x dx

cos3 x dx =

17. 0

π/2 1 2 3 = sin x − sin x = 3 3 0 π/2

1 4

sin2 (x/2) cos2 (x/2)dx =

18. 0

1 = 8

π/3 4

19.

cos2 5θ dθ = −π

1 2

−π

1 2

(1 − cos 2x)dx 0

(1 + cos 10θ)dθ =

sin 4x cos 2x dx =

π/2

π/2 1 x − sin 2x = π/16 2 0

π/6

21.

π/3 1 1 5 7 sin 3x − sin 3x sin 3x(1 − sin 3x) cos 3x dx = =0 15 21 0

0 π

1 8

sin2 x dx =

sin 3x cos 3x dx =

20.

π/2

π/3

π/6

1 2

θ+

1 sin 10θ 10

π =π −π

π/6 1 1 cos 6x (sin 2x + sin 6x)dx = − cos 2x − 4 12 0

= [(−1/4)(1/2) − (1/12)(−1)] − [−1/4 − 1/12] = 7/24

1 sin kx dx = 2

22. 0

23.

2π

1 (1 − cos 2kx)dx = 2

1 tan(2x − 1) + C 2

25. u = e−x , du = −e−x dx; − 26.

1 ln | sin 3x| + C 3

2π 1 1 x− sin 2kx sin 4πk (k = 0) =π− 2k 4k 0

1 24. − ln | cos 5x| + C 5

tan u du = ln | cos u| + C = ln | cos(e−x )| + C 27.

1 ln | sec 4x + tan 4x| + C 4

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Exercise Set 8.3

28. u =

√

353

1 x, du = √ dx; 2 x

√ √ 2 sec u du = 2 ln | sec u + tan u| + C = 2 ln sec x + tan x + C

1 tan3 x + C 3 1 1 5 2 2 30. tan x(1 + tan x) sec 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 u2 du =

29. u = tan x,

1 3 = sec3 x tan x + 4 4 =

sec x dx − 2

sec3 x dx + ln | sec x + tan x|

1 1 5 1 sec3 x tan x − sec x tan x + ln | sec x + tan x| + ln | sec x + tan x| + C 4 4 2 2

3 1 5 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 3 3 = sec x dx − sec3 x dx sec x tan x + 4 4 1 1 3 sec3 x dx = sec x tan x − 4 4 =

(equation (20))

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 2 3 37. sec t(sec t tan t)dt = sec t + C 38. sec4 x(sec x tan x)dx = sec5 x + C 3 5 =

39.

sec4 x dx =

(1 + tan2 x) sec2 x dx =

(sec2 x + tan2 x sec2 x)dx = tan x +

40. Using equation (20), 1 3 5 3 sec x dx = sec x tan x + sec3 x dx 4 4 =

3 3 1 sec3 x tan x + sec x tan x + ln | sec x + tan x| + C 4 8 8

1 tan3 x + C 3

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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 tan4 x dx =

42. Use equation (19) to get

43.

√

tan x(1 + tan2 x) sec2 x dx =

sec1/2 x(sec x tan x)dx =

44.

π/8

(sec2 2x − 1)dx =

45. 0

2 2 tan3/2 x + tan7/2 x + C 3 7

2 sec3/2 x + C 3

π/8 1 = 1/2 − π/8 tan 2x − x 2 0

π/6

sec2 2θ(sec 2θ tan 2θ)dθ =

46.

1 tan3 x − tan x + x + C 3

1 sec3 2θ 6

π/6 = (1/6)(2)3 − (1/6)(1) = 7/6 0

47. u = x/2, π/4 π/4 √ 1 tan4 u − tan2 u − 2 ln | cos u| tan5 u du = = 1/2 − 1 − 2 ln(1/ 2) = −1/2 + ln 2 2 2 0 0 48. u = πx,

1 π

π/4

sec u tan u du = 0

1 sec u π

(csc2 x − 1) csc2 x(csc x cot x)dx =

49.

cos2 3t 1 dt = · sin2 3t cos 3t

π/4

√ = ( 2 − 1)/π

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

1 csc 3t cot 3t dt = − csc 3t + C 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 50.

1 2π 53. (a) sin mx cos nx dx = [sin(m + n)x + sin(m − n)x]dx 2 0 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.

2π

0 2π

1 [cos(m + n)x + cos(m − n)x]dx; 2 0 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; 2 0 0 since m = n, evaluate sin at integer multiples of 2π to get 0. (b)

cos mx cos nx dx =

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Exercise Set 8.3

355

2π

54. (a) 0

1 sin mx cos mx dx = 2

2π 1 sin 2mx dx = − cos 2mx = 0 4m

2π

2π 1 x+ sin 2mx = π (b) 2m 0 0 0 2π 2π 2π 1 1 1 x− (c) sin2 mx dx = (1 − cos 2mx) dx = sin 2mx = π 2 2 2m 0 0

2π

1 cos mx dx = 2

2π

1 (1 + cos 2mx) dx = 2

55. y = tan x, 1 + (y )2 = 1 + tan2 x = sec2 x, π/4 π/4 √ π/4 √ 2 L= sec x dx = sec x dx = ln | sec x + tan x| = ln( 2 + 1) 0

π/4

π/4 1 (2 − sec2 x)dx = π(2x − tan x) = π(π − 2) 2 0

π/4

(1 − tan2 x)dx = π

56. V = π 0

π/4

(cos2 x − sin2 x)dx = π

57. V = π 0

0 π

sin2 x dx =

58. V = π 0

π 2

π/4 1 cos 2x dx = π sin 2x = π/2 2 0

(1 − cos 2x)dx = 0

L 59. With 0 < α < β, D = Dβ −Dα = 2π

π 2

π 1 x − sin 2x = π 2 /2 2 0

L ln | sec x + tan x| sec x dx = 2π

β α

L sec β + tan β ln = 2π sec α + tan α

100 ln(sec 25◦ + tan 25◦ ) = 7.18 cm 2π 100 sec 50◦ + tan 50◦ (b) D = ln = 7.34 cm 2π sec 30◦ + tan 30◦ 61. (a) csc x dx = sec(π/2 − x)dx = − ln | sec(π/2 − x) + tan(π/2 − x)| + C

60. (a) D =

√

2 [sin x cos(π/4) + cos x sin(π/4)] =

dx 1 =√ sin x + cos x 2

√

2 sin(x + π/4),

1 csc(x + π/4)dx = − √ ln | csc(x + π/4) + cot(x + π/4)| + C 2 √ 1 2 + cos x − sin x = − √ ln +C 2 sin x + cos x

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356

Chapter 8

a b a2 + b2 √ sin x + √ cos x = a2 + b2 (sin x cos θ + cos x sin θ) a2 + b2 a2 + b2 √ √ √ where cos θ = a/ a2 + b2 and sin θ = b/ a2 + b2 so a sin x + b cos x = a2 + b2 sin(x + θ) 1 1 dx =√ csc(x + θ)dx = − √ ln | csc(x + θ) + cot(x + θ)| + C and a sin x + b cos x a2 + b2 a2 + b2 √ a2 + b2 + a cos x − b sin x 1 = −√ ln +C 2 2 a sin x + b cos x a +b

63. a sin x + b cos x =

π/2

sinn x dx = −

64. (a) 0

π/2 1 n − 1 π/2 n−2 n − 1 π/2 n−2 + sin x dx = sin x dx sinn−1 x cos x n n n 0 0 0

(b) By repeated application of the formula in Part (a) π/2 π/2 n−3 n−1 sinn x dx = sinn−4 x dx n n − 2 0 0  π/2  n−3 n−5 1 n−1   ··· dx, n even   n n−2 n−4 2 0 = π/2  n − 1 n − 3 n − 5  2   · · · sin x dx, n odd  n−2 n−4 3 n 0  1 · 3 · 5 · · · (n − 1) π   · , n even  2 · 4 · 6···n 2 =    2 · 4 · 6 · · · (n − 1) , n odd 3 · 5 · 7···n

π/2

sin3 x dx =

65. (a) 0

π/2

sin5 x dx =

2 3 2·4 = 8/15 3·5

sin4 x dx =

1·3 π · = 3π/16 2·4 2

sin6 x dx =

1·3·5 π · = 5π/32 2·4·6 2

π/2

(b) 0

(d)

π/2

66. Similar to proof in Exercise 64.

EXERCISE SET 8.4 1. x = 2 sin θ, dx = 2 cos θ dθ, 4 cos2 θ dθ = 2 (1 + cos 2θ)dθ = 2θ + sin 2θ + C 1 = 2θ + 2 sin θ cos θ + C = 2 sin−1 (x/2) + x 4 − x2 + C 2 1 1 sin θ, dx = cos θ dθ, 2 2 1 1 1 1 cos2 θ dθ = (1 + cos 2θ)dθ = θ + sin 2θ + C 2 4 4 8

2. x =

1 1 1 1 θ + sin θ cos θ + C = sin−1 2x + x 1 − 4x2 + C 4 4 4 2

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Exercise Set 8.4

357

3. x = 4 sin θ, dx = 4 cos θ dθ, 16 sin2 θ dθ = 8 (1 − cos 2θ)dθ = 8θ − 4 sin 2θ + C = 8θ − 8 sin θ cos θ + C 1 = 8 sin−1 (x/4) − x 16 − x2 + C 2 4. x = 3 sin θ, dx = 3 cos θ dθ, √ 1 9 − x2 1 1 1 2 csc θ dθ = − dθ = cot θ + C = − +C 2 9 9 9 9x sin θ 5. x = 2 tan θ, dx = 2 sec2 θ dθ, 1 1 1 1 1 1 2 cos θ dθ = (1 + cos 2θ)dθ = dθ = θ+ sin 2θ + C 2 8 sec θ 8 16 16 32 =

x 1 1 x 1 +C θ+ sin θ cos θ + C = tan−1 + 16 16 16 2 8(4 + x2 )

√ √ 6. x = 5 tan θ, dx = 5 sec2 θ dθ, 1 1 sec θ tan θ − ln | sec θ + tan θ| + C1 5 tan2 θ sec θ dθ = 5 (sec3 θ − sec θ)dθ = 5 2 2 √ 5 + x2 + x 5 1 5 1 √ = x 5 + x2 − ln + C1 = x 5 + x2 − ln( 5 + x2 + x) + C 2 2 2 2 5 7. x = 3 sec θ, dx = 3 sec θ tan θ dθ, x 3 tan2 θ dθ = 3 (sec2 θ − 1)dθ = 3 tan θ − 3θ + C = x2 − 9 − 3 sec−1 + C 3 8. x = 4 sec θ, dx = 4 sec θ tan θ dθ, √ 1 1 1 x2 − 16 1 cos θ dθ = dθ = sin θ + C = +C 16 sec θ 16 16 16x 9. x = sin θ, dx = cos θ dθ, 1 − cos2 θ sin θ dθ 3 sin3 θ dθ = 3 = 3 − cos θ + cos3 θ + C = −3 1 − x2 + (1 − x2 )3/2 + C √ √ 10. x = 5 sin θ, dx = 5 cos θ dθ, √ √ 1 1 5 1 25 5 sin3 θ cos2 θ dθ = 25 5 − cos3 θ + cos5 θ + C = − (5 − x2 )3/2 + (5 − x2 )5/2 + C 3 5 3 5 11. x =

2 3 2 sec θ, dx = sec θ tan θ dθ, 3 3 4

1 3 dθ = sec θ 4

cos θ dθ =

3 1 2 sin θ + C = 9x − 4 + C 4 4x

12. t = tan θ, dt = sec2 θ dθ, sec3 θ tan2 θ + 1 dθ = sec θ dθ = (sec θ tan θ + csc θ)dθ tan θ tan θ √ 1 + t2 − 1 2 = sec θ + ln | csc θ − cot θ| + C = 1 + t + ln +C |t| 1 13. x = sin θ, dx = cos θ dθ, dθ = sec2 θ dθ = tan θ + C = x/ 1 − x2 + C 2 cos θ

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358

Chapter 8

√ sec θ 1 1 x2 + 25 csc θ cot θ dθ = − csc θ+C = − +C dθ = 2 25 25 25x tan θ 1 1 2 15. x = 3 sec θ, dx = 3 sec θ tan θ dθ, sec θ dθ = ln | sec θ + tan θ| + C = ln x + x − 9 + C 3 3

14. x = 5 tan θ, dx = 5 sec2 θ dθ,

1 25

16. 1 + 2x2 + x4 = (1 + x2 )2 , x = tan θ, dx = sec2 θ dθ, 1 1 1 1 2 (1 + cos 2θ)dθ = θ + sin 2θ + C dθ = cos θ dθ = 2 2 4 sec2 θ =

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

3 3 sec θ, dx = sec θ tan θ dθ, 2 2 3 sec θ tan θ dθ cos θ 1 x 1 1 1 + C = − csc θ + C = − √ = +C dθ = − 3 2 2 18 18 sin θ 18 27 tan θ sin θ 9 4x2 − 9

17. x =

18. x = 5 sec θ, dx = 5 sec θ tan θ dθ, = 375 sec4 θ dθ = 125 sec2 θ tan θ + 250 sec2 θ dθ = 125 sec2 θ tan θ + 250 tan θ + C = x2 x2 − 25 + 50 x2 − 25 + C 19. ex = sin θ, ex dx = cos θ dθ, 1 1 1 1 1 cos2 θ dθ = (1 + cos 2θ)dθ = θ + sin 2θ + C = sin−1 (ex ) + ex 1 − e2x + C 2 2 4 2 2 20. u = sin θ,

