CHAPTER 7
Principles of Integral Evaluation
EXERCISE SET 7.1
1. u = 4 − 2x, du = −2dx, −
3
2. u = 4 + 2x, du = 2dx,
2
Z
Z
3. u = x , du = 2xdx,
1
2
4. u = x2 , du = 2xdx,
2
2
Z
1
2
Z
√
1
1
u3 du = − u4 + C = − (4 − 2x)4 + C
8
8
u du = u3/2 + C = (4 + 2x)3/2 + C
tan u du = −2 ln | cos u | + C = −2 ln | cos(x2 )| + C
5. u = 2 + cos 3x, du = −3 sin 3xdx,
6. u =
2
1
2
x, du = dx,
3
3
6
Z
1
8. u = ln x, du = dx,
x
−
1
3
Z
du
1
1
= − ln |u| + C = − ln(2 + cos 3x) + C
u
3
3
du
1
1
2
= tan−1 u + C = tan−1 x + C
2
1+u
6
6
3
Z
7. u = ex , du = ex dx,
1
1
tan u + C = tan(x2 ) + C
2
2
sec2 u du =
sinh u du = cosh u + C = cosh ex + C
Z
sec u tan u du = sec u + C = sec(ln x) + C
Z
2
9. u = tan x, du = sec xdx,
Z
du
1
1
= sin−1 u + C = sin−1 (x2 ) + C
2
2
2
1−u
Z
1
1
1
11. u = cos 5x, du = −5 sin 5xdx, −
u5 du = − u6 + C = − cos6 5x + C
5
30
30
2
10. u = x , du = 2xdx,
1
2
eu du = eu + C = etan x + C
√
Z
12. u = sin x, du = cos x dx,
13. u = ex , du = ex dx,
14. u = tan−1 x, du =
√
Z
p
1 + 1 + sin2 x 1 + √1 + u2 du
√
= − ln + C = − ln +C
u
sin x
u u2 + 1
p
p
du
= ln u + u2 + 4 + C = ln ex + e2x + 4 + C
4 + u2
Z
−1
dx,
eu du = eu + C = etan x + C
√
1
1 + x2
Z
1
x − 1, du = √
dx,
2 x−1
2
16. u = x2 + 2x, du = (2x + 2)dx,
1
2
15. u =
17. u =
√
1
x, du = √ dx,
2 x
√
eu du = 2eu + C = 2e
Z
cot u du =
x−1
1
1
ln | sin u| + C = ln sin |x2 + 2x| + C
2
2
Z
2 cosh u du = 2 sinh u + C = 2 sinh
343
+C
√
x+C
344
Chapter 7
Z
dx
18. u = ln x, du =
,
x
19. u =
√
1
1
du
=− +C =−
+C
2
u
u
ln x
Z
1
x, du = √ dx,
2 x
2 du
=2
3u
Z
e−u ln 3 du = −
2 −u ln 3
2 −√x
e
+C =−
3
+C
ln 3
ln 3
Z
20. u = sin θ, du = cos θdθ,
21. u =
Z
22.
sec u tan u du = sec u + C = sec(sin θ) + C
2
2
, du = − 2 dx,
x
x
√
Z
csch2 u du =
1
1
2
coth u + C = coth + C
2
2
x
p
dx
= ln x + x2 − 4 + C
x2 − 4
Z
23. u = e−x , du = −e−x dx,
1
24. u = ln x, du = dx,
x
26. u = x
−1/2
−
cos u du = sin u + C = sin(ln x) + C
√
ex dx
=
1 − e2x
1
, du = − 3/2 dx,
2x
1
2
27. u = x2 , du = 2xdx,
28. 2u = ex , 2du = ex dx,
1 2 + u 1 2 + e−x du
+C
=
−
+
C
=
−
ln
ln
4 − u2
4 2 − u
4 2 − e−x Z
Z
25. u = ex , du = ex dx,
2
1
2
−
Z
Z
Z
−
1
du
=
csc u
2
√
Z
√
du
= sin−1 u + C = sin−1 ex + C
1 − u2
2 sinh u du = −2 cosh u + C = −2 cosh(x−1/2 ) + C
Z
1
1
sin u du = − cos u + C = − cos(x2 ) + C
2
2
2du
= sin−1 u + C = sin−1 (ex /2) + C
4 − 4u2
2
29. 4−x = e−x ln 4 , u = −x2 ln 4, du = −2x ln 4 dx = −x ln 16 dx,
Z
1 u
1 −x2 ln 4
1 −x2
1
eu du = −
e +C =−
e
+C =−
4
+C
−
ln 16
ln 16
ln 16
ln 16
30. 2πx = eπx ln 2 ,
Z
2πx dx =
1 πx ln 2
1
e
+C =
2πx + C
π ln 2
π ln 2
Z
1
1
31. (a) u = sin x, du = cos x dx,
u du = u2 + C = sin2 x + C
2
2
Z
Z
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
=
cosh 2x
cosh2 x + sinh2 x
sech2 x
=
1 + tanh2 x
32. (a) sech 2x =
(now multiply top and bottom by sech2 x)
Exercise Set 7.2
345
Z
(b)
Z
sech2x dx =
Z
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
Z
Z
x
e
(d)
sech x dx = 2
dx = 2 tan−1 (ex ) + C
2x
e +1
(c) sech x =
sec2 x
1
1
=
=
2
tan x
cos x tan x
cos x sin x
Z
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
=
= csc(π/2 − x), so
(c) sec x =
cos
x
sin(π/2
− x)
Z
Z
1
sec x dx = csc(π/2 − x) dx = − ln tan(π/2 − x) + C
2
33. (a)
EXERCISE SET 7.2
1. u = x, dv = e−2x dx, du = dx, v = − 21 e−2x ;
Z
Z
1 −2x
1
1
1 −2x
−2x
xe
dx = − xe
+
e
dx = − xe−2x − e−2x + C
2
2
2
4
1
2. u = x, dv = e dx, du = dx, v = e3x ;
3
3x
3. u = x2 , dv = ex dx, du = 2x dx, v = ex ;
Z
For
Z
Z
Z
1
1
xe dx = xe3x −
3
3
3x
x2 ex dx = x2 ex − 2
Z
Z
e3x dx =
1 3x 1 3x
xe − e + C
3
9
xex dx.