1 √ du = sin−1 2 − u2

sin θ √ 2

21. x = sin θ, dx = cos θ dθ, π/2 1 1 1 5 sin3 θ cos2 θ dθ = 5 − cos3 θ + cos5 θ = 5(1/3 − 1/5) = 2/3 3 5 0 0 22. x = sin θ, dx = cos θ dθ, π/6 π/6 1 1 3 sec θ, dθ = sec θ tan θ + ln | sec θ + tan θ| 2 2 0 0 1 2 1 1 1 1 1 2 √ √ + ln( √ + √ = = + ln 3 2 3 3 2 3 4 3 3

π/3

23. x = sec θ, dx = sec θ tan θ dθ, π/4

24. x =

√

2 sec θ, dx =

√

1 dθ = sec θ

2 sec θ tan θ dθ, 2 0

π/4

π/3

cos θ dθ = sin θ π/4

√ √ = ( 3 − 2)/2

π/4

π/4 tan2 θ dθ = 2 tan θ − 2θ = 2 − π/2 0

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Exercise Set 8.4

359

√ √ 25. x = 3 tan θ, dx = 3 sec2 θ dθ, √3/2 1 π/3 cos3 θ 1 − u2 1 π/3 sec θ 1 π/3 1 − sin2 θ 1 dθ = du (u = sin θ) dθ = cos θ dθ = 9 π/6 tan4 θ 9 π/6 sin4 θ 9 π/6 9 1/2 u4 sin4 θ √ √3/2 √3/2 1 1 1 10 3 + 18 1 −4 −2 − 3+ (u − u )du = = = 9 1/2 9 3u u 1/2 243 √ √ 26. x = 3 tan θ, dx = 3 sec2 θ dθ, √ π/3 √ π/3 √ π/3 3 tan3 θ 3 3 3 1 − cos2 θ sin θ dθ sin θ dθ = dθ = 3 3 0 sec θ 3 0 3 0 √ √ π/3 √ 1 1 1 3 3 1 3 − cos θ + cos θ − + − −1 + = 5 3/72 = = 3 3 3 2 24 3 0 27. u = x2 + 4, du = 2x dx, 1 1 1 1 du = ln |u| + C = ln(x2 + 4) + C; or x = 2 tan θ, dx = 2 sec2 θ dθ, 2 u 2 2 √ x2 + 4 tan θ dθ = ln | sec θ| + C1 = ln + C1 = ln(x2 + 4)1/2 − ln 2 + C1 2 1 = ln(x2 + 4) + C with C = C1 − ln 2 2 x 2 tan2 θ dθ = 2 tan θ − 2θ + C = x − 2 tan−1 + C; alternatively 28. x = 2 tan θ, dx = 2 sec2 θ dθ, 2 2 x dx x dx = dx − 4 = x − 2 tan−1 + C x2 + 4 x2 + 4 2 1 1 x2 + 1 , , 1 + (y )2 = 1 + 2 = x x x2 2 2 x +1 dx; x = tan θ, dx = sec2 θ dθ, L= x2 1 tan−1 (2) tan−1 (2) tan−1 (2) sec3 θ tan2 θ + 1 L= dθ = sec θ dθ = (sec θ tan θ + csc θ)dθ tan θ tan θ π/4 π/4 π/4 √ tan−1 (2)

√ √ √ 5 1 − = sec θ + ln | csc θ − cot θ| = 5 + ln 2 + ln | 2 − 1| − 2 2 π/4 √ √ √ 2+2 2 √ = 5 − 2 + ln 1+ 5

29. y =

30. y = 2x, 1 + (y )2 = 1 + 4x2 , 1 1 1 L= 1 + 4x2 dx; x = tan θ, dx = sec2 θ dθ, 2 2 0 tan−1 2 1 1 1 sec θ dθ = sec θ tan θ + ln | sec θ + tan θ| 2 2 2 0 0 √ √ 1 √ 1 1√ 1 = ( 5)(2) + ln | 5 + 2| = 5 + ln(2 + 5) 4 4 2 4

1 L= 2

tan−1 2

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360

Chapter 8

31. y = 2x, 1 + (y )2 = 1 + 4x2 , 1 1 1 S = 2π x2 1 + 4x2 dx; x = tan θ, dx = sec2 θ dθ, 2 2 0 −1 tan−1 2 π π tan 2 S= tan2 θ sec3 θ dθ = (sec2 θ − 1) sec3 θ dθ = 4 0 4 0 tan−1 2 1 π 1 1 3 sec θ tan θ − sec θ tan θ − ln | sec θ + tan θ| = = 8 8 4 4 0

32. V = π

π/2

π sin θ cos θ dθ = 4 2

V =π 0

π/2

π sin 2θ dθ = 8

1 dx = tan−1 (x − 2) + C (x − 2)2 + 1 x−1 1 −1 +C 35. dx = sin 2 4 − (x − 1)2 1 1 dx = 16(x + 1/2)2 + 1 4

37.

(sec5 θ − sec3 θ)dθ

√ √ π [18 5 − ln(2 + 5)] 32

1 (x − 3)2 + 1

π (1 − cos 4θ)dθ = 8

tan−1 2

π/2

33.

36.

1 − y 2 dy; y = sin θ, dy = cos θ dθ,

π 4

34.

1 1 − (x − 1)2

1 1 dx = tan−1 (4x + 2) + C (x + 2)2 + 1 4

dx = ln x − 3 + (x − 3)2 + 1 + C

x dx, let u = x + 1, (x + 1)2 + 1 1 u u−1 1 du = ln(u2 + 1) − tan−1 u + C du = − 2 2 2 u +1 u +1 2 u +1 =

39.

1 ln(x2 + 2x + 2) − tan−1 (x + 1) + C 2

4 − (x + 1)2 dx, let x + 1 = 2 sin θ, = 4 cos2 θ dθ = 2(1 + cos 2θ) dθ 1 x+1 −1 + (x + 1) 3 − 2x − x2 + C = 2θ + sin 2θ + C = 2 sin 2 2

40.

ex (ex + 1/2)2 + 3/4 1 u2 + 3/4

dx, let u = ex + 1/2,

√ du = sinh−1 (2u/ 3) + C = sinh−1 √

2ex + 1 √ 3

3 tan θ, 2 √ 2 e2x + ex + 1 2ex + 1 √ sec θ dθ = ln | sec θ + tan θ| + C = ln + √ + C1 3 3 = ln(2 e2x + ex + 1 + 2ex + 1) + C x

Alternate solution: let e + 1/2 =

π/2 π2 = 16 0

dx = sin−1 (x − 1) + C

38.

1 θ − sin 4θ 4

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Exercise Set 8.5

361

1 1 dx = 2 2(x + 1) + 5 2

41.

1 1 dx = √ tan−1 2/5(x + 1) + C 2 (x + 1) + 5/2 10

42.

2x + 3 dx, let u = x + 1/2, 4(x + 1/2)2 + 4 2u + 2 1 u 1 1 1 du = + du = ln(u2 + 1) + tan−1 u + C 4u2 + 4 2 u2 + 1 u2 + 1 4 2 1 1 ln(x2 + x + 5/4) + tan−1 (x + 1/2) + C 4 2 2 2 1 1 −1 x − 2 √ dx = dx = sin = π/6 2 4x − x2 4 − (x − 2)2 1 1 =

43. 1

44.

4x − x2 dx =

4 − (x − 2)2 dx, let x − 2 = 2 sin θ,

π/2

π/2 cos2 θ dθ = 2θ + sin 2θ = 2π

−π/2

45. u = sin2 x, du = 2 sin x cos x dx; 1 1 2 1 u 1 − u2 + sin−1 u + C = sin x 1 − sin4 x + sin−1 (sin2 x) + C 1 − u2 du = 4 4 2 46. u = x sin x, du = (x cos x + sin x) dx; 1 1 1 1 1 + u2 du = u 1 + u2 + sinh−1 u + C = x sin x 1 + x2 sin2 x + sinh−1 (x sin x) + C 2 2 2 2 47. (a) x = 3 sinh u, dx = 3 cosh u du, du = u + C = sinh−1 (x/3) + C (b) x = 3 tan θ, dx = 3 sec2 θ dθ, sec θ dθ = ln | sec θ + tan θ| + C = ln x2 + 9/3 + x/3 + C √ but sinh−1 (x/3) = ln x/3 + x2 /9 + 1 = ln x/3 + x2 + 9/3 so the results agree. 48. x = cosh u, dx = sinh u du, 1 1 1 2 sinh u du = (cosh 2u − 1)du = sinh 2u − u + C 2 4 2 1 1 1 1 = sinh u cosh u − u + C = x x2 − 1 − cosh−1 x + C 2 2 2 2 √ because cosh u = x, and sinh u = cosh2 u − 1 = x2 − 1

EXERCISE SET 8.5 1. 3. 5.

A B + (x − 3) (x + 4) C 2x − 3 A B = + 2+ − 1) x x x−1

x2 (x

C Dx + E A B + 2+ 3+ 2 x x x x +2

5 A B C = + + x(x − 2)(x + 2) x x−2 x+2

C A B + + 2 x + 2 (x + 2) (x + 2)3

A Bx + C + 2 x−1 x +6

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362

7. 9.

10.

11.

12.

13.

Chapter 8

Cx + D Ax + B + 2 2 x +5 (x + 5)2

Bx + C Dx + E A + 2 + 2 x−2 x +1 (x + 1)2

A B 1 1 1 = + ; A = , B = − so (x − 4)(x + 1) x−4 x+1 5 5 1 1 1 1 1 1 x − 4 1 dx − dx = ln |x − 4| − ln |x + 1| + C = ln +C x−4 5 x+1 5 5 5 x + 1 5 1 A B 1 1 = + ; A = − , B = so (x + 1)(x − 7) x+1 x−7 8 8 1 1 1 1 1 1 x − 7 1 +C dx + dx = − ln |x + 1| + ln |x − 7| + C = ln − 8 x−7 8 8 8 x + 1 8 x+1 11x + 17 A B = + ; A = 5, B = 3 so (2x − 1)(x + 4) 2x − 1 x + 4 1 1 5 5 dx + 3 dx = ln |2x − 1| + 3 ln |x + 4| + C 2x − 1 x+4 2 A B 5x − 5 = + ; A = 1, B = 2 so (x − 3)(3x + 1) x − 3 3x + 1 1 2 1 dx + 2 dx = ln |x − 3| + ln |3x + 1| + C x−3 3x + 1 3 2x2 − 9x − 9 A B C = + + ; A = 1, B = 2, C = −1 so x(x + 3)(x − 3) x x+3 x−3 x(x + 3)2 1 1 1 +C dx + 2 dx − dx = ln |x| + 2 ln |x + 3| − ln |x − 3| + C = ln x x+3 x−3 x−3 Note that the symbol C has been recycled; to save space this recycling is usually not mentioned.

A B C 1 1 1 = + + ; A = −1, B = , C = so x(x + 1)(x − 1) x x+1 x−1 2 2 1 1 1 1 1 1 1 dx + dx + dx = − ln |x| + ln |x + 1| + ln |x − 1| + C − x 2 x+1 2 x−1 2 2 1 (x + 1)(x − 1) 1 |x2 − 1| = ln + C = ln +C 2 2 x 2 x2 1 1 1 x2 − 8 =x−3+ , x−3+ dx = x2 − 3x + ln |x + 3| + C 15. x+3 x+3 x+3 2 1 x2 + 1 2 2 dx = x2 + x + 2 ln |x − 1| + C 16. =x+1+ , x+1+ x−1 x−1 x−1 2 14.

17.

18.

3x2 − 10 A 12x − 22 12x − 22 B = ; A = 12, B = 2 so =3+ 2 , + x2 − 4x + 4 x − 4x + 4 (x − 2)2 x − 2 (x − 2)2 1 1 dx = 3x + 12 ln |x − 2| − 2/(x − 2) + C 3dx + 12 dx + 2 x−2 (x − 2)2 x2 A B 3x − 2 3x − 2 = + ; A = −1, B = 4 so =1+ 2 , − 3x + 2 x − 3x + 2 (x − 1)(x − 2) x−1 x−2 1 1 dx − dx + 4 dx = x − ln |x − 1| + 4 ln |x − 2| + C x−1 x−2 x2

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Exercise Set 8.5

19.

x5 + x2 + 2 x2 + x + 2 = x2 + 1 + , 3 x −x x3 − x x2 + x + 2 A B C = + + ; A = −2, B = 1, C = 2 so x(x + 1)(x − 1) x x+1 x−1 1 2 2 dx + dx + dx (x2 + 1)dx − x x+1 x−1 (x + 1)(x − 1)2 1 1 +C = x3 + x − 2 ln |x| + ln |x + 1| + 2 ln |x − 1| + C = x3 + x + ln 3 3 x2

20.

x5 − 4x3 + 1 1 = x2 + 3 , x3 − 4x x − 4x 1 A B C 1 1 1 = + + ; A = − , B = + , C = + so x(x + 2)(x − 2) x x+2 x−2 4 8 8 1 1 1 1 1 1 x2 dx − dx + dx + dx 4 x 8 x+2 8 x−2 =

21.

22.

23.

24.

25.

26.