xex dx use u = x, dv = ex dx, du = dx, v = ex to get
xex dx = xex − ex + C1 so
Z
x2 ex dx = x2 ex − 2xex + 2ex + C
1
4. u = x2 , dv = e−2x dx, du = 2x dx, v = − e−2x ;
2
Z
For xe−2x dx use u = x, dv = e−2x dx to get
Z
1
x2 e−2x dx = − x2 e−2x +
2
Z
Z
1
1
1
1
xe−2x dx = − xe−2x +
e−2x dx = − xe−2x − e−2x + C
2
2
2
4
Z
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
Z
Z
1
1
1
1
x sin 3x dx = − x cos 3x +
cos 3x dx = − x cos 3x + sin 3x + C
3
3
3
9
Z
xe−2x dx
346
Chapter 7
1
6. u = x, dv = cos 2x dx, du = dx, v = sin 2x;
2
Z
Z
1
1
1
1
x cos 2x dx = x sin 2x −
sin 2x dx = x sin 2x + cos 2x + C
2
2
2
4
Z
Z
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
Z
For
x sin x dx use u = x, dv = sin x dx to get
Z
Z
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;
Z
Z
Z
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
Z
Z
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
9. u = ln x, dv = x dx, du = dx, v = x2 ;
x
2
Z
1
1
x ln x dx = x2 ln x −
2
2
Z
x dx =
1
1 2
x ln x − x2 + C
2
4
√
1
2
10. u = ln x, dv = x dx, du = dx, v = x3/2 ;
x
3
Z
Z
√
2
2
2
4
x ln x dx = x3/2 ln x −
x1/2 dx = x3/2 ln x − x3/2 + C
3
3
3
9
Z
Z
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
Z
Z
Use u = ln x, dv = dx to get
ln x dx = x ln x − dx = x ln x − x + C1 so
Z
(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
Z
√
ln x
√ dx = 2 x ln x−2
x
Z
√
√
1
√ dx = 2 x ln x−4 x+C
x
Z
Z
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
Z
Z 3x
2
2
but
dx =
1+
dx = x + ln(3x − 2) + C1 so
3x − 2
3x − 2
3
Z
2
ln(3x − 2)dx = x ln(3x − 2) − x − ln(3x − 2) + C
3
Z
Z
2x
x2
2
2
dx,
v
=
x;
ln(x
+
4)dx
=
x
ln(x
+
4)
−
2
dx
2
2
x +4
x +4
Z
Z x2
4
x
but
dx =
1− 2
dx = x − 2 tan−1 + C1 so
2
x +4
x +4
2
Z
x
ln(x2 + 4)dx = x ln(x2 + 4) − 2x + 4 tan−1 + C
2
14. u = ln(x2 + 4), dv = dx, du =
Exercise Set 7.2
347
√
15. u = sin−1 x, dv = dx, du = 1/ 1 − x2 dx, v = x;
Z
Z
p
p
sin−1 x dx = x sin−1 x − x/ 1 − x2 dx = x sin−1 x + 1 − x2 + C
16. u = cos−1 (2x), dv = dx, du = − √
Z
cos−1 (2x)dx = x cos−1 (2x) +
2
dx, v = x;
1 − 4x2
Z
√
1p
2x
dx = x cos−1 (2x) −
1 − 4x2 + C
2
2
1 − 4x
3
17. u = tan−1 (3x), dv = dx, du =
dx, v = x;
1 + 9x2
Z
Z
1
3x
tan−1 (3x)dx = x tan−1 (3x) −
dx = x tan−1 (3x) − ln(1 + 9x2 ) + C
1 + 9x2
6
Z
Z
1 2
1
1 2
1
x2
−1
−1
dx,
v
=
dx
18. u = tan−1 x, dv = x dx, du =
x
;
x
tan
x
dx
=
x
tan
x
−
2
1+x
2
2
2
1 + x2
Z
Z x2
1
but
dx =
1−
dx = x − tan−1 x + C1 so
1 + x2
1 + x2
Z
1
1
1
x tan−1 x dx = x2 tan−1 x − x + tan−1 x + C
2
2
2
Z
Z
x
x
x
x
19. u = e , dv = sin x dx, du = e dx, v = − cos x; e sin x dx = −e cos x + ex cos x dx.
Z
ex cos x dx use u = ex , dv = cos x dx to get
For
Z
ex sin x dx = −ex cos x + ex sin x −
Z
x
Z
e sin x dx = e (sin x − cos x) + C1 ,
2
20. u = e3x , dv = cos 2x dx, du = 3e3x dx, v =
Z
e3x cos 2x dx =
Z
Z
ex sin x dx =
Z
e
e
3x
1
3
9
cos 2x dx = e3x sin 2x + e3x cos 2x −
2
4
4
e
3x
ex sin x dx so
1 x
e (sin x − cos x) + C
2
e3x sin 2x dx. Use u = e3x , dv = sin 2x dx to get
1
3
sin 2x dx = − e3x cos 2x +
2
2
Z
Z
1
sin 2x;
2
3x
Z
13
4
3
1 3x
e sin 2x −
2
2
ex cos x = ex sin x −
ex sin x dx,
Z
x
Z
e3x cos 2x dx, so
Z
e3x cos 2x dx,
1
cos 2x dx = e3x (2 sin 2x + 3 cos 2x) + C1 ,
4
Z
e3x cos 2x dx =
1 3x
e (2 sin 2x + 3 cos 2x) + C
13
cos(ln x)
dx, v = x;
x
Z
Z
sin(ln x)dx = x sin(ln x) − cos(ln x)dx. Use u = cos(ln x), dv = dx to get
21. u = sin(ln x), dv = dx, du =
Z
Z
cos(ln x)dx = x cos(ln x) +
sin(ln x)dx so
348
Chapter 7
Z
Z
sin(ln x)dx = x sin(ln x) − x cos(ln x) −
Z
sin(ln x)dx =
sin(ln x)dx,
1
x[sin(ln x) − cos(ln x)] + C
2
1
22. u = cos(ln x), dv = dx, du = − sin(ln x)dx, v = x;
x
Z
Z
cos(ln x)dx = x cos(ln x) + sin(ln x)dx. Use u = sin(ln x), dv = dx to get
Z
Z
sin(ln x)dx = x sin(ln x) −
cos(ln x)dx so
Z
Z
cos(ln x)dx = x cos(ln x) + x sin(ln x) −
Z
cos(ln x)dx =
cos(ln x)dx,
1
x[cos(ln x) + sin(ln x)] + C
2
23. u = x, dv = sec2 x dx, du = dx, v = tan x;
Z
Z
Z
sin x
2
x sec x dx = x tan x − tan x dx = x tan x −
dx = x tan x + ln | cos x| + C
cos x
24. u = x, dv = tan2 x dx = (sec2 x − 1)dx, du = dx, v = tan x − x;
Z
Z
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
25. u = x2 , dv = xex dx, du = 2x dx, v =
Z
2
1
x e dx = x2 ex −
2
3 x2
Z
2
xex dx =
1 x2
e ;
2
1 2 x2 1 x2
x e − e +C
2
2
1
1
dx, du = (x + 1)ex dx, v = −
;
2
(x + 1)
x+1
Z
Z
xex
xex
xex
ex
dx
=
−
+
ex dx = −
+ ex + C =
+C
2
(x + 1)
x+1
x+1
x+1
26. u = xex , dv =
27. u = x, dv = e2x dx, du = dx, v =
Z
0
2
xe2x dx =
1 2x
xe
2
2
−
0
1
2
Z
0
2
1 2x
e ;
2
1
e2x dx = e4 − e2x
4
1
28. u = x, dv = e−5x dx, du = dx, v = − e−5x ;
5
2
0
1
= e4 − (e4 − 1) = (3e4 + 1)/4
4
Exercise Set 7.2
Z
1
xe
−5x
0
349
1
1
dx = − xe−5x
5
1
+
5
0
1
Z
1
1
= − e−5 − e−5x
5
25
e−5x dx
0
1
0
1
1
= − e−5 − (e−5 − 1) = (1 − 6e−5 )/25
5
25
1
1
dx, v = x3 ;
x
3
e
e
Z e
Z e
1
1
1
1
1
1
x2 ln x dx = x3 ln x −
x2 dx = e3 − x3 = e3 − (e3 − 1) = (2e3 + 1)/9
3
3
3
9
3
9
1
1
1
1
29. u = ln x, dv = x2 dx, du =
1
1
1
dx, du = dx, v = − ;
x2
x
x
e
Z e
ln x
1
1
dx
=
−
dx
ln
x
+
√ x2
√
x2
x
e
e
e
√
√
1
1
1
1
1
1
1
3 e−4
√ − +√ =
= − + √ ln e −
+
=
−
e
x √e
e 2 e e
2e
e
e
30. u = ln x, dv =
Z
e
√
e
1
dx, v = x;
x+2
1
Z 1
Z 1
Z 1
2
x
dx = ln 3 + ln 1 −
1−
dx
ln(x + 2)dx = x ln(x + 2)
−
x+2
−1
−1
−1 x + 2
−1
1
= ln 3 − [x − 2 ln(x + 2)]
= ln 3 − (1 − 2 ln 3) + (−1 − 2 ln 1) = 3 ln 3 − 2
31. u = ln(x + 2), dv = dx, du =
−1
32. u = sin−1 x, dv = dx, du = √
√
Z
√3/2
3/2
−1
sin
−1
x dx = x sin
1
dx, v = x;
1 − x2
−
x
0
√
Z
0
0
3/2
√
√
√3/2
3
3 p
x
−1
√
sin
+ 1 − x2
dx =
2
2
1 − x2
0
√
√ 3 π
π 3 1
1
−
=
+ −1=
2 3
2
6
2
√
1
√
dθ, v = θ;
2θ θ − 1
4
Z 4
√
√ 4 1Z 4
√
√
1
−1
−1
−1
−1
√
sec
θdθ = θ sec
θ −
dθ = 4 sec 2 − 2 sec
2− θ−1
2 2
θ−1
2
2
2
π
π √
√
5π
=4
−2
− 3+1=
− 3+1
3
4
6
33. u = sec−1
θ, dv = dθ, du =
1
1
34. u = sec−1 x, dv = x dx, du = √
dx, v = x2 ;
2
x x2 − 1
2
Z 2
Z
1
1 2
x
√
x sec−1 x dx = x2 sec−1 x −
dx
2−1
2
2
x
1
1
1
2
√
1
1p 2
= [(4)(π/3) − (1)(0)] −
x − 1 = 2π/3 − 3/2
2
2
1
350
Chapter 7
1
35. u = x, dv = sin 2x dx, du = dx, v = − cos 2x;
2
π
π
Z π
Z 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
Z
π
36.