1 1 1 3 1 x − ln |x| + ln |x + 2| + ln |x − 2| + C 3 4 8 8

B C 2x2 + 3 A + = + ; A = 3, B = −1, C = 5 so x(x − 1)2 x x − 1 (x − 1)2 1 1 1 dx − dx + 5 dx = 3 ln |x| − ln |x − 1| − 5/(x − 1) + C 3 x x−1 (x − 1)2 A B C 3x2 − x + 1 = + 2+ ; A = 0, B = −1, C = 3 so 2 x (x − 1) x x x−1 1 1 dx = 1/x + 3 ln |x − 1| + C − dx + 3 2 x x−1 2x2 − 10x + 4 B C A + + = ; A = 1, B = 1, C = −2 so 2 (x + 1)(x − 3) x + 1 x − 3 (x − 3)2 2 1 1 2 dx = ln |x + 1| + ln |x − 3| + dx + dx − + C1 x+1 x−3 (x − 3)2 x−3 2x2 − 2x − 1 C A B = + 2+ ; A = 3, B = 1, C = −1 so 2 x (x − 1) x x x−1 1 1 1 1 dx + dx = 3 ln |x| − − ln |x − 1| + C dx − 3 x x2 x−1 x x2 B A C + = + ; A = 1, B = −2, C = 1 so (x + 1)3 x + 1 (x + 1)2 (x + 1)3 2 1 1 2 1 dx − − dx + dx = ln |x + 1| + +C 2 3 x+1 (x + 1) (x + 1) x + 1 2(x + 1)2 B A C 2x2 + 3x + 3 + = + ; A = 2, B = −1, C = 2 so 3 2 (x + 1) x + 1 (x + 1) (x + 1)3 1 1 1 1 1 dx + 2 dx = 2 ln |x + 1| + +C 2 dx − − x+1 (x + 1)2 (x + 1)3 x + 1 (x + 1)2

363

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364

27.

28.

Chapter 8

A Bx + C 2x2 − 1 = + 2 ; A = −14/17, B = 12/17, C = 3/17 so 2 (4x − 1)(x + 1) 4x − 1 x +1 2x2 − 1 7 6 3 dx = − ln |4x − 1| + ln(x2 + 1) + tan−1 x + C 2 (4x − 1)(x + 1) 34 17 17

29.

30.

31.

32.

1 A Bx + C 1 1 = + 2 ; A = , B = − , C = 0 so + 2) x x +2 2 2

x(x2

1 x2 1 1 1 2 ln |x| − ln(x ln + 2) + C = dx = +C x3 + 2x 2 4 4 x2 + 2

x3 + 3x2 + x + 9 Ax + B Cx + D = 2 + 2 ; A = 0, B = 3, C = 1, D = 0 so 2 2 (x + 1)(x + 3) x +1 x +3 3 x + 3x2 + x + 9 1 dx = 3 tan−1 x + ln(x2 + 3) + C (x2 + 1)(x2 + 3) 2 x3 + x2 + x + 2 Ax + B Cx + D = 2 + 2 ; A = D = 0, B = C = 1 so 2 2 (x + 1)(x + 2) x +1 x +2 3 x + x2 + x + 2 1 dx = tan−1 x + ln(x2 + 2) + C 2 (x2 + 1)(x2 + 2) x x3 − 2x2 + 2x − 2 =x−2+ 2 , 2 x +1 x +1 3 1 1 x − 3x2 + 2x − 3 dx = x2 − 2x + ln(x2 + 1) + C x2 + 1 2 2 x4 + 6x3 + 10x2 + x x = x2 + 2 , x2 + 6x + 10 x + 6x + 10 x u−3 x dx = dx = du, x2 + 6x + 10 (x + 3)2 + 1 u2 + 1 =

u=x+3

1 ln(u2 + 1) − 3 tan−1 u + C1 2

x4 + 6x3 + 10x2 + x 1 1 dx = x3 + ln(x2 + 6x + 10) − 3 tan−1 (x + 3) + C x2 + 6x + 10 3 2 1 1 A B 33. Let x = sin θ to get dx, and = + ; A = −1/6, 2 x + 4x − 5 (x + 5)(x − 1) x+5 x−1 1 1 1 1 1 x − 1 1 − sin θ 1 + C = ln + C. dx + dx = ln B = 1/6 so we get − 6 x+5 6 x−1 6 x + 5 6 5 + sin θ so

34. Let x = et ; then

et dt = e2t − 4

1 dx, −4

A 1 B = + ; A = −1/4, B = 1/4 so x+2 x−2 (x + 2)(x − 2) 1 1 1 et − 2 1 1 1 x − 2 + C = ln t + C. − dx + dx = ln 4 x+2 4 x−2 4 x + 2 4 e + 2

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Exercise Set 8.5

365 2

35. V = π 0

x4 x4 18x2 − 81 dx, 4 =1+ 4 , 2 2 2 (9 − x ) x − 18x + 81 x − 18x2 + 81

18x2 − 81 A C B D 18x2 − 81 = = + ; + + 2 2 2 2 2 (9 − x ) (x + 3) (x − 3) x + 3 (x + 3) x − 3 (x − 3)2 9 9 9 9 A = − , B = , C = , D = so 4 4 4 4 2 9 19 9 9/4 9 9/4 V = π x − ln |x + 3| − =π + ln |x − 3| − − ln 5 4 x+3 4 x−3 0 5 4

5 ex dx du = , 36. Let u = e to get x (1 + ex ) e u(1 + u) − ln 5 − ln 5 1/5 5 5 B du 1 A ; A = 1, B = −1; = (ln u − ln(1 + u)) = ln 5 = + 1+u u(1 + u) u 1/5 u(1 + u) 1/5 ln 5

37.

dx = 1 + ex

ln 5

Cx + D x2 + 1 Ax + B + = 2 ; A = 0, B = 1, C = D = −2 so + 2x + 3)2 x + 2x + 3 (x2 + 2x + 3)2 x2 + 1 2x + 2 1 dx − dx dx = (x2 + 2x + 3)2 (x + 1)2 + 2 (x2 + 2x + 3)2 (x2

x+1 1 = √ tan−1 √ + 1/(x2 + 2x + 3) + C 2 2 38.

Cx + D x5 + x4 + 4x3 + 4x2 + 4x + 4 Ax + B Ex + F + 2 = 2 + 2 ; (x2 + 2)3 x +2 (x + 2)2 (x + 2)3 A = B = 1, C = D = E = F = 0 so √ x+1 1 1 dx = ln(x2 + 2) + √ tan−1 (x/ 2) + C 2 x +2 2 2

39. x4 − 3x3 − 7x2 + 27x − 18 = (x − 1)(x − 2)(x − 3)(x + 3), B C D 1 A + + + ; = x−1 x−2 x−3 x+3 (x − 1)(x − 2)(x − 3)(x + 3) A = 1/8, B = −1/5, C = 1/12, D = −1/120 so dx 1 1 1 1 = ln |x − 1| − ln |x − 2| + ln |x − 3| − ln |x + 3| + C 4 3 2 x − 3x − 7x + 27x − 18 8 5 12 120 40. 16x3 − 4x2 + 4x − 1 = (4x − 1)(4x2 + 1), 1 A Bx + C = + 2 ; A = 4/5, B = −4/5, C = −1/5 so (4x − 1)(4x2 + 1) 4x − 1 4x + 1 1 1 dx 1 ln(4x2 + 1) − tan−1 (2x) + C = ln |4x − 1| − 16x3 − 4x2 + 4x − 1 5 10 10 2

41. Let u = x , du = 2x dx, 0

42.

x 1 dx = 4 x +1 2

1 1 du = tan−1 u 2 1+u 2

1 = 0

π 1π = . 24 8

B 1 1 1 A + ;A = ,B = so = 2 −x a−x a+x 2a 2a a + x 1 1 1 1 1 +C + dx = (− ln |a − x| + ln |a + x| ) + C = ln 2a a−x a+x 2a 2a a − x a2

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366

Chapter 8

43. If the polynomial has distinct roots r1 , r2 , r1 = r2 , then the partial fraction decomposition will B A , , and they will give logarithms and no inverse tangents. If contain terms of the form x − r1 x − r2 B A , will appear, and neither there are two roots not distinct, say x = r, then the terms x − r (x − r)2 will give an inverse tangent term. The only other possibility is no real roots, and the integrand 1 , which will yield an inverse tangent, speciﬁcally can be written in the form b 2 b2 a x + 2a + c − 4a b −1 for some constant A. A x+ of the form tan 2a 44. Since there are no inverse tangent terms, the roots are real. Since there are no logarithmic terms, 1 there are no terms of the form , so the only terms that can arise in the partial fraction x−r 1 . Therefore the original quadratic had a multiple root. decomposition form is one like (x − a)2 45. Yes, for instance the integrand

1 , whose integral is precisely tan−1 x + C. x2 + 1

EXERCISE SET 8.6 1. Formula (60):

3x + ln |−1 + 3x| + C 9

1 x 3. Formula (65): ln +C 5 5 + 2x 5. Formula (102):

1 (x − 1)(2x + 3)3/2 + C 5

√ 1 4 − 3x − 2 +C 7. Formula (108): ln √ 2 4 − 3x + 2 1 x + 4 +C 9. Formula (69): ln 8 x − 4

2. Formula (62):

1 4 + ln |4 − 5x| + C 25 4 − 5x

1 − 5x 1 +C 4. Formula (66): − − 5 ln x x 6. Formula (105):

√ 2 (−x − 4) 2 − x + C 3 √ −1

8. Formula (108): tan

3x − 4 +C 2

1 x − 3 10. Formula (70): ln +C 6 x + 3

x 2 3 x − 3 − ln x + x2 − 3 + C 2 2 √ x2 − 5 12. Formula (94): − + ln(x + x2 − 5) + C x x 2 13. Formula (95): x + 4 − 2 ln(x + x2 + 4) + C 2 √ x2 − 2 x 9 x 14. Formula (90): − 9 − x2 + sin−1 + C +C 15. Formula (74): 2x 2 2 3 11. Formula (73):

√ 16. Formula (80): −

4 − x2 x − sin−1 + C x 2

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Exercise Set 8.6

17. Formula (79):

367

√

2 + √4 − x2 4 − x2 − 2 ln +C x √

18. Formula (117): − 20. Formula (40): −

6x − x2 +C 3x

19. Formula (38): −

1 1 sin(7x) + sin x + C 14 2

1 1 cos(7x) + cos(3x) + C 14 6

21. Formula (50):

x4 [4 ln x − 1] + C 16

23. Formula (42):

e−2x (−2 sin(3x) − 3 cos(3x)) + C 13

22. Formula (50): −2

ln x + 2 √ x

ex (cos(2x) + 2 sin(2x)) + C 5 u du 4 1 1 2x 2x 2x +C = + ln 4 − 3e 25. u = e , du = 2e dx, Formula (62): 2 (4 − 3u)2 18 4 − 3e2x

24. Formula (43):

26. u = cos 2x, du = −2 sin 2xdx, Formula (65): − √ 2 3 27. u = 3 x, du = √ dx, Formula (68): 3 2 x

28. u = sin 4x, du = 4 cos 4xdx, Formula (68):

29. u = 2x, du = 2dx, Formula (76):

1 2

√

du 1 cos 2x = − ln +C 2u(3 − u) 6 3 − cos 2x

√ du 1 −1 3 x tan +C = u2 + 4 3 2 1 4

du sin 4x 1 = tan−1 +C 9 + u2 12 3

du 1 = ln 2x + 4x2 − 9 + C 2 u2 − 9

√

√ 2x2 , du = 2 2xdx, Formula (72): √ 1 x2 4 3 √ u2 + 3 du = 2x + 3 + √ ln 2x2 + 2x4 + 3 + C 4 2 2 4 2

30. u =

31. u = 2x2 , du = 4xdx, u2 du = 16x5 dx, Formula (81): √ u2 du 1 x2 1 √ =− 2 − 4x4 + sin−1 ( 2x2 ) + C 2 4 4 4 2−u 1 du √ =− 3 − 4x2 + C 32. u = 2x, du = 2dx, Formula (83): 2 3x u2 3 − u2 1 1 33. u = ln x, du = dx/x, Formula (26): sin2 u du = ln x + sin(2 ln x) + C 2 4 1 1 1 cos2 u du = − e−2x − sin 2e−2x + C 34. u = e−2x , du = −2e−2x , Formula (27): − 2 4 8 1 1 35. u = −2x, du = −2dx, Formula (51): ueu du = (−2x − 1)e−2x + C 4 4

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368

Chapter 8

1 1 (u ln u−u)+C = (3x+1)[ln(3x+1)−1]+C 3 3 sin 3x 1 du 1 1 +C 37. u = sin 3x, du = 3 cos 3x dx, Formula (67): = + ln 3 u(u + 1)2 3 1 + sin 3x 1 + sin 3x

1 36. u = 3x+1, du = 3 dx, Formula (11): 3

1 38. u = ln x, du = dx, Formula (105): x 39. u = 4x2 , du = 8xdx, Formula (70):

√ 1 8

1 40. u = 2e , du = 2e dx, Formula (69): 2 x

ln u du =

√ u du 1 (2 ln x + 1) 4 ln x − 1 + C = 12 4u − 1

2 4x − 1 du 1 +C = ln u2 − 1 16 4x2 + 1 2ex + √3 du 1 √ +C = √ ln 3 − u2 4 3 2ex − 3