(x + x cos x)dx =
0
1 2
x
2
π
Z
+
π
x cos x dx =
0
0
π2
+
2
Z
π
x cos x dx;
0
u = x, dv = cos x dx, du = dx, v = sin x
π Z π
π
Z π
Z 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
xdx, du = √
dx, v = x3/2 ;
3
2 x(1 + x)
3
Z 3
Z 3
√
√
√
2
1
x
x tan−1 xdx = x3/2 tan−1 x −
dx
3
3
1
+
x
1
1
1
Z √ 3 1 3
1
2 3/2
−1
x −
1−
dx
= x tan
3
3 1
1+x
1
3
√
√
1
1
2 3/2
−1
x tan
x − x + ln |1 + x| = (2 3π − π/2 − 2 + ln 2)/3
=
3
3
3
1
37. u = tan−1
x, dv =
2x
dx, v = x;
x2 + 1
2 Z 2
Z 2
Z 2
1
2x2
2
2
dx = 2 ln 5 − 2
dx
1− 2
ln(x + 1)dx = x ln(x + 1) −
2
x +1
0
0
0 x +1
0
2
−1
= 2 ln 5 − 2(x − tan x) = 2 ln 5 − 4 + 2 tan−1 2
38. u = ln(x2 + 1), dv = dx, du =
0
39. true
40. false; choose u = ln x
41. false; ex is not a factor of the integrand
42. true; the column of p(x) eventually has zero entries
43. t =
√
√
Z
e
√
Z
e
x, t2 = x, dx = 2t dt
Z
x
dx = 2
tet dt; u = t, dv = et dt, du = dt, v = et ,
x
dx = 2tet − 2
Z
√
√
et dt = 2(t − 1)et + C = 2( x − 1)e x + C
√
44. Z
t = x, t2 = x, dx
Z = 2t dt
√
cos x dx = 2 t cos t dt; u = t, dv = cos tdt, du = dt, v = sin t,
Z
cos
√
Z
x dx = 2t sin t − 2
√
√
√
sin tdt = 2t sin t + 2 cos t + C = 2 x sin x + 2 cos x + C
Exercise Set 7.2
351
45. Let f1 (x), f2 (x), f3 (x) denote successive antiderivatives of f (x),
so that f30 (x) = f2 (x), f20 (x) = f1 (x), f10 (x) = f (x). Let p(x) = ax2 + bx + c.
Repeated
Differentiation
Repeated
Antidifferentiation
ax2 + bx + c
PP +
PP
PP
2ax + b
PP
P−
PP
P
2a
PP
+
PP
PP
0
Z
Then
f (x)
f1 (x)
f2 (x)
f3 (x)
p(x)f (x) dx = (ax2 + bx + c)f1 (x) − (2ax + b)f2 (x) + 2af3 (x) + C. Check:
d
[(ax2 +bx + c)f1 (x) − (2ax + b)f2 (x) + 2af3 (x)]
dx
= (2ax + b)f1 (x) + (ax2 + bx + c)f (x) − 2af2 (x) − (2ax + b)f1 (x) + 2af2 (x) = p(x)f (x)
Z
46. Let I denote
ex cos x dx. Then (Method 1)
Repeated
Differentiation
Repeated
Antidifferentiation
ex
ex
ex
cos x
PP+
P
PP−
P
PP sin x
P
PP − cos x
P
and thus I = ex (sin x + cos x) − I, so I =
On the other hand (Method 2)
Differentiation
cos x
1 x
e (sin x + cos x) + C.
2
Repeated
Antidifferentiation
ex
PP +
PP
− sin x
PPex
PP −
PP
− cos x
PPex
and thus I = ex (sin x + cos x) − I, so I =
1 x
e (sin x + cos x) + C, as before.
2
352
Chapter 7
47.
Repeated
Differentiation
Repeated
Antidifferentiation
3x2 − x + 2
e−x
PP +
PP
6x − 1
PP −e−x
PP −
PP
6
PP e−x
PP +
PP
0
PP −e−x
Z
48.
(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
PP +
PP
PP − cos x
2x + 1
PP −
PP
PP − sin x
2
PP +
PP
PP cos x
0
Z
(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
49.
Repeated
Differentiation
Repeated
Antidifferentiation
4x4
sin 2x
PP +
PP
PP 1
16x3
− cos 2x
2
PP −
PP
1
PP
48x2
− sin 2x
4
PP +
PP
PP 1
96x
cos 2x
8
PP −
PP
PP 1
96
sin 2x
16
PP +
PP
PP 1
0
− cos 2x
32
Z
4x4 sin 2x dx = (−2x4 + 6x2 − 3) cos 2x + (4x3 − 6x) sin 2x + C
Exercise Set 7.2
50.
353
Repeated
Differentiation
x3
Repeated
Antidifferentiation
√
2x + 1
HH
+
H
HH
2
3x
HH
−
H
HH
6x
HH
+
H
HH
6
HH
−
H
HH
0
Z
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
51.
Repeated
Differentiation
Repeated
Antidifferentiation
eax
sin bx
HH +
HH
H − 1 cos bx
aeax
b
HH −
HH
H − 1 sin bx
a2 eax
b2
1
a
a2
eax sin bx dx = − eax cos bx + 2 eax sin bx − 2 I, so
b
b
b
eax
I= 2
(a sin bx − b cos bx) + C
a + b2
Z
I=
e−3θ
52. From Exercise 51 with a = −3, b = 5, x = θ, answer = √ (−3 sin 5θ − 5 cos 5θ) + C
34
53. (a) We perform a single integration by parts:
Z
u = cos x, dv = sin x dx, du = − sin x dx, v = − cos x,
2
sin x cos x dx = − cos x−
Z
sin x cos x dx.
Thus
Z
Z
1
2 sin x cos x dx = − cos2 x + C, sin x cos x dx = − cos2 x + C
2
Z
Z
1
1
Alternatively, u = sin x, du = cos x dx, sin x cos x dx = u du = u2 + C = sin2 x + C
2
2
(b) Since sin2 x + cos2 x = 1, they are equal (although the symbol ’C’ refers to different constants
in the two equations).
p
x
54. (a) u = x2 , dv = √
, du = 2x dx, v = x2 + 1,
x2 + 1
1 Z 1 p
1
Z 1
3
p
√
x
2
1√
2
√
2x x2 + 1 dx = 2 − (x2 + 1)3/2 = −
dx = x2 x2 + 1 −
2+
3
3
3
x2 + 1
0
0
0
0
354
Chapter 7
√
p
x
dx,
(b) u = x2 + 1, du = √
2
x +1
√
1
1√
2√
2− 2− +1=−
2+
=
3
3
3
2
2
(u − 1) du =
1
55. (a) A =
1
2
.
3
1
e
Z
h
(ln x)2 dx = π (x(ln x)2 − 2x ln x + 2x)
(b) V = π
π/2
56. A =
0
Z
1
(x − x sin x)dx = x2
2
π
57. V = 2π
π/2
Z
π/2
−
0
0
= π(e − 2)
π/2
π2
− (−x cos x + sin x)
= π 2 /8 − 1
x sin x dx =
8
0
π
x sin x dx = 2π(−x cos x + sin x) = 2π 2
0
Z
ie
1
1
Z
√2
1 3
u −u
3
1
e
ln x dx = (x ln x − x) = 1
e
Z
Z
0
π/2
π/2
58. V = 2π
= π(π − 2)
x cos x dx = 2π(cos x + x sin x)
0
0
Z
59. distance =
π
t3 sin tdt;
0
Repeated
Differentiation
Repeated
Antidifferentiation
t3
sin t
HH +
HH
3t2
H − cos t
H
HH
−
HH − sin t
6t
HH
+
H
HH cos t
6
HH
−
H
HH sin t
0
Z
π
π
t3 sin t dx = [(−t3 cos t + 3t2 sin t + 6t cos t − 6 sin t)] = π 3 − 6π
0
0
1
cos(kωt); the integrand is an even function of t so
kω
π/ω
Z π/ω
Z π/ω
Z π/ω
2
1
t sin(kωt) dt = 2
t sin(kωt) dt = − t cos(kωt)
+2
cos(kωt) dt
kω
kω
−π/ω
0
0
0
π/ω
2π(−1)k+1
2
2π(−1)k+1
=
+
sin(kωt)
=
2
2
2
kω
k ω
kω 2
0
60. u = 2t, dv = sin(kωt)dt, du = 2dt, v = −
Exercise Set 7.2
355
Z
3
sin2 x dx
61. (a)
4
3
1
1
− sin x cos x + x + C
4
2
2
3
3
sin x cos x + x + C
8
8
π/2
Z π/2
Z π/2
1
4
sin3 x dx
(b)
sin5 x dx = − sin4 x cos x
+
5
5
0
0
0 #
"
Z
π/2
4
1
2 π/2
2
=
sin x dx
− sin x cos x
+
5
3
3 0
0
π/2
8
8
= − cos x
=
15
15
0
Z
Z
62. (a)
1
sin x dx = − sin3 x cos x +
4
1
= − sin3 x cos x +
4
1
= − sin3 x cos x −
4
4
1
4
cos4 x sin x +
5
5
cos5 x dx =
Z
cos3 x dx =
1
4 1
2
cos4 x sin x +
cos2 x sin x + sin x + C
5
5 3
3
1
cos4 x sin x +
5
Z
1
(b)
cos6 x dx = cos5 x sin x +
6
4
8
cos2 x sin x +
sin x + C
15
15
Z
5
cos4 x dx
6
Z
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
=
63. u = sinn−1 x, dv = sin x dx, du = (n − 1) sinn−2 x cos x dx, v = − cos x;
Z
Z
sinn x dx = − sinn−1 x cos x + (n − 1) sinn−2 x cos2 x dx
n−1
= − sin
n
Z
n
n−1
sin x dx = − sin
sinn x dx = −
sinn−2 x (1 − sin2 x)dx
Z
sinn−2 x dx − (n − 1)
x cos x + (n − 1)
= − sinn−1 x cos x + (n − 1)
Z
Z
Z
Z
sinn x dx,
sinn−2 x dx,
x cos x + (n − 1)
1
n−1
sinn−1 x cos x +
n
n
Z
sinn−2 x dx
64. (a) u = secn−2 x, dv = sec2 x dx, du = (n − 2) secn−2 x tan x dx, v = tan x;
Z
Z
secn x dx = secn−2 x tan x − (n − 2) secn−2 x tan2 x dx
= secn−2 x tan x − (n − 2)
n−2
= sec
Z
Z
x tan x − (n − 2)
secn−2 x (sec2 x − 1)dx
n
sec x dx + (n − 2)
Z
secn−2 x dx,
356
Chapter 7
Z
n
(n − 1)
Z
n−2
x tan x + (n − 2)
sec x dx = sec
1
n−2
sec x dx =
secn−2 x tan x +
n−1
n−1
Z
n
Z
tann x dx =
(b)
=
Z
Z
tan4 x dx =
Z
tann−2 x (sec2 x − 1) dx =
1
tann−1 x −
n−1
Z
1
tan3 x −
3
Z
secn−2 x dx
tann−2 x sec2 x dx −
Z
tann−2 x dx
tann−2 x dx
(c) u = xn , dv = ex dx, du = nxn−1 dx, v = ex ;
65. (a)
Z
secn−2 x dx,
tan2 x dx =
Z
xn ex dx = xn ex − n
1
tan3 x − tan x +
3
Z
Z
dx =
xn−1 ex dx
1
tan3 x − tan x + x + C
3
Z
2
1
2
1
sec2 x tan x +
sec2 x dx = sec2 x tan x + tan x + C
3
3
3
3
Z
Z
Z
3 x
3 x
2 x
3 x
2 x
x
(c)
x e dx = x e − 3 x e dx = x e − 3 x e − 2 xe dx
Z
sec4 x dx =
(b)
Z
= x3 ex − 3x2 ex + 6 xex − ex dx = x3 ex − 3x2 ex + 6xex − 6ex + C
66. (a) u = 3x,
Z
Z
Z
Z
1
1 2 u
2
1
u2 eu du =
u2 eu − 2 ueu du =
u e −
ueu − eu du
x2 e3x dx =
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
Z
xe−
√
x
−1
Z
u3 eu du,
dx = 2
0
0
Z
3 u
Z
3 u
u e du = u e − 3
Z
2 u
u
u e du = u e − 3 u e − 2 ue du
2 u
3 u
Z
= u3 eu − 3u2 eu + 6 ueu − eu du = u3 eu − 3u2 eu + 6ueu − 6eu + C,
Z
−1
3 u
3
2
u e du = 2(u − 3u + 6u − 6)e
2
0
u
−1
= 12 − 32e−1
0
67. u = x, dv = f 00 (x)dx, du = dx, v = f 0 (x);
Z
1
−1
1
Z
x f 00 (x)dx = xf 0 (x)
−
−1
1
f 0 (x)dx
−1
1
= f 0 (1) + f 0 (−1) − f (x)
= f 0 (1) + f 0 (−1) − f (1) + f (−1)
−1
Exercise Set 7.3
357
Z
68. (a)
Z
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.
Z
Z
Z
(b) u(v + C1 ) − (v + C1 )du = uv + C1 u − v du − C1 u = uv − v du
dx
69. u = ln(x + 1), dv = dx, du =
, v = x + 1;
x+1
Z
Z
Z
Z
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
Z
Z
Z
Z 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
70. u = ln(3x − 2), dv = dx, du =
1
1
71. u = tan−1 x, dv = x dx, du =
dx, v = (x2 + 1)
1 + x2
2
Z
Z
Z
Z
1
1
x tan−1 x dx = u dv = uv − v du = (x2 + 1) tan−1 x −
dx
2
2
=
1 2
1
(x + 1) tan−1 x − x + C
2
2
1
1
1
72. u =
, dv = dx, du = −
dx, v = ln x
ln x
x
x(ln x)2
Z
Z
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.
EXERCISE SET 7.3
Z
Z
1
1
1
u3 du = − cos4 x + C
2. u = sin 3x,
u5 du =
sin6 3x + C
4
3
18
Z
1
1
1
sin2 5θ =
(1 − cos 10θ) dθ = θ −
sin 10θ + C
2
2
20
Z
1
1
1
cos2 3x dx =
(1 + cos 6x)dx = x +
sin 6x + C
2
2
12
Z
1
1
3
sin aθ dθ = sin aθ(1 − cos2 aθ) dθ = − cos aθ −
cos3 aθ + C (a 6= 0)
a
3a
Z
cos3 at dt = (1 − sin2 at) cos at dt
1. u = cos x, −
Z
3.
Z
4.
Z
5.
Z
6.
Z
=
Z
cos at dt −
sin2 at cos at dt =
1
1
sin at −
sin3 at + C (a 6= 0)
a
3a
358
Chapter 7
7. u = sin ax,
Z
8.
1
a
Z
u du =
sin3 x cos3 x dx =
Z
sin3 x(1 − sin2 x) cos x dx
Z
(sin3 x − sin5 x) cos x dx =
1
1
sin4 x − sin6 x + C
4
6
Z
(sin2 t − sin4 t) cos t dt
=
Z
9.
Z
sin2 t cos3 t dt =
1
sin2 ax + C, a 6= 0
2a
sin2 t(1 − sin2 t) cos t dt =
1
1
sin3 t − sin5 t + C
3
5
Z
Z
10.
sin3 x cos2 x dx = (1 − cos2 x) cos2 x sin x dx
=
Z
1
1
(cos2 x − cos4 x) sin x dx = − cos3 x + cos5 x + C
3
5
Z
Z
Z
1
1
1
1
sin2 x cos2 x dx =
sin2 2x dx =
(1 − cos 4x)dx = x −
sin 4x + C
11.