41. u = 2ex , du = 2ex dx, Formula (74): √ √ 1 1 3 1 3 3 − u2 du = u 3 − u2 + sin−1 (u/ 3) + C = ex 3 − 4e2x + sin−1 (2ex / 3) + C 2 4 4 2 4 42. u = 3x, du = 3dx, Formula (80): √ √ √ 4 − u2 du 4 − u2 4 − 9x2 −1 3 = −3 − 3 sin (u/2) + C = − − 3 sin−1 (3x/2) + C 2 u u x 43. u = 3x, du = 3dx, Formula (112): 1 5 u−5 5 5 1 25 u− +C u − u2 du = u − u2 + sin−1 3 3 6 6 3 216 5 18x − 5 18x − 5 25 −1 2 = +C 5x − 9x + sin 36 216 5 44. u =

√

5x, du =

√

5 dx, Formula (117): √ √ (u/ 5) − u2 du x − 5x2 √ √ =− + C = −2 +C x u/(2 5) u (u/ 5) − u2

45. u = 2x, du = 2dx, Formula (44): u sin u du = (sin u − u cos u) + C = sin 2x − 2x cos 2x + C 46. u =

√

x, u = x, 2udu = dx, Formula (45): 2

√ 47. u = − x, u2 = x, 2udu = dx, Formula (51): 2

u cos u du = 2 cos

√

√ √ x + 2 x sin x + C

√ √ ueu du = −2( x + 1)e− x + C

48. u = 2 + x2 , du = 2x dx, Formula (11): 1 1 1 1 ln u du = (u ln u − u) + C = (2 + x2 ) ln(2 + x2 ) − (2 + x2 ) + C 2 2 2 2 49. x2 + 6x − 7 = (x + 3)2 − 16; u = x + 3, du = dx, Formula (70): 1 x − 1 du 1 u − 4 + C = +C = ln ln u2 − 16 8 u + 4 8 x + 7

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Exercise Set 8.6

369

50. x2 + 2x − 3 = (x + 1)2 − 4, u = x + 1, du = dx, Formula (77): 1 4 − u2 du = u 4 − u2 + 2 sin−1 (u/2) + C 2 1 = (x + 1) 3 − 2x − x2 + 2 sin−1 ((x + 1)/2) + C 2 51. x2 − 4x − 5 = (x − 2)2 − 9, u = x − 2, du = dx, Formula (77): u+2 u u du du √ √ du = +2 √ = − 9 − u2 + 2 sin−1 + C 2 2 2 3 9−u 9−u 9−u x − 2 +C = − 5 + 4x − x2 + 2 sin−1 3 52. x2 + 6x + 13 = (x + 3)2 + 4, u = x + 3, du = dx, Formula (71): (u − 3) du 3 1 3 1 = ln(u2 + 4) − tan−1 (u/2) + C = ln(x2 + 6x + 13) − tan−1 ((x + 3)/2) + C u2 + 4 2 2 2 2 √ 53. u = x − 2, x = u2 + 2, dx = 2u du; 2 4 2 4 2 2 2u (u + 2)du = 2 (u4 + 2u2 )du = u5 + u3 + C = (x − 2)5/2 + (x − 2)3/2 + C 5 3 5 3 √ 54. u = x + 1, x = u2 − 1, dx = 2u du; √ 2 2 2 (u2 − 1)du = u3 − 2u + C = (x + 1)3/2 − 2 x + 1 + C 3 3 √ 55. u = x3 + 1, x3 = u2 − 1, 3x2 dx = 2u du; 2 2 5 2 3 2 3 2 2 (u4 − u2 )du = u2 (u2 − 1)du = u − u +C = (x + 1)5/2 − (x3 + 1)3/2 + C 3 15 9 15 9 3 √ 56. u = x3 − 1, x3 = u2 + 1, 3x2 dx = 2u du; 1 2 2 2 du = tan−1 u + C = tan−1 x3 − 1 + C 2 3 u +1 3 3 57. u = x1/3 , x = u3 , dx = 3u2 du; 3u2 u 1 1 du = 3 du = 3 + du u3 − u u2 − 1 2(u + 1) 2(u − 1) =

3 3 ln |x1/3 + 1| + ln |x1/3 − 1| + C 2 2

58. u = x1/6 , x = u6 , dx = 6u5 du; 6u5 u3 1 2 du = 6 u −u+1− du du = 6 u3 + u2 u+1 u+1 = 2x1/2 − 3x1/3 + 6x1/6 − 6 ln(x1/6 + 1) + C 59. u = x1/4 , x = u4 , dx = 4u3 du; 4

1 du = 4 u(1 − u)

x1/4 1 1 du = 4 ln +C + u 1−u |1 − x1/4 |

60. u = x1/2 , x = u2 , dx = 2u du; 2 √ √ 2u2 u +1−1 du = 2 du = 2u − 2 tan−1 u + C = 2 x − 2 tan−1 x + C u2 + 1 u2 + 1

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370

Chapter 8

61. u = x1/6 , x = u6 , dx = 6u5 du; 1 u3 du = 6 u2 + u + 1 + du = 2x1/2 + 3x1/3 + 6x1/6 + 6 ln |x1/6 − 1| + C 6 u−1 u−1 √ 62. u = x, x = u2 , dx = 2u du; 2 √ √ 2 u +u du = −2 u+2+ du = −x − 4 x − 4 ln | x − 1| + C −2 u−1 u−1 √ 63. u = 1 + x2 , x2 = u2 − 1, 2x dx = 2u du, x dx = u du; 1 (u2 − 1)du = (1 + x2 )3/2 − (1 + x2 )1/2 + C 3 64. u = (x + 3)1/5 , x = u5 − 3, dx = 5u4 du; 5 15 5 (u8 − 3u3 )du = (x + 3)9/5 − (x + 3)4/5 + C 9 4 2 1 1 du = ln | tan(x/2) + 1| + C du = 65. u+1 2u 1 − u2 1 + u2 1+ + 1 + u2 1 + u2 1 2 1 66. du = du 2u 1 + u2 u2 + u + 1 2+ 1 + u2 2 2 tan(x/2) + 1 1 −1 √ du = √ tan = +C (u + 1/2)2 + 3/4 3 3 1 1 dθ = du = − + C = − cot(θ/2) + C 67. u = tan(θ/2), 1 − cos θ u2 u 68. u = tan(x/2), 2 1 1 2 2 du = du = dz (z = u + 4/3) 3u2 + 8u − 3 3 (u + 4/3)2 − 25/9 3 z 2 − 25/9 1 tan(x/2) − 1/3 1 z − 5/3 + C = ln +C = ln z + 5/3 5 tan(x/2) + 3 5 69. u = tan(x/2),

1 2

70. u = tan(x/2), 71. 2

1 − u2 1 du = u 2

(1/u − u)du =

1 1 ln | tan(x/2)| − tan2 (x/2) + C 2 4

1 − u2 du = −u + 2 tan−1 u + C = x − tan(x/2) + C 1 + u2

x 1 1 t (Formula (65), a = 4, b = −1) dt = ln t(4 − t) 4 4−t 2 x 1 x 1 x x 1 ln − ln 1 = ln , ln = 0.5, ln = 2, = 4 4−x 4 4−x 4 4−x 4−x

x = e2 , x = 4e2 − e2 x, x(1 + e2 ) = 4e2 , x = 4e2 /(1 + e2 ) ≈ 3.523188312 4−x

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Exercise Set 8.6

x √ 1 (Formula (108), a = −1, b = 2) dt = 2 tan−1 2t − 1 1 t 2t − 1 1 √ √ = 2 tan−1 2x − 1 − tan−1 1 = 2 tan−1 2x − 1 − π/4 , √ √ √ 2(tan−1 2x − 1 − π/4) = 1, tan−1 2x − 1 = 1/2 + π/4, 2x − 1 = tan(1/2 + π/4),

72.

371

√

x = [1 + tan2 (1/2 + π/4)]/2 ≈ 6.307993516

73. A =

25 − x2 dx =

4 x 1 25 x 25 − x2 + sin−1 2 2 5 0

=6+

74. A =

1 3

u2 − 4 du =

1 √ 3 32 − 2 ln(6 + = 3

4 25 sin−1 ≈ 17.59119023 2 5

9x2 − 4 dx; u = 3x,

2/3

(Formula (74), a = 5)

1 3 √

6 1 2 u u − 4 − 2 ln u + u2 − 4 (Formula (73), a2 = 4) 2 2 √ √ 2 32) + 2 ln 2 = 4 2 − ln(3 + 2 2) ≈ 4.481689467 3

1 dx; u = 4x, 25 − 16x2 0 u + 5 4 1 4 1 1 1 A= du = = ln ln 9 ≈ 0.054930614 (Formula (69), a = 5) 2 4 0 25 − u 40 u−5 0 40

75. A =

76. A =

√

4 x ln x dx = x3/2 9 =

π/2

77. V = 2π

3 ln x − 1 2

4 (12 ln 4 − 7) ≈ 4.282458815 9

π/2 x cos x dx = 2π(cos x + x sin x) = π(π − 2) ≈ 3.586419094

√ 4π (3x + 8)(x − 4)3/2 x x − 4 dx = 15 =

79. V = 2π

−3

V = 2π

80. V = 2π 1

(Formula (102), a = −4, b = 1) 4

1024 π ≈ 214.4660585 15

−3 ue du = 2πe (u − 1) = 2π(1 − 4e−3 ) ≈ 5.031899801 u

xe−x dx; u = −x,

(Formula (45))

78. V = 2π

(Formula (50), n = 1/2) 1

0 5

5 π 2 x ln x dx = x (2 ln x − 1) 2 1

= π(25 ln 5 − 12) ≈ 88.70584621

(Formula (50), n = 1)

(Formula (51))

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372

Chapter 8

1 + 16x2 dx; u = 4x,

81. L = 0

1 4

1 + u2 du =

82. L =

1 4 √

83. S = 2π

−1

S = −2π

84. S = 2π 1

√

x2 + 9 dx = x

(sin x) 1 + cos2 x dx; u = cos x,

S=π

(Formula (72), a2 = 1)

3 3 + √x2 + 9 x2 + 9 − 3 ln x 1 1 √ √ √ 3 + 10 √ ≈ 3.891581644 (Formula (89), a = 3) = 3 2 − 10 + 3 ln 1+ 2

1 + 9/x2 dx =

8 u 1 1 + u2 + ln u + 1 + u2 2 2 0 √ 1 65 + ln(8 + 65) ≈ 8.409316783 8

1 u 1 2 2 du = 4π 1 + du = 4π 1 + u + ln u + 1 + u a2 = 1 2 2 0 0

√ √ = 2π 2 + ln(1 + 2) ≈ 14.42359945 (Formula (72)) 1

4√ 4 1 x +1 1 + 1/x4 dx = 2π dx; u = x2 , 3 x x 1 √ √ 16 u2 + 1 u2 + 1 du = π − + ln u + u2 + 1 u2 u 1 √ √ √ 257 16 + 257 √ =π 2− ≈ 9.417237485 (Formula (93), a2 = 1) + ln 16 1+ 2

20 cos6 u sin3 u du

85. (a) s(t) = 2 + 0

=−

s(t)

(b) 4

40 166 20 sin2 t cos7 t − cos7 t + 9 63 63

3 2 1 t 3

a(u) du = −

86. (a) v(t) = 0

s(t) = 10 +

1 −t 1 1 3 1 1 e cos 2t + e−t sin 2t + e−t cos 6t − e−t sin 6t + − 10 5 74 37 10 74

v(u) du 0

=−

343866 2 35 −t 6 −t 16 3 −t e cos 2t − e−t sin 2t + e cos 6t + e sin 6t + t+ 50 25 2738 1369 185 34225

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Exercise Set 8.6

373 s(t)

(b) 4 3 2 1

t 3

87. (a)

1 + tan(x/2) 1 + u 2 1 +C dx = + C = ln sec x dx = du = ln cos x 1 â&#x2C6;&#x2019; u2 1 â&#x2C6;&#x2019; u 1 â&#x2C6;&#x2019; tan(x/2) cos(x/2) + sin(x/2) cos(x/2) + sin(x/2) + C = ln 1 + sin x + C = ln cos(x/2) â&#x2C6;&#x2019; sin(x/2) cos(x/2) + sin(x/2) cos x

= ln |sec x + tan x| + C Ď&#x20AC; x x tan + tan 1 + tan x 4 2 = 2 (b) tan + = x Ď&#x20AC; x 4 2 1 â&#x2C6;&#x2019; tan tan 1 â&#x2C6;&#x2019; tan 4 2 2 1 dx = 1/u du = ln | tan(x/2)| + C but 88. csc x dx = sin x Ď&#x20AC;

ln | tan(x/2)| =

1 (1 â&#x2C6;&#x2019; cos x)/2 1 1 â&#x2C6;&#x2019; cos x 1 sin2 (x/2) ln = ln = ln ; also, 2 cos2 (x/2) 2 (1 + cos x)/2 2 1 + cos x