4
8
8
32
Z
Z
Z
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
Z
Z
Z
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
Z
Z
1
1
1
sin 2x cos 3x dx =
(sin 5x − sin x)dx = − cos 5x + cos x + C
2
10
2
Z
Z
1
1
1
sin 3θ cos 2θdθ =
(sin 5θ + sin θ)dθ = − cos 5θ − cos θ + C
2
10
2
Z
Z
1
1
[sin(3x/2) + sin(x/2)]dx = − cos(3x/2) − cos(x/2) + C
sin x cos(x/2)dx =
2
3
Z
3
u = cos x, − u1/3 du = − cos4/3 x + C
4
=
13.
14.
15.
16.
Z
π/2
cos3 x dx =
17.
0
Z
π/2
(1 − sin2 x) cos x dx
0
π/2
2
1
3
=
= sin x − sin x
3
3
0
Z
π/2
sin2 (x/2) cos2 (x/2)dx =
18.
0
=
Z
19.
0
π/3
sin4 3x cos3 3x dx =
Z
0
1
4
Z
1
8
π/2
sin2 x dx =
0
x−
1
sin 2x
2
1
8
Z
π/2
(1 − cos 2x)dx
0
π/2
= π/16
0
π/3
sin4 3x(1 − sin2 3x) cos 3x dx =
1
1
sin5 3x −
sin7 3x
15
21
π/3
=0
0
Exercise Set 7.3
Z
359
π
cos2 5θ dθ =
20.
−π
Z
π/6
21.
0
1
2
π
Z
(1 + cos 10θ)dθ =
−π
1
sin 4x cos 2x dx =
2
π/6
Z
0
1
2
θ+
1
sin 10θ
10
π
=π
−π
π/6
1
1
(sin 2x + sin 6x)dx = − cos 2x −
cos 6x
4
12
0
= [(−1/4)(1/2) − (1/12)(−1)] − [−1/4 − 1/12] = 7/24
Z
2π
22.
0
23.
1
sin kx dx =
2
2
−x
, du = −e
−x
√
Z
dx; −
tan u du = ln | cos u| + C = ln | cos(e−x )| + C
29. u = tan x,
Z
30.
Z
31.
Z
32.
Z
33.
Z
34.
27.
1
x, du = √ dx;
2 x
Z
u2 du =
1
tan3 x + C
3
Z
tan 4x(1 + tan2 4x) sec2 4x dx =
tan4 θ(1 + tan2 θ) sec2 θ dθ =
(tan5 x + tan7 x) sec2 x dx =
Z
(sec2 θ − 1)2 sec θ tan θdθ =
2
Z
(sec x − 1) sec x dx =
=
1
3
sec3 x tan x +
4
4
Z
Z
1
1
tan6 x + tan8 x + C
6
8
(tan 4x + tan3 4x) sec2 4x dx =
1
1
tan2 4x +
tan4 4x + C
8
16
1
1
tan5 θ + tan7 θ + C
5
7
sec4 x(sec2 x − 1) sec x tan x dx =
2
1
ln | sec 4x + tan 4x| + C
4
√
√ 2 sec u du = 2 ln | sec u + tan u| + C = 2 ln sec x + tan x + C
Z
tan5 x(1 + tan2 x) sec2 x dx =
Z
2π
1
1
x−
sin 2kx
sin 4πk (k 6= 0)
=π−
2k
4k
0
1
24. − ln | cos 5x| + C
5
1
ln | sin 3x| + C
3
28. u =
35.
0
1
(1 − cos 2kx)dx =
2
1
tan(2x − 1) + C
2
25. u = e
26.
2π
Z
Z
(sec6 x − sec4 x) sec x tan x dx =
(sec4 θ − 2 sec2 θ + 1) sec θ tan θdθ =
5
3
Z
(sec x − 2 sec x + sec x)dx =
sec3 x dx − 2
Z
1
1
sec7 x − sec5 x + C
7
5
1
2
sec5 θ − sec3 θ + sec θ + C
5
3
5
sec x dx − 2
Z
3
sec x dx +
sec3 x dx + ln | sec x + tan x|
1
5 1
1
3
= sec 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
Z
sec x dx
360
Chapter 7
Z
2
Z
[sec x − 1] sec x dx = [sec5 x − sec3 x]dx
Z
Z
1
3
3
3
=
sec x tan x +
sec x dx − sec3 x dx
4
4
Z
1
1
3
sec3 x dx
= sec x tan x −
4
4
36.
3
(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
Z
Z
1
1
38.
sec4 x(sec x tan x)dx = sec5 x + C
37.
sec2 t(sec t tan t)dt = sec3 t + C
3
5
=
Z
Z
4
39.
sec x dx =
2
Z
2
(sec2 x + tan2 x sec2 x)dx = tan x +
(1 + tan x) sec x dx =
1
tan3 x + C
3
40. Using equation (20),
Z
Z
3
1
sec3 x dx
sec5 x dx = sec3 x tan x +
4
4
=
1
3
3
sec3 x tan x + sec x tan x + ln | sec x + tan x| + C
4
8
8
41. u = 4x, use equation (19) to get
Z
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
Z
42. Use equation (19) to get
43.
Z √
Z
tan x(1 + tan2 x) sec2 x dx =
sec1/2 x(sec x tan x)dx =
44.
π/8
Z
45.
0
π/6
2
sec3/2 x + C
3
1
(sec 2x − 1)dx =
tan 2x − x
2
π/8
0
= 1/2 − π/8
0
1
sec 2θ(sec 2θ tan 2θ)dθ = sec3 2θ
6
2
46.
1
tan3 x − tan x + x + C
3
2
2
tan3/2 x + tan7/2 x + C
3
7
2
Z
tan4 x dx =
π/6
= (1/6)(2)3 − (1/6)(1) = 7/6
0
47. u = x/2,
Z
π/4
2
0
1
tan u du =
tan4 u − tan2 u − 2 ln | cos u|
2
48. u = πx,
Z
49.
5
2
1
π
Z
π/4
sec u tan u du =
0
2
π/4
√
= 1/2 − 1 − 2 ln(1/ 2) = −1/2 + ln 2
0
π/4
√
1
sec u
= ( 2 − 1)/π
π
0
(csc x − 1) csc x(csc x cot x)dx =
Z
1
1
(csc4 x − csc2 x)(csc x cot x)dx = − csc5 x + csc3 x + C
5
3
Exercise Set 7.3
Z
50.
Z
51.
Z
52.
361
1
cos2 3t
·
dt =
2
sin 3t cos 3t
Z
1
csc 3t cot 3t dt = − csc 3t + C
3
(csc2 x − 1) cot x dx =
Z
Z
csc x(csc x cot x)dx −
1
cos x
dx = − csc2 x − ln | sin x| + C
sin x
2
1
(cot2 x + 1) csc2 x dx = − cot3 x − cot x + C
3
53. true
54. false; the method would work but is tedious. Better to use the identity cos2 x = 1 − sin2 x and the
substitution u = sin x.
55. false
56. true
2π
Z
2π
cos(m + n)x cos(m − n)x
[sin(m+n)x+sin(m−n)x]dx = −
−
2(m + n)
2(m − n) 0
0
2π
= 0, cos(m − n)x
= 0.
1
2
sin mx cos nx dx =
57. (a)
0
2π
but cos(m + n)x
2π
Z
0
0
2π
Z
Z
2π
Z
2π
1
[cos(m + n)x + cos(m − n)x]dx;
2 0
since m 6= n, evaluate sin at integer multiples of 2π to get 0.
Z
Z 2π
1 2π
(c)
[cos(m − n)x − cos(m + n)x] dx;
sin mx sin nx dx =
2 0
0
since m 6= n, evaluate sin at integer multiples of 2π to get 0.
(b)
cos mx cos nx dx =
0
2π
Z
58. (a)
0
2π
Z
(b)
0
1
sin mx cos mx dx =
2
1
cos2 mx dx =
2
Z
2π
0
2π
1
cos 2mx = 0
sin 2mx dx = −
4m
0
0
1
(1 + cos 2mx) dx =
2
1
x+
sin 2mx
2m
2π
=π
0
2π
Z
Z 2π
1 2π
1
1
2
(c)
sin mx dx =
(1 − cos 2mx) dx =
x−
sin 2mx = π
2 0
2
2m
0
0
59. y 0 = tan x, 1 + (y 0 )2 = 1 + tan2 x = sec2 x,
π/4
Z
√
L=
sec2 x dx =
0
Z
π/4
π/4
√
sec x dx = ln | sec x + tan x|
= ln( 2 + 1)
0
Z
π/4
(1 − tan2 x)dx = π
60. V = π
0
Z
0
Z
π/4
1
(2 − sec2 x)dx = π(2x − tan x)
= π(π − 2)
2
0
π/4
0
π/4
2
2
Z
(cos x − sin x)dx = π
61. V = π
0
0
π/4
1
cos 2x dx = π sin 2x
2
#π/4
= π/2
0
362
Chapter 7
Z
π
62. V = π
sin2 x dx =
0
π
2
Z
π
(1 − cos 2x)dx =
0
L
63. With 0 < α < β, D = Dβ −Dα =
2π
Z
β
α
π
2
x−
1
sin 2x
2
π
= π 2 /2
0
L
sec x dx =
ln | sec x + tan x|
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 =
= 7.34 cm
ln 2π
sec 30◦ + tan 30◦ 64. (a) D =
Z
65. (a)
Z
csc x dx =
sec(π/2 − x)dx = − ln | sec(π/2 − x) + tan(π/2 − x)| + C
= − ln | csc x + cot x| + C
| csc x − cot x|
1
= ln
= ln | csc x − cot x|,
| csc x + cot x|
| csc2 x − cot2 x|
sin x 1
cos x +
=
ln
− ln | csc x + cot x| = − ln 1 + cos x sin x
sin x 2 sin(x/2) cos(x/2) = ln | tan(x/2)|
= ln 2 cos2 (x/2)
(b) − ln | csc x + cot x| = ln
66. sin x + cos x =
=
Z
i
√ h √
√
2 (1/ 2) sin x + (1/ 2) cos x
√
2 [sin x cos(π/4) + cos x sin(π/4)] =
1
dx
=√
sin x + cos x
2
√
2 sin(x + π/4),
Z
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 p
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)
67. a sin x + b cos x =
Z
and
p
dx
1
=√
2
a sin x + b cos x
a + b2
Z
1
ln | csc(x + θ) + cot(x + θ)| + C
+ b2
√
a2 + b2 + a cos x − b sin x 1
= −√
ln +C
2
2
a
sin
x
+
b
cos
x
a +b
csc(x + θ)dx = − √
a2
Exercise Set 7.4
Z
68. (a)
0
π/2
363
1
sin x dx = − sinn−1 x cos x
n
n
π/2
0
n−1
+
n
Z
π/2
n−2
sin
0
n−1
x dx =
n
Z
π/2
sinn−2 x dx
0
(b) By repeated application of the formula in Part (a)
Z π/2
Z π/2
n−1
n−3
n
sin x dx =
sinn−4 x dx
n
n−2
0
0
 Z π/2

n−3
n−5
1
n−1


dx, n even
···


n
n−2
n−4
2
0
=
Z π/2


n−1
n−3
n−5
2


sin x dx, n odd
···

n
n−2
n−4
3
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
Z π/2
Z π/2
1·3 π
2
(b)
· = 3π/16
sin4 x dx =
69. (a)
sin3 x dx =
3
2
·4 2
0
0
Z π/2
Z π/2
2·4
1·3·5 π
5
(c)
sin x dx =
= 8/15
(d)
· = 5π/32
sin6 x dx =
3
·
5
2·4·6 2
0
0
70. Similar to proof in Exercise 68.
EXERCISE SET 7.4
1. x = 2 sin θ, dx = 2 cos θ dθ,
Z
Z
4 cos2 θ dθ = 2 (1 + cos 2θ)dθ = 2θ + sin 2θ + C
1 p
= 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
Z
Z
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 p
θ + sin θ cos θ + C = sin−1 2x + x 1 − 4x2 + C
4
4
4
2
3. x = 4 sin θ, dx = 4 cos θ dθ,
Z
Z
16 sin2 θ dθ = 8 (1 − cos 2θ)dθ = 8θ − 4 sin 2θ + C = 8θ − 8 sin θ cos θ + C
1 p
= 8 sin−1 (x/4) − x 16 − x2 + C
2
4. x = 3 sin θ, dx = 3 cos θ dθ,
√
Z
Z
1
1
1
1
9 − x2
2
dθ =
csc θ dθ = − cot θ + C = −
+C
2
9
9
9
9x
sin θ
364
Chapter 7
5. x = 2 tan θ, dx = 2 sec2 θ dθ,
Z
Z
Z
1
1
1
1
1
1
2
dθ
=
cos
θ
dθ
=
(1 + cos 2θ)dθ =
θ+
sin 2θ + C
2
8
sec θ
8
16
16
32
=
1
1
x
1
x
θ+
sin θ cos θ + C =
tan−1 +
+C
16
16
16
2 8(4 + x2 )
√
√
6. x = 5 tan θ, dx = 5 sec2 θ dθ,
Z
Z
1
1
2
3
5 tan θ sec θ dθ = 5 (sec θ − sec θ)dθ = 5
sec θ tan θ − ln | sec θ + tan θ| + C1
2
2
√
p
p
2
5
5
5+x +x
1
1 p
√
= 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,
Z
Z
p
x
2
3 tan 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,
√
Z
Z
1
1
1
1
x2 − 16
dθ =
cos θ dθ =
sin θ + C =
+C
16
sec θ
16
16
16x
9. x = sin θ, dx = cos θ dθ,
Z
Z
3
3 sin θ dθ = 3
1 − cos2 θ sin θ dθ
p
= 3 − cos θ + cos3 θ + C = −3 1 − x2 + (1 − x2 )3/2 + C
√
5 sin θ, dx = 5 cos θ dθ,
√ Z
√
1
1
5
1
3
2
3
5
25 5 sin θ cos θ dθ = 25 5 − cos θ + cos θ + C = − (5 − x2 )3/2 + (5 − x2 )5/2 + C
3
5
3
5
10. x =
11. x =
√
2
2
3
sec θ, dx = sec θ tan θ dθ,
3
3
4
Z
3
1
dθ =
sec θ
4
Z
cos θ dθ =
3
1 p 2
9x − 4 + C
sin θ + C =
4
4x
12. t = tan θ, dt = sec2 θ dθ,
Z
Z
Z
sec3 θ
tan2 θ + 1
dθ =
sec θ dθ = (sec θ tan θ + csc θ)dθ
tan θ
tan θ
√
p
1 + t2 − 1
= sec θ + ln | csc θ − cot θ| + C = 1 + t2 + ln
+C
|t|
Z
13. x = sin θ, dx = cos θ dθ,
1
dθ =
cos2 θ
1
14. x = 5 tan θ, dx = 5 sec θ dθ,
25
2
Z
sec2 θ dθ = tan θ + C = x/
p
1 − x2 + C
√
Z
sec θ
1
x2 + 25
1
dθ
=
csc
θ
cot
θ
dθ
=
−
csc
θ+C
=
−
+C
25
25
25x
tan2 θ
Z
15. x = 3 sec θ, dx = 3 sec θ tan θ dθ,
Z
1
1p 2
sec θ dθ = ln | sec θ + tan θ| + C = ln x +
x − 9 + C
3
3
Exercise Set 7.4
365