1 â&#x2C6;&#x2019; cos2 x 1 â&#x2C6;&#x2019; cos x 1 1 1 â&#x2C6;&#x2019; cos x = = â&#x2C6;&#x2019; ln | csc x + cot x| = so ln 2 2 1 + cos x (1 + cos x) (csc x + cot x) 2 1 + cos x â&#x2C6;&#x161; 89. Let u = tanh(x/2) then cosh(x/2) = 1/ sech(x/2) = 1/ 1 â&#x2C6;&#x2019; tanh2 (x/2) = 1/ 1 â&#x2C6;&#x2019; u2 , â&#x2C6;&#x161; sinh(x/2) = tanh(x/2) cosh(x/2) = u/ 1 â&#x2C6;&#x2019; u2 , so sinh x = 2 sinh(x/2) cosh(x/2) = 2u/(1 â&#x2C6;&#x2019; u2 ), cosh x = cosh2 (x/2) + sinh2 (x/2) = (1 + u2 )/(1 â&#x2C6;&#x2019; u2 ), x = 2 tanhâ&#x2C6;&#x2019;1 u, dx = [2/(1 â&#x2C6;&#x2019; u2 )]du; 1 dx 2u + 1 2 tanh(x/2) + 1 2 2 â&#x2C6;&#x161; = + C = â&#x2C6;&#x161; tanâ&#x2C6;&#x2019;1 + C. du = â&#x2C6;&#x161; tanâ&#x2C6;&#x2019;1 â&#x2C6;&#x161; 2 cosh x + sinh x u2 + u + 1 3 3 3 3

1 1 1 du = sinâ&#x2C6;&#x2019;1 u + C = sinâ&#x2C6;&#x2019;1 (x4 ) + C. 2 4 4 1â&#x2C6;&#x2019;u 30 32 32 30 91. (cos x sin x â&#x2C6;&#x2019; cos x sin x)dx = cos30 x sin30 x(cos2 x â&#x2C6;&#x2019; sin2 x)dx sin31 2x 1 sin30 2x cos 2x dx = = 30 +C 2 31(231 ) 1 90. Let u = x to get 4 4

â&#x2C6;&#x161;

â&#x2C6;&#x161; â&#x2C6;&#x161; â&#x2C6;&#x161; 2 1 2 x â&#x2C6;&#x2019; x â&#x2C6;&#x2019; 4dx = â&#x2C6;&#x161; ( x + 2 â&#x2C6;&#x2019; x â&#x2C6;&#x2019; 2)dx = [(x + 2)3/2 â&#x2C6;&#x2019; (x â&#x2C6;&#x2019; 2)3/2 ] + C 92. 3 2 1 1 1 1 1 dx = â&#x2C6;&#x2019; du = â&#x2C6;&#x2019; ln |u| + C = â&#x2C6;&#x2019; ln |1 + xâ&#x2C6;&#x2019;9 | + C 93. 10 â&#x2C6;&#x2019;9 x (1 + x ) 9 u 9 9

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374

Chapter 8

B Cx + D Ex + F A + + 2 + 2 x+4 x−5 x +1 (x + 1)2 2 x−2 3 3 + − − (b) − x + 4 x − 5 x2 + 1 (x2 + 1)2 x 1 3 2 −1 −1 (c) −3 ln |x + 4| + 2 ln |x − 5| + 2 tan x − ln(x + 1) − + tan x + C 2 2 x2 + 1

94. (a) (x + 4)(x − 5)(x2 + 1)2 ;

EXERCISE SET 8.7 1. exact value = 14/3 ≈ 4.666666667 (a) 4.667600663, |EM | ≈ 0.000933996 (b) 4.664795679, |ET | ≈ 0.001870988 (c) 4.666651630, |ES | ≈ 0.000015037 3. exact value = 2 (a) 2.008248408, |EM | ≈ 0.008248408 (b) 1.983523538, |ET | ≈ 0.016476462 (c) 2.000109517, |ES | ≈ 0.000109517 5. exact value = e−1 − e−4 ≈ 0.3495638023 (a) 0.3482563710, |EM | ≈ 0.0013074313 (b) 0.3521816066, |ET | ≈ 0.0026178043 (c) 0.3495793657, |ES | ≈ 0.0000155634

2. exact value = 2 (a) 1.998377048, |EM | ≈ 0.001622952 (b) 2.003260982, |ET | ≈ 0.003260982 (c) 2.000072698, |ES | ≈ 0.000072698 4. exact value = sin(1) ≈ 0.841470985 (a) 0.841821700, |EM | ≈ 0.000350715 (b) 0.840769642, |ET | ≈ 0.000701343 (c) 0.841471453, |ES | ≈ 0.000000468 1 ln 5 ≈ 0.804718956 2 (a) 0.801605339, |EM | ≈ 0.003113617 (b) 0.811019505, |ET | ≈ 0.006300549 (c) 0.805041497, |ES | ≈ 0.000322541

6. exact value =

√

1 15 x − 1, f (x) = − (x − 1)−3/2 , f (4) (x) = − (x − 1)−7/2 ; K2 = 1/4, K4 = 15/16 4 16 27 27 (1/4) = 0.002812500 (b) |ET | ≤ (1/4) = 0.005625000 (a) |EM | ≤ 2400 1200 243 (c) |ES | ≤ (15/16) ≈ 0.000126563 180 × 104

7. f (x) =

√ 3 105 −9/2 x 8. f (x) = 1/ x, f (x) = x−5/2 , f (4) (x) = ; K2 = 3/4, K4 = 105/16 4 16 27 27 (a) |EM | ≤ (3/4) = 0.008437500 (b) |ET | ≤ (3/4) = 0.016875000 2400 1200 243 (105/16) ≈ 0.000885938 (c) |ES | ≤ 180 × 104 9. f (x) = sin x, f (x) = − sin x, f (4) (x) = sin x; K2 = K4 = 1 π3 (1) ≈ 0.012919282 2400 π5 (c) |ES | ≤ (1) ≈ 0.000170011 180 × 104

(a) |EM | ≤

(b) |ET | ≤

π3 (1) ≈ 0.025838564 1200

10. f (x) = cos x, f (x) = − cos x, f (4) (x) = cos x; K2 = K4 = 1 1 1 (a) |EM | ≤ (1) ≈ 0.000416667 (b) |ET | ≤ (1) ≈ 0.000833333 2400 1200 1 (1) ≈ 0.000000556 (c) |ES | ≤ 180 × 104

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Exercise Set 8.7

375

11. f (x) = e−x , f (x) = f (4) (x) = e−x ; K2 = K4 = e−1 9 −1 (e ) ≈ 0.001226265 800 27 (e−1 ) ≈ 0.000049664 (c) |ES | ≤ 2 × 105

(a) |EM | ≤

(b) |ET | ≤

9 −1 (e ) ≈ 0.002452530 400

12. f (x) = 1/(2x + 1), f (x) = 8(2x + 1)−3 , f (4) (x) = 384(2x + 1)−5 ; K2 = 8, K4 = 384 8 (8) ≈ 0.026666667 2400 32 (c) |ES | ≤ (384) ≈ 0.006826667 180 × 104

(a) |EM | ≤

13. (a) n > (c) n > 14. (a) n >

(27)(1/4) (24)(5 × 10−4 )

1/2

≈ 23.7; n = 24

(243)(15/16) (180)(5 × 10−4 ) (27)(3/4) (24)(5 × 10−4 )

1/4

≈ 41.1; n = 42

(π 3 )(1) (b) n > (12)(10−3 )

(b) n >

(1)(1) (12)(10−3 )

(8)(e−1 ) (b) n > (12)(10−6 )

1/2

(32)(384) (c) n > (180)(10−6 )

≈ 33.5; n = 34

1/2 ≈ 58.1; n = 59

1/2 ≈ 50.8; n = 51

1/2 ≈ 9.1; n = 10

≈ 1.5; n = 2

1/2 (8)(e−1 ) 17. (a) n > ≈ 350.2; n = 351 (24)(10−6 ) 1/4 (32)(e−1 ) (c) n > ≈ 15.99; n = 16 (180)(10−6 ) (8)(8) (24)(10−6 )

1/2

1/4

(27)(3/4) (12)(5 × 10−4 )

≈ 6.5; n = 7

(1)(1) (c) n > (180)(10−3 )

(27)(1/4) (12)(5 × 10−4 )

≈ 11.5; n = 12

1/2

18. (a) n >

(b) n >

1/4

1/2 (π 3 )(1) 15. (a) n > ≈ 35.9; n = 36 (24)(10−3 ) 1/4 (π 5 )(1) (c) n > ≈ 6.4; n = 8 (180)(10−3 ) (1)(1) (24)(10−3 )

8 (8) ≈ 0.053333333 1200

≈ 7.1; n = 8

(b) n >

1/2

(243)(105/16) (c) n > (180)(5 × 10−4 )

16. (a) n >

(b) |ET | ≤

≈ 1632.99; n = 1633

1/4 ≈ 90.9; n = 91

(b) n >

(8)(8) (12)(10−6 )

1/2 ≈ 495.2; n = 496

1/2 ≈ 2309.4; n = 2310

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376

Chapter 8

19. g(X0 ) = aX02 + bX0 + c = 4a + 2b + c = f (X0 ) = 1/X0 = 1/2; similarly 9a + 3b + c = 1/3, 16a + 4b + c = 1/4. Three equations in three unknowns, with solution a = 1/24, b = −3/8, c = 13/12, g(x) = x2 /24 − 3x/8 + 13/12. 4 4 2 25 x 3x 13 dx = g(x) dx = − + 24 8 12 36 2 2 25 1 1 4 1 ∆x [f (X0 ) + 4f (X1 ) + f (X2 )] = + + = 3 3 2 3 4 36 20. Suppose g(x) = ax2 + bx + c passes through the points (0, f (0)) = (0, 0), (m, f (m)) = (1/6, 1/4), and (1/3, f (2m)) = (1/3, 3/4). Then g(0) = c = 0, 1/4 = g(1/6) = a/36 + b/6, and 3/4 = g(1/3) = a/9 + b/3, with solution a = 9/2, b = 3/4 or g(x) = 9x2 /2 + 3x/4. Then (∆x/3)(Y0 + 4Y1 O + Y2 ) = (1/18)(0 + 4(1/4) + 3/4) = 7/72, and 1/3 1/3 3 2 g(x) dx = (3/2)x + (3/8)x = 1/18 + 1/24 = 7/72 0

21. 0.746824948, 0.746824133

22. 1.236098366, 1.236067977

23. 1.511518747, 1.515927142

24. 2.418388347, 2.418399152

25. 0.805376152, 0.804776489

26. 1.536963087, 1.544294774

27. (a) 3.142425985, |EM | ≈ 0.000833331 (b) 3.139925989, |ET | ≈ 0.001666665 (c) 3.141592614, |ES | ≈ 0.000000040

28. (a) 3.152411433, |EM | ≈ 0.010818779 (b) 3.104518326, |ET | ≈ 0.037074328 (c) 3.127008159, |ES | ≈ 0.014584495

29. S14 = 0.693147984, |ES | ≈ 0.000000803 = 8.03 × 10−7 ; the method used in Example 6 results in a value of n which ensures that the magnitude of the error will be less than 10−6 , this is not necessarily the smallest value of n. 30. (a) underestimates, because the graph of cos x2 is concave down on the interval (0, 1) (b) overestimates, because the graph of cos x2 is concave up on the interval (3/2, 2) 31. f (x) = x sin x, f (x) = 2 cos x − x sin x, |f (x)| ≤ 2| cos x| + |x| | sin x| ≤ 2 + 2 = 4 so K2 ≤ 4, 1/2 (8)(4) ≈ 115.5; n = 116 (a smaller n might suﬃce) n> (24)(10−4 ) 32. f (x) = ecos x , f (x) = (sin2 x)ecos x − (cos x)ecos x , |f (x)| ≤ ecos x (sin2 x + | cos x|) ≤ 2e so 1/2 (1)(2e) K2 ≤ 2e, n > ≈ 47.6; n = 48 (a smaller n might suﬃce) (24)(10−4 ) √ 3 33. f (x) = x x, f (x) = √ , lim+ |f (x)| = +∞ 4 x x→0 √ √ x sin x + cos x , lim+ |f (x)| = +∞ 34. f (x) = sin x, f (x) = − x→0 4x3/2 √

π/2

35. L = −π/2

√

1 + sin2 x dx ≈ 3.820187623

36. L = 1

1 + 1/x4 dx ≈ 2.146822803

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Exercise Set 8.7

377

v dt ≈

37. 0

38.

15 [0 + 4(35.2) + 2(60.1) + 4(79.2) + 2(90.9) + 4(104.1) + 114.4] ≈ 1078 ft (3)(6)

0.02

0.08

0.20

0.40

0.60

0.70

0.60

8 [0 + 4(0.02) + 2(0.08) + 4(0.20) + 2(0.40) + 4(0.60) + 2(0.70) + 4(0.60) + 0] (3)(8)

a dt ≈ 0

≈ 2.7 cm/s

180

v dt ≈

39. 0

180 [0.00 + 4(0.03) + 2(0.08) + 4(0.16) + 2(0.27) + 4(0.42) + 0.65] = 37.9 mi (3)(6)

1800

(1/v)dx ≈

40. 0

1 1800 4 2 4 2 4 1 ≈ 0.71 s + + + + + + (3)(6) 3100 2908 2725 2549 2379 2216 2059

41. V =

πr dy = π 0

r2 dy ≈ π

16 [(8.5)2 + 4(11.5)2 + 2(13.8)2 + 4(15.4)2 + (16.8)2 ] (3)(4)

≈ 9270 cm3 ≈ 9.3 L

600

h dx ≈

42. A = 0

600 [0 + 4(7) + 2(16) + 4(24) + 2(25) + 4(16) + 0] = 9000 ft2 , (3)(6)

V = 75A ≈ 75(9000) = 675, 000 ft3 43. (a) The maximum value of |f (x)| is approximately 3.8442 (b) n = 18 (c) 0.9047406684 44. (a) The maximum value of |f (x)| is approximately 1.46789 (b) n = 8 (c) 1.112830350 45. (a) The maximum value of |f (4) (x)| is approximately 42.5518. (b) n = 4 (c) 0.9045241594 46. (a) The maximum value of |f (4) (x)| is approximately 7.02267. (b) n = 4 (c) 1.111442702 b−a [y0 + y1 + . . . + yn−2 + yn−1 ] n b−a [y1 + y2 + . . . + yn−1 + yn ] Right endpoint approximation ≈ n b−a1 [y0 + 2y1 + 2y2 + . . . + 2yn−2 + 2yn−1 + yn ] Average of the two = n 2

47. (a) Left endpoint approximation ≈

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378

Chapter 8

yk + yk+1 . If we sum from k = 0 2 to k = n − 1 then we get the right hand side of (2).