16. 1 + 2x2 + x4 = (1 + x2 )2 , x = tan θ, dx = sec2 θ dθ,
Z
Z
Z
1
1
1
1
2
(1 + cos 2θ)dθ = θ + sin 2θ + C
dθ
=
cos
θ
dθ
=
sec2 θ
2
2
4
=
1
1
1
x
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
Z
Z
3
1
1 1
sec θ tan θ dθ
cos θ
1
x
=
dθ = −
+C
+ C = − csc θ + C = − √
2
18
18 sin θ
18
27 tan3 θ
sin2 θ
9 4x2 − 9
17. x =
18. x = 5 sec θ, dx = 5 sec θ tan θ dθ,
Z
= 375 sec4 θ dθ
Z
= 125 sec2 θ tan θ + 250 sec2 θ dθ
= 125psec2 θ tan θ + p
250 tan θ + C
2
2
= x x − 25 + 50 x2 − 25 + C
19. ex = sin θ, ex dx = cos θ dθ,
Z
Z
1
1
1
1 p
1
2
(1 + cos 2θ)dθ = θ + sin 2θ + C = sin−1 (ex ) + ex 1 − e2x + C
cos θ dθ =
2
2
4
2
2
Z
sin θ
1
√
20. u = sin θ,
du = sin−1 √
+C
2
2 − u2
21. x = sin θ, dx = cos θ dθ,
π/2
Z 1
1
1
3
3
5
2
= 5(1/3 − 1/5) = 2/3
5
sin θ cos θ dθ = 5 − cos θ + cos θ
3
5
0
0
22. x = sin θ, dx = cos θ dθ,
π/6
Z π/6
1
1
sec3 θ dθ =
sec θ tan θ + ln | sec θ + tan θ|
2
2
0
0
1 2 1
1
2
1
1 1
√ √ + ln( √ + √
=
= + ln 3
2 3 3 2
3 4
3
3
Z
π/3
23. x = sec θ, dx = sec θ tan θ dθ,
π/4
24. x =
√
2 sec θ, dx =
√
1
dθ =
sec θ
Z
2 sec θ tan θ dθ, 2
0
√
√
π/4
Z
π/3
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
25. x = 3 tan θ, dx = 3 sec2 θ dθ,
Z
Z
Z
Z √3/2
1 π/3 sec θ
1 π/3 cos3 θ
1 π/3 1 − sin2 θ
1
1 − u2
dθ
=
dθ
=
cos
θ
dθ
=
du (u = sin θ)
9 π/6 tan4 θ
9 π/6 sin4 θ
9 π/6
9 1/2
u4
sin4 θ
√
√3/2
Z √3/2
1
1
1
1
10 3 + 18
−4
−2
=
(u − u )du =
− 3+
=
9 1/2
9
3u
u 1/2
243
366
Chapter 7
√
√
26. x = 3 tan θ, dx = 3 sec2 θ dθ,
√ Z π/3
√ Z π/3
√ Z π/3
3
tan3 θ
3
3
3
sin θ dθ =
1 − cos2 θ sin θ dθ
dθ =
3 0
sec3 θ
3 0
3 0
√ √ π/3
√
3
3
1
1
1
1
3
=
− cos θ + cos θ
=
− +
− −1 +
= 5 3/72
3
3
3
2 24
3
0
27. true
28. false; −π/2 ≤ θ ≤ π/2
29. false; x = a sec θ
1
30. true; A = 4
2
1
Z
p
1−
x2
Z
dx; let x = sin θ, 0 ≤ θ ≤ π/2, and A = 2
π/2
cos2 θ dθ = π/2
0
0
31. u = x2 + 4, du = 2x dx,
Z
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
√
Z
tan θ dθ = ln | sec θ| + C1 = ln
=
1
ln(x2 + 4) + C with C = C1 − ln 2
2
Z
32. x = 2 tan θ, dx = 2 sec2 θ dθ,
Z
x2 + 4
+ C1 = ln(x2 + 4)1/2 − ln 2 + C1
2
x2
dx =
2
x +4
Z
Z
dx − 4
2 tan2 θ dθ = 2 tan θ − 2θ + C = x − 2 tan−1
x
+ C; alternatively
2
dx
x
= x − 2 tan−1 + C
x2 + 4
2
1
1
x2 + 1
, 1 + (y 0 )2 = 1 + 2 =
,
x
x
x2
Z 2r 2
x +1
dx;
x = tan θ, dx = sec2 θ dθ,
L=
x2
1
33. y 0 =
Z tan−1 (2)
tan2 θ + 1
L=
sec θ dθ =
(sec θ tan θ + csc θ)dθ
tan θ
π/4
π/4
π/4
!
√
tan−1 (2)
h√
i
√
√
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
Z
tan−1 (2)
sec3 θ
dθ =
tan θ
Z
tan−1 (2)
34. y 0 = 2x, 1 + (y 0 )2 = 1 + 4x2 ,
Z
L=
0
1
p
1
1
1 + 4x2 dx; x = tan θ, dx = sec2 θ dθ,
2
2
Exercise Set 7.4
1
L=
2
367
tan−1 2
Z
1
sec θ dθ =
2
3
0
1
1
sec θ tan θ + ln | sec θ + tan θ|
2
2
tan−1 2
0
√
√
1 √
1
1√
1
= ( 5)(2) + ln | 5 + 2| =
5 + ln(2 + 5)
4
4
2
4
35. y 0 = 2x, 1 + (y 0 )2 = 1 + 4x2 ,
1
Z
x2
S = 2π
0
S=
π
4
p
1
1
1 + 4x2 dx; x = tan θ, dx = sec2 θ dθ,
2
2
tan−1 2
Z
tan2 θ sec3 θ dθ =
0
π
4
tan−1 2
Z
(sec2 θ − 1) sec3 θ dθ =
0
π
4
Z
tan−1 2
(sec5 θ − sec3 θ)dθ
0
tan−1 2
√
√
π 1
1
1
π
3
=
sec θ tan θ − sec θ tan θ − ln | sec θ + tan θ|
[18 5 − ln(2 + 5)]
=
4 4
8
8
32
0
Z
36. V = π
1
y2
p
1 − y 2 dy; y = sin θ, dy = cos θ dθ,
0
Z
V =π
0
π/2
π
sin θ cos θ dθ =
4
2
2
Z
0
π/2
π
sin 2θ dθ =
8
2
Z
1
dx = tan−1 (x − 2) + C
(x − 2)2 + 1
Z
1
p
dx = sin−1 (x − 1) + C
1 − (x − 1)2
Z
1
p
dx = sin−1
4 − (x − 1)2
37.
38.
39.
Z
x−1
2
Z
π/2
0
π
(1 − cos 4θ)dθ =
8
+C
Z
1
dx =
16(x + 1/2)2 + 1
Z
p
1
p
dx = ln x − 3 + (x − 3)2 + 1 + C
(x − 3)2 + 1
40.
41.
1
1
dx = tan−1 (4x + 2) + C
2
(4x + 2) + 1
4
Z
42.
x
dx, let u = x + 1,
(x + 1)2 + 1
Z
Z u−1
u
1
1
du
=
−
du = ln(u2 + 1) − tan−1 u + C
u2 + 1
u2 + 1 u2 + 1
2
=
43.
Z p
Z
=
1
ln(x2 + 2x + 2) − tan−1 (x + 1) + C
2
4 − (x + 1)2 dx,
2
4 cos θ dθ =
let x + 1 = 2 sin θ,
Z
2(1 + cos 2θ) dθ
p
1
x+1
= 2θ + sin 2θ + C = 2 sin−1
+ (x + 1) 3 − 2x − x2 + C
2
2
1
θ − sin 4θ
4
π/2
π2
=
16
0
368
Chapter 7
ex
Z
44.
p
Z
1
p
dx, let u = ex + 1/2,
(ex + 1/2)2 + 3/4
−1
u2 + 3/4
du = sinh
√
−1
(2u/ 3) + C = sinh
2ex + 1
√
3
+C
√
3
tan θ,
2
!
√
Z
2 e2x + ex + 1 2ex + 1
√
+ √
+ C1
sec θ dθ = ln | sec θ + tan θ| + C = ln
3
3
p
= ln(2 e2x + ex + 1 + 2ex + 1) + C
x
Alternate solution: let e + 1/2 =
Z
1
1
dx =
2
2(x + 1) + 5
2
45.
Z
p
1
1
dx = √ tan−1 2/5(x + 1) + C
2
(x + 1) + 5/2
10
Z
46.
2x + 3
dx, let u = x + 1/2,
4(x + 1/2)2 + 4
Z Z
1
u
1
1
1
2u + 2
du =
+
du = ln(u2 + 1) + tan−1 u + C
4u2 + 4
2
u2 + 1 u2 + 1
4
2
=
2
Z
47.