(b) Area of trapezoid = (xk+1 − xk )

y 3 2 1 yk xk

yk+1

x k+1

48. right endpoint, trapezoidal, midpoint, left endpoint 49. Given g(x) = Ax2 + Bx + C, suppose ∆x = 1 and m = 0. Then set Y0 = g(−1), Y1 = g(0), Y2 = g(1). Also Y0 = g(−1) = A − B + C, Y1 = g(0) = C, Y2 = g(1) = A + B + C, with solution C = Y1 , B = 12 (Y2 − Y0 ), and A = 12 (Y0 + Y2 ) − Y1 . 1 1 2 2 1 1 g(x) dx = 2 (Ax2 + C) dx = A + 2C = (Y0 + Y2 ) − Y1 + 2Y1 = (Y0 + 4Y1 + Y2 ), Then 3 3 3 3 −1 0 which is exactly what one gets applying the Simpson’s Rule. The general case with the interval (m − ∆x, m + ∆x) and values Y0 , Y1 , Y2 , can be converted by x−m the change of variables z = . Set g(x) = h(z) = h((x − m)/∆x) to get dx = ∆x dz and ∆x m+∆x

h(z) dz =

∆x

−1

m−∆x

g(x) dx. Finally, Y0 = g(m − ∆x) = h(−1),

Y1 = g(m) = h(0), Y2 = g(m + ∆x) = h(1). 50. From Exercise 49 we know, for i = 0, 1, . . . , n − 1 that x2i+2 1b−a [y2i + 4y2i+1 + y2i+2 ], gi (x) dx = 3 2n x2i b−a is the width of the partition and acts as ∆x in Exercise 49. 2n Summing over all the subintervals note that y0 and y2n are only listed once; so b n−1 x2i+2 f (x) dx = gi (x) dx because

i=0

x2i

1b−a [y0 + 4y1 + 2y2 + 4y3 + . . . + 4y2n−2 + 2y2n−1 + y2n ] 3 2n

EXERCISE SET 8.8 1. (a) (b) (c) (d) (e) (f )

improper; infinite discontinuity at x = 3 continuous integrand, not improper improper; infinite discontinuity at x = 0 improper; infinite interval of integration improper; infinite interval of integration and infinite discontinuity at x = 1 continuous integrand, not improper

2. (a) improper if p > 0 (b) improper if 1 ≤ p ≤ 2 (c) integrand is continuous for all p, not improper

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Exercise Set 8.8

10.

11.

12.

lim

→+∞

379

1 − e−2x 2

1 1 lim (−e−2 + 1) = 2 →+∞ 2

= 0

1 1 2 ln(1 + x ) = lim [ln(1 + .2 ) − ln 2] = +∞, divergent lim →+∞ 2 →+∞ 2 −1 x = lim 2 coth−1 3 − 2 coth−1 . = 2 coth−1 3

−1

lim −2 coth

→+∞

2 1 lim − e−x →+∞ 2

1 − 2 −e + 1 = 1/2 →+∞ 2

= lim 0

1 lim − →+∞ 2 ln2 x

= lim − →+∞

1 1 1 2 + 2 = 2 2 ln .

√ √ lim 2 ln x = lim (2 ln . − 2 ln 2) = +∞, divergent √

→+∞

lim −

→−∞

1 4(2x − 1)2

1 x tan−1 →−∞ 3 3 1 3x e →−∞ 3

0 = lim

→−∞

lim

1 1 π . = [π/4 − (−π/2)] = π/4 − tan−1 →−∞ 3 4 3 3

= lim

→−∞

1 [−1 + 1/(2. − 1)2 ] = −1/4 4

= lim

lim

1 1 1 3 = − e 3 3 3

0 1 1 1 x ln(3 − 2e ) = ln 3 lim − ln(3 − 2e ) = lim →−∞ →−∞ 2 2 2

+∞

13.

+∞

√

∞

so −∞

x x2 + 2

√

x dx and −∞

diverges.

14.

x dx converges if −∞

1 x dx = lim x2 →+∞ 2

dx = lim

→+∞

+∞

x dx both converge; it diverges if either (or both) 0

1 = lim .2 = +∞ so →+∞ 2

+∞

x dx is divergent. −∞

√ x2 + 2 = lim ( .2 + 2 − 2) = +∞ 0

→+∞

x dx is divergent. x2 + 2

x 1 1 1 dx = lim − = lim [−1/(.2 + 3) + 1/3] = , 2 2 2 →+∞ (x + 3) 2(x + 3) 0 →+∞ 2 6 0 0 ∞ x x similarly dx = −1/6 so dx = 1/6 + (−1/6) = 0 2 + 3)2 2 + 3)2 (x (x −∞ −∞

15.

+∞

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380

Chapter 8

16. 0

−∞

+∞

−∞

17.

e−t π π −1 −t −1 − = , − tan dt = lim − tan (e ) = lim (e ) + →+∞ →+∞ 1 + e−2t 4 4 0 0

π π e−t −1 −t −1 − − dt = lim − tan (e ) = lim (e ) = , + tan →−∞ →−∞ 1 + e−2t 4 4

+∞

e−t π π π dt = + = −2t 1+e 4 4 2

lim − −

→4

18.

19.

20.

21.

lim

→0+

1 x−4

3 2/3 x 2

3 = lim+ (4 − .2/3 ) = 6 →0 2

→π/2−

lim − ln(cos .) = +∞, divergent

→π/2−

√ lim −2 4 − x = lim 2(− 4 − . + 2) = 4 √

→4−

−1

lim sin

→1−

x = lim sin−1 . = π/2 0

→1−

1 √ √ 2 2) = − 8 lim − 9 − x = lim (− 8 + 9 − . + +

→−3

24.

1 1 − = +∞, divergent = lim− − .−4 4 →4

lim − ln(cos x) =

22.

23.

lim

→π/3+

→−3

√

π/2 1 − 2 cos x =

lim + (1 −

→π/3

lim − ln(1 − tan x) =

→π/4−

25. 0

26. 0

27. 0

√

1 − 2 cos .) = 1

lim − ln(1 − tan .) = +∞, divergent

→π/4−

dx = lim ln |x − 2| = lim (ln |. − 2| − ln 2) = −∞, divergent x−2 →2− →2− 0 2 2 dx dx = lim −1/x = lim (−1/2 + 1/.) = +∞ so is divergent 2 + x2 →0+ →0 −2 x 3 x−1/3 dx = lim+ x2/3 →0 2

3 = lim+ (4 − .2/3 ) = 6, →0 2

3 3 x−1/3 dx = lim− x2/3 = lim− (.2/3 − 1) = −3/2 2 2 →0 →0 −1 −1 8 so x−1/3 dx = 6 + (−3/2) = 9/2 −1

28. 0

dx = lim 3(x − 1)1/3 →1− (x − 1)2/3

= lim 3[(. − 1)1/3 − (−1)1/3 ] = 3 0

→1−

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Exercise Set 8.8

381

+∞

1 dx = x2

+∞ 1 1 dx + dx where a > 0; take a = 1 for convenience, 2 2 x x 0 0 a 1 +∞ 1 1 dx = lim (−1/x) = lim (1/. − 1) = +∞ so dx is divergent. 2 + + x x2 →0 →0 0

+∞

dx √ = x x2 − 1

29. Deﬁne

30. Deﬁne 1

dx √ + x x2 − 1

+∞

dx √ where a > 1, x x2 − 1

take a = 2 for convenience to get 2 2 dx −1 √ = lim+ sec x = lim+ (π/3 − sec−1 .) = π/3, 2 →1 →1 1 x x −1 +∞ +∞ dx dx √ √ = lim sec−1 x = π/2 − π/3 so = π/2. 2−1 2−1 →+∞ x x x x 2 1 2 31. 0

+∞

32. +∞

33.

+∞

e−x dx = 1 − e−x

√

→+∞

√ 2 3

+∞

36. A = 0

39. 1 +

dy dx

→+∞

√

du = 1 − u2

√ 2 3

= lim 2(1 − 1

√

→0+

√

= 2 lim

du 1 u = 2 lim tan−1 →+∞ 2 u2 + 4 2

√ du √ = lim 2 u u →0+

1 − e− = 2

= lim tan−1 →+∞

du = lim sin−1 u →1 1 − u2

= lim sin−1 . = →1

1 xe−3x dx = lim − (3x + 1)e−3x →+∞ 9

= 1/3 0

(b) 2.804364

4 − x2/3 4 =1+ = 2/3 ; the arc length is 2/3 x x

2 x1/3

(d) 0.504067 8 = 12

2/3

dx = 3x

=1+

ln x dx = x ln x − x + C,

ln x dx = lim

→0+

but lim . ln . = lim →0+

π 2

41.

√ . π − tan−1 3 = 2 6

.) = 2

x2 4 = ; the arc length is 4 − x2 4 − x2 4 2 −1 x √ dx = lim dx = lim 2 sin = 2 sin−1 1 = π 4 − x2 2 0 →2− 0 →2− 4 − x2

dy 40. 1 + dx 2 0

1 −x e (sin x − cos x) = 1/2 →+∞ 2 0

37. (a) 2.726585

−e−u

e−x cos x dx = lim

+∞

e−x dx = − 1 − e−2x

lim

e−u du = 2 lim

√

35.

dx =2 x(x + 4)

34.

+∞

dx = 2

√

√ x

e− √

+∞

→0+

1 ln x dx = lim (x ln x − x) = lim (−1 − . ln . + .), →0+

ln . = lim (−.) = 0 so 1/. →0+

→0+

ln x dx = −1 0

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382

Chapter 8

42.

ln x ln x 1 − + C, dx = − 2 x x x +∞ ln x 1 ln . 1 ln x ln x − − + 1 , − − dx = lim dx = lim = lim →+∞ 1 x2 →+∞ x2 x x 1 →+∞ . . 1 +∞ ln . 1 ln x but lim =1 = lim = 0 so →+∞ . →+∞ . x2 1

+∞

43.

−3x

dx = lim

→+∞

+∞

44. A = 4

−3x

dx = lim

→+∞

8 x−2 dx = lim 2 ln →+∞ x2 − 4 x+2

1 − e−3x 3

= lim

→+∞

1 1 − e−3 + 3 9

1 3

.−2 1 = 2 ln 3 = lim 2 ln − ln →+∞ .+2 3

π −2x 45. (a) V = π e dx = − = π/2 lim e 2 →+∞ 0 0 +∞ e−x 1 + e−2x dx, let u = e−x to get (b) S = 2π 0 0 1

√ √ u 1 S = −2π 1 + u2 du = 2π 1 + u2 + ln u + 1 + u2 = π 2 + ln(1 + 2) 2 2 1 0

+∞

−2x

47. (a) For x ≥ 1, x2 ≥ x, e−x ≤ e−x +∞ −x e dx = lim e−x dx = lim −e−x = lim (e−1 − e− ) = 1/e (b) 2

→+∞

+∞

→+∞

e−x dx is convergent and is ≤ 1/e. 2

48. (a) If x ≥ 0 then ex ≥ 1,

ex 1 ≤ 2x + 1 2x + 1

dx 1 = lim ln(2x + 1) = +∞ →+∞ 0 2x + 1 →+∞ 2 0 +∞ (c) By Parts (a) and (b) and Exercise 46(a),

(b)

lim

49. V = lim

→+∞

A = lim

→+∞

ex dx is divergent. 2x + 1

(π/x2 ) dx = lim −(π/x) = lim (π − π/.) = π →+∞

→+∞

2π(1/x) 1 + 1/x4 dx;

use Exercise 46(a) with f (x) = 2π/x, g(x) = (2π/x) 1 + 1/x4 and a = 1 to see that the area is inﬁnite. √ +∞ x3 + 1 for x ≥ 2, 50. (a) 1 ≤ 1dx = +∞ x 2 +∞ +∞ x dx 1 dx ≤ (b) = lim − = 1/24 →+∞ x5 + 1 x4 3x3 2 2 2 ∞ +∞ +∞ xex xex dx (c) dx ≥ ≥ = +∞ 2x + 1 2x + 1 2x + 1 0 1 1

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Exercise Set 8.8

383

1 , above the x-axis, and to the 2 1 + ∞x 1 . Solving for right of the y-axis is given by 1 + x2 0 1 1−y 1−y y= dy. y , the area is also given by the improper integral y 0

51. The area under the curve y =

52. (b) u =

√

+∞

x, 0

√ +∞ +∞ cos x √ dx = 2 cos u du; cos u du diverges by Part (a). x 0 0

dx 1 1 x cos θ dθ = 2 sin θ + C = √ = +C 2 2 2 3/2 2 r r (r + x ) r r2 + x2 x 2πN Ir 2πN I √ lim lim (./ r2 + .2 − a/ r2 + a2 ) so u = = →+∞ r 2 r 2 + x2 a k kr →+∞

53. Let x = r tan θ to get

2πN I (1 − a/ r2 + a2 ). kr

M to get 2RT 3/2 −2 M M 1 2RT 8RT 2 4 = =√ (a) v¯ = √ 2 2RT M πM π 2RT π 3/2 √ −5/2 M M 3 π 3RT 4 3RT 2 √ so vrms = = (b) vrms = 8 2RT M M π 2RT

54. Let a2 =

55. (a) Satellite’s weight = w(x) = k/x2 lb when x = distance from center of Earth; w(4000) = 6000 4000+ 10 so k = 9.6 × 10 and W = 9.6 × 1010 x−2 dx mi·lb.

4000

+∞

10 −2

9.6 × 10 x

(b) 4000

dx = lim −9.6 × 10 /x 10

→+∞

= 2.4 × 107 mi·lb

4000

1 −st 1 56. (a) L{1} = e dt = lim − e = →+∞ s s 0 0 +∞ +∞ 1 −(s−2)t 1 −st 2t −(s−2)t 2t e e e dt = e dt = lim − = (b) L{e } = →+∞ s−2 s−2 0 0 0 +∞ e−st 1 (−s sin t − cos t) = 2 (c) L{sin t} = e−st sin t dt = lim 2 →+∞ s s + 1 +1 0 0 +∞ e−st s (−s cos t + sin t) = 2 e−st cos t dt = lim 2 (d) L{cos t} = →+∞ s + 1 s +1 0 0

+∞

−st

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384

Chapter 8

+∞

57. (a) L{f (t)} =

−st

→+∞

+∞

(b) L{f (t)} = (c) L{f (t)} = 3

58.

= 0

→+∞

+∞

1 s2

t2 e−st dt = lim −(t2 /s + 2t/s2 + 2/s3 )e−st

−st

dt = lim −(t/s + 1/s )e 2

1 e−st dt = lim − e−st →+∞ s

100

1000

10,000

0.8862269

= 0

2 s3

e−3s = s

+∞ 2 2 2 e−ax dx = √ e−u du = π/a a 0 0 +∞ +∞ √ √ 2 2 2 2 2 (b) x = 2σu, dx = 2σ du, √ e−x /2σ dx = √ e−u du = 1 π 0 2πσ 0

59. (a) u =

60. (a)

√

+∞ −x2

(b)

dx =

+∞

a dx, 2

e−x dx ≈ 0.8862;

√

ax, du =

−x2

√

π/2 ≈ 0.8862

dx +

+∞ −x2

dx so E =

+∞ −x2

+∞

dx <

xe−x dx = 2

1 −9 e < 7 × 10−5 2

1 dx ≈ 1.047; π/3 ≈ 1.047 +1 0 +∞ 4 +∞ 1 1 1 (b) dx = dx + dx so 6+1 6+1 6+1 x x x 0 0 4 +∞ +∞ 1 1 1 dx < dx = < 2 × 10−4 E= 6 6 x +1 x 5(4)5 4 4

61. (a)

+∞

62. If p = 0, then

(1)dx = lim x = +∞, →+∞

0 +∞

if p = 0, then

1 e dx = lim epx →+∞ p

63. If p = 1, then

if p = 1, then 0

1 = lim (ep − 1) = →+∞ p

−1/p, +∞,

p<0 . p>0

1 dx = lim ln x = +∞; x →0+ 1−p 1 dx x 1/(1 − p), p < 1 1−p = lim = lim [(1 − . )/(1 − p)] = . +∞, p>1 xp →0+ 1 − p →0+

√ 64. u = 1 − x, u2 = 1 − x, 2u du = −dx; 1 0

√ 1 √ 2 − u2 du = 2 2 − u2 du = u 2 − u2 + 2 sin−1 (u/ 2) = 2 + π/2 −2 1

cos(u2 )du ≈ 1.809

65. 2 0

66. −2

sin(1 − u2 )du = 2 1

sin(1 − u2 )du ≈ 1.187 0

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Review Exercises, Chapter 8

+∞

385

−t

e dt = lim −e

67. (a) Γ(1) =

→+∞

+∞

(b) Γ(x + 1) =

= lim (−e− + 1) = 1 →+∞

tx e−t dt; let u = tx , dv = e−t dt to get

0 x −t

+∞

Γ(x + 1) = −t e

+∞

x−1 −t

x −t

e dt = −t e

+∞ + xΓ(x) 0

tx = 0 (by multiple applications of L’Hˆ opital’s rule) t→+∞ t→+∞ et so Γ(x + 1) = xΓ(x) lim tx e−t = lim

(c) Γ(2) = (1)Γ(1) = (1)(1) = 1, Γ(3) = 2Γ(2) = (2)(1) = 2, Γ(4) = 3Γ(3) = (3)(2) = 6 It appears that Γ(n) = (n − 1)! if n is a positive integer. +∞ +∞ √ √ √ 2 1 −1/2 −t (d) Γ = t e dt = 2 e−u du (with u = t) = 2( π/2) = π 2 0 0 1 1√ 3 3√ 3 5 1 3 = Γ = = Γ = (e) Γ π, Γ π 2 2 2 2 2 2 2 4 68. (a) t = − ln x, x = e−t , dx = −e−t dt, 1 0 (ln x)n dx = − (−t)n e−t dt = (−1)n 0

+∞ n

+∞

tn e−t dt = (−1)n Γ(n + 1)

1/n

1/n−1

dt, (b) t = x , x = t , dx = (1/n)t +∞ +∞ n e−x dx = (1/n) t1/n−1 e−t dt = (1/n)Γ(1/n) = Γ(1/n + 1) 0

69. (a)

cos θ − cos θ0 = 2 sin2 (θ0 /2) − sin2 (θ/2) = 2(k 2 − k 2 sin2 φ) = 2k 2 cos2 φ √ 1 1 = 2 k cos φ; k sin φ = sin(θ/2) so k cos φ dφ = cos(θ/2) dθ = 1 − sin2 (θ/2) dθ 2 2 1 2k cos φ = 1 − k 2 sin2 φ dθ, thus dθ = dφ and hence 2 1 − k 2 sin2 φ 8L π/2 1 L π/2 1 2k cos φ √ dφ = 4 dφ · T = 2 g 0 g 0 2k cos φ 1 − k 2 sin φ 1 − k 2 sin2 φ

(b) If L = 1.5 ft and θ0 = (π/180)(20) = π/9, then √ π/2 3 dφ ≈ 1.37 s. T = 2 0 2 1 − sin (π/18) sin2 φ

REVIEW EXERCISES, CHAPTER 8 1. u = 4 + 9x, du = 9 dx,

2. u = πx, du = π dx,

1 π

1 9

u1/2 du =

cos u du =

2 (4 + 9x)3/2 + C 27

1 1 sin u + C = sin πx + C π π

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386

Chapter 8

2 u1/2 du = − cos3/2 θ + C 3 du dx , = ln |u| + C = ln |ln x| + C 4. u = ln x, du = x u 1 1 5. u = tan(x2 ), u2 du = tan3 (x2 ) + C 2 6

3. u = cos θ, −

√ 6. u = x, x = u2 , dx = 2u du, 3 3 3 9 u2 3 −1 u du = 2 du = 2u − 6 tan 2 1 − =6− π 2 2 u +9 3 0 2 0 u +9 0 √ 7. (a) With u = x: √ 1 1 du = 2 sin−1 (u/ 2) + C = 2 sin−1 ( x/2) + C; dx = 2 √ √ √ 2 x 2−x 2−u √ with u = 2 − x : √ √ √ 1 1 dx = −2 √ du = −2 sin−1 (u/ 2) + C = −2 sin−1 ( 2 − x/ 2) + C1 ; √ √ 2 x 2−x 2−u completing the square: 1 dx = sin−1 (x − 1) + C. 1 − (x − 1)2 (b) In the three diﬀer by a constant, in particular results in Part (a) √ √ the antiderivatives 2 sin−1 ( x/2) = π − 2 sin−1 ( 2 − x/ 2) = π/2 + sin−1 (x − 1). 8. (a) u = x2 , dv = √ 0

√

x3 x2 + 1

x x2

dx, du = 2x dx, v =

+1 1 dx = x2 x2 + 1 − 2 0

√

2 2 − (x2 + 1)3/2 3

√

x2 + 1;

x(x2 + 1)1/2 dx

1 = 0

√

√ 2 √ 2 − [2 2 − 1] = (2 − 2)/3 3

(b) u2 = x2 + 1, x2 = u2 − 1, 2x dx = 2u du, x dx = u du; 1 1 √2 2 x3 x2 u −1 √ √ dx = x dx = u du 2 2 u x +1 x +1 0 0 1 √ √2 2 √ 1 3 2 = (u − 1)du = = (2 − 2)/3 u −u 3 1 1 9. u = x, dv = e−x dx, du = dx, v = −e−x ; xe−x dx = −xe−x + e−x dx = −xe−x − e−x + C 1 10. u = x, dv = sin 2x dx, du = dx, v = − cos 2x; 2 1 1 1 1 cos 2x dx = − x cos 2x + sin 2x + C x sin 2x dx = − x cos 2x + 2 2 2 4

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Review Exercises, Chapter 8

387

2x 2 dx, v = x; ln(2x + 3)dx = x ln(2x + 3) − dx 11. u = ln(2x + 3), dv = dx, du = 2x + 3 2x + 3 2x 3 3 but dx = x − ln(2x + 3) + C1 so dx = 1− 2x + 3 2x + 3 2 3 ln(2x + 3)dx = x ln(2x + 3) − x + ln(2x + 3) + C 2 2 12. u = tan−1 (2x), dv = dx, du = dx, v = x; 1 + 4x2 2x 1 tan−1 (2x)dx = x tan−1 (2x) − dx = x tan−1 (2x) − ln(1 + 4x2 ) + C 1 + 4x2 4 13.

Repeated Diﬀerentiation

Repeated Antidiﬀerentiation

8x4

cos 2x + 1 sin 2x 2

32x3 −

1 − cos 2x 4

96x

1 − sin 2x 8

192x −

1 cos 2x 16

192 +

1 sin 2x 32

8x4 cos 2x dx = (4x4 − 12x2 + 6) sin 2x + (8x3 − 12x) cos 2x + C 14. distance =

t2 e−t dt; u = t2 , dv = e−t dt, du = 2tdt, v = −e−t ,

distance = −t2 e−t

−5

distance = −25e

5 +2 0

0 −t

− 2te

te−t dt; u = 2t, dv = e−t dt, du = 2dt, v = −e−t ,

5 +2 0

e−t dt = −25e−5 − 10e−5 − 2e−t

5 0

= −25e−5 − 10e−5 − 2e−5 + 2 = −37e−5 + 2 1 1 1 2 (1 − cos 10θ)dθ = θ − sin 10θ + C 15. sin 5θ dθ = 2 2 20 16. sin3 2x cos2 2x dx = (1 − cos2 2x) cos2 2x sin 2x dx = 17.

1 sin x cos 2x dx = 2

1 1 (cos2 2x − cos4 2x) sin 2x dx = − cos3 2x + cos5 2x + C 6 10 1 1 (sin 3x − sin x)dx = − cos 3x + cos x + C 6 2

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388

Chapter 8

18. 0

π/6

1 sin 2x cos 4x dx = 2

π/6

π/6 1 1 (sin 6x − sin 2x)dx = − cos 6x + cos 2x 12 4 0

= [(−1/12)(−1) + (1/4)(1/2)] − [−1/12 + 1/4] = 1/24 19. u = 2x, 1 1 1 4 4 sin 2x dx = sin u du = − sin3 u cos u + 2 2 4 1 1 3 − sin u cos u + = − sin3 u cos u + 8 2 8

3 4 1 2

sin u du

1 3 3 = − sin3 u cos u − sin u cos u + u + C 16 16 8 3 3 1 sin 2x cos 2x + x + C = − sin3 2x cos 2x − 8 16 8 20. u = x2 , 1 1 x cos5 (x2 )dx = cos5 u du = (cos u)(1 − sin2 u)2 du 2 2 1 1 2 cos u du − cos u sin u du + cos u sin4 u du = 2 2 =

1 1 1 sin u − sin3 u + sin5 u + C 2 3 10

1 1 1 sin(x2 ) − sin3 (x2 ) + sin5 (x2 ) + C 2 3 10

21. x = 3 sin θ, dx = 3 cos θ dθ, 9 9 9 9 9 (1 − cos 2θ)dθ = θ − sin 2θ + C = θ − sin θ cos θ + C 9 sin2 θ dθ = 2 2 4 2 2 9 1 = sin−1 (x/3) − x 9 − x2 + C 2 2 22. x = 4 sin θ, dx = 4 cos θ dθ, √ 1 16 − x2 1 1 1 2 csc θ dθ = − cot θ + C = − +C dθ = 2 16 16 16 16x sin θ 23. x = sec θ, dx = sec θ tan θ dθ,

sec θ dθ = ln | sec θ + tan θ| + C = ln x + x2 − 1 + C

24. x = 5 sec θ, dx = 5 sec θ tan θ dθ, 25 25 25 sec3 θ dθ = sec θ tan θ + ln | sec θ + tan θ| + C1 2 2 1 25 ln |x + x2 − 25| + C = x x2 − 25 + 2 2

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Review Exercises, Chapter 8

389

25. x = 3 tan θ, dx = 3 sec2 θ dθ, 9 tan2 θ sec θ dθ = 9 sec3 θ dθ − 9 sec θ dθ 9 9 sec θ tan θ − ln | sec θ + tan θ| + C 2 2 1 1 9 1 = x 9 + x2 − ln | 9 + x2 + x| + C 2 2 3 3 =

26. 2x = tan θ, 2 dx = sec2 θ dθ, 2 sec θ csc θ dθ = (sec θ tan θ + csc θ) dθ = sec θ − ln | csc θ + cot θ| + C √ 1 + 4x2 1 = 1 + 4x2 − ln + +C 2x 2x 27.

28.

29.

30.

1 A B 1 1 = + ; A = − , B = so (x + 4)(x − 1) x+4 x−1 5 5 1 1 1 1 1 1 x − 1 1 +C dx + dx = − ln |x + 4| + ln |x − 1| + C = ln − 5 x+4 5 x−1 5 5 5 x + 4 1 A B 1 1 = + ; A = , B = − so (x + 1)(x + 7) x+1 x+7 6 6 1 1 1 1 1 1 1 x + 1 +C dx − dx = ln |x + 1| − ln |x + 7| + C = ln x+1 6 x+7 6 6 6 x + 7 6 x2 + 2 6 =x−2+ , x+2 x+2

6 x−2+ x+2

32.

dx =

1 2 x − 2x + 6 ln |x + 2| + C 2

B C x2 + x − 16 A + + = ; A = −1, B = 2, C = −1 so 2 (x + 1)(x − 3) x + 1 x − 3 (x − 3)2 1 1 1 dx − dx + 2 dx − x+1 x−3 (x − 3)2 = − ln |x + 1| + 2 ln |x − 3| +

31.

1 (x − 3)2 1 + C = ln + +C x−3 |x + 1| x−3

A C B x2 = + ; A = 1, B = −4, C = 4 so + (x + 2)3 x + 2 (x + 2)2 (x + 2)3 1 1 4 1 2 dx + 4 dx = ln |x + 2| + +C dx − 4 − x+2 (x + 2)2 (x + 2)3 x + 2 (x + 2)2 A Bx + C 1 = + 2 ; A = 1, B = −1, C = 0 so x(x2 + 1) x x +1 1 x2 1 1 2 ln(x ln + 1) + C = dx = ln |x| − +C x3 + x 2 2 x2 + 1

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390

Chapter 8

33. (a) With x = sec θ: √ 1 x2 − 1 dx = cot θ dθ = ln | sin θ| + C = ln + C; valid for |x| > 1. x3 − x |x| (b) With x = sin θ: 1 1 dx = − dθ = − 2 csc 2θ dθ x3 − x sin θ cos θ

√

= − ln | csc 2θ − cot 2θ| + C = ln | cot θ| + C = ln

1 − x2 + C, 0 < |x| < 1. |x|

A B 3−x 3−x C = + 2+ ; A = −4, B = 3, C = 4 dx, 2 3 2 x (x + 1) x x x+1 1 x +x 2 3 A = −4 ln |x| − + 4 ln |x + 1| x 1 2

34. A =

= (−4 ln 2 −

3 3 3 3 + 4 ln 3) − (−4 ln 1 − 3 + 4 ln 2) = − 8 ln 2 + 4 ln 3 = + 4 ln 2 2 2 4

35. #40

36. #52

37. #113

38. #108

39. #28

40. #71

41. exact value = 14/3 ≈ 4.666666667 (a) 4.667600663, |EM | ≈ 0.000933996 (b) 4.664795679, |ET | ≈ 0.001870988 (c) 4.666651630, |ES | ≈ 0.000015037 1 ln 5 ≈ 0.804718956 2 (a) 0.801605339, |EM | ≈ 0.003113617 (b) 0.811019505, |ET | ≈ 0.006300549 (c) 0.805041497, |ES | ≈ 0.000322541

42. exact value =

43. f (x) =

√

1 15 x + 1, f (x) = − (x + 1)−3/2 , f (4) (x) = − (x + 1)−7/2 ; K2 = 1/4, K4 = 15/16 4 16

27 (1/4) = 0.002812500 2400 243 (15/16) ≈ 0.000126563 (c) |ES | ≤ 180 × 104

(a) |EM | ≤

(b) |ET | ≤

27 (1/4) = 0.005625000 1200

44. f (x) = 1/(2x + 3), f (x) = 8(2x + 3)−3 , f (4) (x) = 384(2x + 3)−5 ; K2 = 8, K4 = 384 8 (8) ≈ 0.026666667 2400 32 (c) |ES | ≤ (384) ≈ 0.006826667 180 × 104

(a) |EM | ≤

45. (a) n > (c) n >

(27)(1/4) (24)(5 × 10−4 )

(b) |ET | ≤

1/2

(243)(15/16) (180)(5 × 10−4 )

≈ 23.7; n = 24

1/4 ≈ 7.1; n = 8

(b) n >

8 (8) ≈ 0.053333333 1200

(27)(1/4) (12)(5 × 10−4 )

1/2 ≈ 33.5; n = 34

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Review Exercises, Chapter 8

391

1/2 (8)(8) 46. (a) n > ≈ 1632.99; n = 1633 (24)(10−6 ) 1/4 (32)(384) (c) n > ≈ 90.9; n = 92 (180)(10−6 )

47.

48.

49.

(8)(8) (b) n > (12)(10−6 )

1/2 ≈ 2309.4; n = 2310

lim (−e−x ) = lim (−e− + 1) = 1

→+∞

1 x lim tan−1 →−∞ 2 2

1 1 π −1 . = [π/4 − (−π/2)] = 3π/8 = lim − tan →−∞ 2 4 2 2

√ lim− −2 9 − x = lim− 2(− 9 − . + 3) = 6 √

→9

50. 0

→9

1 dx = 2x − 1

1/2

1 dx + 2x − 1

1 1 1 dx = lim ln(2x − 1) + lim ln(2x − 1) + C 2x − 1 →1/2− 2 →1/2+ 2

1/2

neither limit exists hence the integral diverges

+∞

51. A = e

52. V = 2π

ln x − 1 ln x dx = lim c − →+∞ x2 x

+∞

−x

xe 0

−x

dx = 2π lim −e →+∞

but lim e− (. + 1) = lim →+∞

+∞

53. 0

54. (a) (c) (e) (g) (i)

→+∞

= 1/e e

(x + 1) = 2π lim 1 − e− (. + 1) 0

→+∞

.+1 1 = lim = 0 so V = 2π →+∞ e e

dx 1 1 π −1 tan (x/a) = lim tan−1 (./a) = = 1, a = π/2 = lim 2 2 →+∞ →+∞ x +a a a 2a 0

integration by parts, u = x, dv = sin x dx reduction formula u-substitution: u = x3 + 1 integration by parts: dv = dx, u = tan−1 x u-substitution: u = 4 − x2

(b) (d) (f ) (h)

u-substitution: u = sin x u-substitution: u = tan x u-substitution: u = x + 1 trigonometric substitution: x = 2 sin θ

√ √ 55. x = 3 tan θ, dx = 3 sec2 θ dθ, 1 1 1 1 x cos θ dθ = sin θ + C = √ +C dθ = 3 sec θ 3 3 3 3 + x2 1 56. u = x, dv = cos 3x dx, du = dx, v = sin 3x; 3 1 1 1 1 sin 3x dx = x sin 3x + cos 3x + C x cos 3x dx = x sin 3x − 3 3 3 9 1 1 1 57. Use Endpaper Formula (31) to get tan7 θ dθ = tan6 θ − tan4 θ + tan2 θ + ln | cos θ| + C. 6 4 2 √ cos θ 1 1 58. dθ, let u = sin θ − 3, du = √ f tan−1 [(sin θ − 3)/ 3] + C 2 2 (sin θ − 3) + 3 u +3 3

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392

Chapter 8

sin 2t(1 − sin 2t) cos 2t dt =

sin 2t cos 2t dt =

59.

(sin2 2t − sin4 2t) cos 2t dt

1 1 sin3 2t − sin5 2t + C 6 10

1 dx = lim (x − 3)2 →3−

60.

1 1 dx = lim − (x − 3)2 x−3 →3−

which is clearly divergent, 0

so the integral diverges. 1 61. u = e2x , dv = cos 3x dx, du = 2e2x dx, v = sin 3x; 3 1 2x 2 2x e2x sin 3x dx. Use u = e2x , dv = sin 3x dx to get e cos 3x dx = e sin 3x − 3 3 2 1 2x 2x e sin 3x dx = − e cos 3x + e2x cos 3x dx so 3 3 1 2x 2 2x 4 2x e cos 3x dx = e sin 3x + e cos 3x − e2x cos 3x dx, 3 9 9 13 9

1 cos 3x dx = e2x (3 sin 3x + 2 cos 3x) + C1 , 9

e2x cos 3x dx =

1 2x e (3 sin 3x + 2 cos 3x) + C 13

√ √ 62. x = (1/ 2) sin θ, dx = (1/ 2) cos θ dθ, π/2 π/2 π/2 1 1 3 1 √ cos3 θ sin θ cos4 θ dθ = √ + cos2 θ dθ 4 −π/2 2 −π/2 2 4 −π/2 π/2 3 1 π/2 3 1 1 3π = √ + dθ = √ π = √ cos θ sin θ 2 2 2 4 2 4 2 8 2 −π/2 −π/2 63.

1 A B C 1 1 1 = + + ;A=− ,B= ,C= so (x − 1)(x + 2)(x − 3) x−1 x+2 x−3 6 15 10 1 1 1 1 1 1 − dx + dx + dx 6 x−1 15 x+2 10 x−3 1 1 1 ln |x + 2| + ln |x − 3| + C = − ln |x − 1| + 6 15 10 2 2 sin θ, dx = cos θ dθ, 3 3 π/6 π/6 π/6 1 1 1 1 1 3 sec θ dθ = dθ = sec θ tan θ + ln | sec θ + tan θ| 24 0 cos3 θ 24 0 48 48 0 √ √ √ √ 1 1 2 1 [(2/ 3)(1/ 3) + ln |2/ 3 + 1/ 3|] = = + ln 3 48 3 2 48

64. x =

65. u = 2 0

√

x − 4, x = u2 + 4, dx = 2u du, 2 2 4 2u2 −1 du = 2u − 4 tan (u/2) = 4 − π 1− 2 du = 2 u2 + 4 u +4 0 0

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Review Exercises, Chapter 8

66. u =

√

393

2u du, u2 + 1 1 1 1 2u2 π −1 du = 2 du = (2u − 2 tan 1 − u) =2− 2+1 u2 + 1 u 2 0 0 ex − 1, ex = u2 + 1, x = ln(u2 + 1), dx =

√

2u ex + 1, ex = u2 − 1, x = ln(u2 − 1), dx = 2 du, u −1 √ x e +1−1 2 1 1 √ x du = ln |u − 1| − ln |u + 1| + C = ln du = − +C u2 − 1 u−1 u+1 e +1+1

67. u =

68.

A Bx + C 1 = + 2 ; A = 1, B = C = −1 so x(x2 + x + 1) x x +x+1 −x − 1 x+1 u + 1/2 dx = − dx = − du, 2 2 x +x+1 (x + 1/2) + 3/4 u2 + 3/4 √ 1 1 = − ln(u2 + 3/4) − √ tan−1 (2u/ 3) + C1 2 3

u = x + 1/2

x(x2

2x + 1 dx 1 1 +C = ln |x| − ln(x2 + x + 1) − √ tan−1 √ + x + 1) 2 3 3

1 dx, v = x; 1 − x2 1/2 1/2 1/2 1/2 x 1 1 √ sin−1 x dx = x sin−1 x − dx = sin−1 + 1 − x2 2 2 1 − x2 0 0 0 0 √ π 3 3 1 π + −1= + −1 = 2 6 4 12 2 1 1 70. tan3 4x(1 + tan2 4x) sec2 4x dx = (tan3 4x + tan5 4x) sec2 4x dx = tan4 4x + tan6 4x + C 16 24 x+3 71. dx, let u = x + 1, (x + 1)2 + 1 u+2 2 √ du = u(u2 + 1)−1/2 + √ du = u2 + 1 + 2 sinh−1 u + C u2 + 1 u2 + 1 = x2 + 2x + 2 + 2 sinh−1 (x + 1) + C

69. u = sin−1 x, dv = dx, du = √

Alternate solution: let x + 1 = tan θ, (tan θ + 2) sec θ dθ = sec θ tan θ dθ + 2 sec θ dθ = sec θ + 2 ln | sec θ + tan θ| + C =

x2 + 2x + 2 + 2 ln( x2 + 2x + 2 + x + 1) + C.

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394

Chapter 8

72. Let x = tan θ to get

1 dx. − x2

C 1 A B = + 2+ ; A = −1, B = −1, C = 1 so x2 (x − 1) x x x−1 1 1 1 1 dx − dx = − ln |x| + + ln |x − 1| + C dx + − x x2 x−1 x x − 1 tan θ − 1 1 + C = cot θ + ln |1 − cot θ| + C = + ln + C = cot θ + ln x x tan θ 73.

74.

lim −

→+∞

1 2(x2 + 1)

1 bx tan−1 →+∞ ab a

= lim a

→+∞

−

= lim

lim

→+∞

1 1 1 = + 2(.2 + 1) 2(a2 + 1) 2(a2 + 1)

1 b. π tan−1 = ab a 2ab