1
4
Z
48.
1
1
ln(x2 + x + 5/4) + tan−1 (x + 1/2) + C
4
2
1
√
dx =
4x − x2
Z
p
Z
4x − x2 dx =
0
2
1
1
−1
p
dx = sin
4 − (x − 2)2
x−2
2
2
= π/6
1
4
p
4 − (x − 2)2 dx, let x − 2 = 2 sin θ,
0
Z
π/2
π/2
= 2π
cos θ dθ = 2θ + sin 2θ
2
4
−π/2
−π/2
49. u = sin2 x, du = 2 sin x cos x dx;
Z
i
i
1h p
1h 2 p
1 p
1 − u2 du =
u 1 − u2 + sin−1 u + C =
sin x 1 − sin4 x + sin−1 (sin2 x) + C
2
4
4
50. u = x sin x, du = (x cos x + sin x) dx;
Z p
p
1 p
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
Z
51. (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,
Z
p
sec θ dθ = ln | sec θ + tan θ| + C = ln
x2 + 9/3 + x/3 + C
p
√
but sinh−1 (x/3) = ln x/3 + x2 /9 + 1 = ln x/3 + x2 + 9/3 so the results agree.
52. x = cosh u, dx = sinh u du,
Exercise Set 7.5
Z
369
Z
1
1
sinh 2u − u + C
4
2
p
1
1
1
1
= sinh u cosh u − u + C = x x2 − 1 − cosh−1 x + C
2
2
2
2
p
√
because cosh u = x, and sinh u = cosh2 u − 1 = x2 − 1
1
sinh u du =
2
2
(cosh 2u − 1)du =
EXERCISE SET 7.5
1.
3.
A
B
+
(x − 3) (x + 4)
2x − 3
C
A
B
= + 2+
− 1)
x
x
x−1
x2 (x
2.
5
A
B
C
= +
+
x(x − 2)(x + 2)
x
x−2 x+2
4.
C
A
B
+
+
2
x + 2 (x + 2)
(x + 2)3
5.
A
C
Dx + E
B
+ 2+ 3+ 2
x
x
x
x +2
6.
A
Bx + C
+ 2
x−1
x +6
7.
Cx + D
Ax + B
+ 2
x2 + 5
(x + 5)2
8.
A
Bx + C
Dx + E
+ 2
+ 2
x−2
x +1
(x + 1)2
9.
10.
11.
12.
13.
1
A
B
1
1
=
+
; A = , B = − so
(x − 4)(x + 1)
x−4 x+1
5
5
Z
Z
1
1
1
1
1
1
1 x − 4 dx −
dx = ln |x − 4| − ln |x + 1| + C = ln +C
5
x−4
5
x+1
5
5
5
x + 1
1
A
B
1
1
=
+
; A = − , B = so
(x + 1)(x − 7)
x+1 x−7
8
8
Z
Z
1
1
1
1
1
1 x − 7 1
dx +
dx = − ln |x + 1| + ln |x − 7| + C = ln +C
−
8
x+1
8
x−7
8
8
8
x + 1
11x + 17
A
B
=
+
; A = 5, B = 3 so
(2x − 1)(x + 4)
2x − 1 x + 4
Z
Z
5
1
1
5
dx + 3
dx = ln |2x − 1| + 3 ln |x + 4| + C
2x − 1
x+4
2
5x − 5
A
B
=
+
; A = 1, B = 2 so
(x − 3)(3x + 1)
x − 3 3x + 1
Z
Z
1
1
2
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
Z
Z
Z
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.
370
14.
Chapter 7
1
A
B
C
1
1
= +
+
; A = −1, B = , C = so
x(x + 1)(x − 1)
x
x+1 x−1
2
2
Z
Z
Z
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 |x2 − 1|
(x + 1)(x − 1) 1
+ C = ln
+C
= ln 2
2
x
2
x2
Z x−3+
1
x+3
15.
x2 − 8
1
=x−3+
,
x+3
x+3
2
x2 + 1
=x+1+
,
x−1
x−1
Z x+1+
2
x−1
16.
17.
18.
dx =
1 2
x − 3x + ln |x + 3| + C
2
dx =
1 2
x + x + 2 ln |x − 1| + C
2
A
3x2 − 10
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
Z
Z
Z
1
1
dx = 3x + 12 ln |x − 2| − 2/(x − 2) + C
dx + 2
3dx + 12
x−2
(x − 2)2
x2
3x − 2
3x − 2
A
B
=1+ 2
,
=
+
; A = −1, B = 4 so
2
x − 3x + 2
x − 3x + 2 (x − 1)(x − 2)
x−1 x−2
Z
Z
Z
1
1
dx −
dx + 4
dx = x − ln |x − 1| + 4 ln |x − 2| + C
x−1
x−2
19. Z
u = x2 − 3x − 10, du = (2x − 3) dx,
du
= ln |u| + C = ln |x2 − 3x − 10| + C
u
2
20. u =
Z 3x + 2x − 1, du = (6x + 2) dx,
1
du
1
1
= ln |u| + C = ln |3x2 + 2x − 1| + C
2
u
2
2
21.
x2 + x + 2
x5 + x2 + 2
= x2 + 1 +
,
3
x −x
x3 − x
A
B
C
x2 + x + 2
= +
+
; A = −2, B = 1, C = 2 so
x(x + 1)(x − 1)
x
x+1 x−1
Z
Z
Z
Z
2
1
2
(x2 + 1)dx −
dx +
dx +
dx
x
x+1
x−1
=
22.
(x + 1)(x − 1)2 1 3
1
+C
x + x − 2 ln |x| + ln |x + 1| + 2 ln |x − 1| + C = x3 + x + ln 3
3
x2
1
x5 − 4x3 + 1
= x2 + 3
,
3
x − 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
Z
Z
Z
Z
1
1
1
1
1
1
x2 dx −
dx +
dx +
dx
4
x
8
x+2
8
x−2
=
1 3 1
1
1
x − ln |x| + ln |x + 2| + ln |x − 2| + C
3
4
8
8
Exercise Set 7.5
23.
24.
25.
26.
27.
28.
29.
30.
A
2x2 + 3
B
C
= +
; A = 3, B = −1, C = 5 so
+
2
x(x − 1)
x
x − 1 (x − 1)2
Z
Z
Z
1
1
1
3
dx = 3 ln |x| − ln |x − 1| − 5/(x − 1) + C
dx −
dx + 5
x
x−1
(x − 1)2
C
3x2 − x + 1
A
B
= + 2+
; A = 0, B = −1, C = 3 so
x2 (x − 1)
x
x
x−1
Z
Z
1
1
−
dx + 3
dx = 1/x + 3 ln |x − 1| + C
x2
x−1
A
2x2 − 10x + 4
B
C
=
; A = 1, B = 1, C = −2 so
+
+
2
(x + 1)(x − 3)
x + 1 x − 3 (x − 3)2
Z
Z
Z
2
1
1
2
dx = ln |x + 1| + ln |x − 3| +
dx +
dx −
+ C1
2
x+1
x−3
(x − 3)
x−3
2x2 − 2x − 1
A
B
C
= + 2+
; A = 3, B = 1, C = −1 so
x2 (x − 1)
x
x
x−1
Z
Z
Z
1
1
1
1
dx +
dx
−
dx = 3 ln |x| − − ln |x − 1| + C
3
x
x2
x−1
x
x2
A
B
C
=
+
+
; A = 1, B = −2, C = 1 so
(x + 1)3
x + 1 (x + 1)2
(x + 1)3
Z
Z
Z
1
2
1
2
1
dx −
dx +
dx = ln |x + 1| +
−
+C
x+1
(x + 1)2
(x + 1)3
x + 1 2(x + 1)2
2x2 + 3x + 3
A
B
C
=
+
+
; A = 2, B = −1, C = 2 so
3
2
(x + 1)
x + 1 (x + 1)
(x + 1)3
Z
Z
Z
1
1
1
1
1
2
dx −
dx
+
2
dx = 2 ln |x + 1| +
−
+C
2
3
x+1
(x + 1)
(x + 1)
x + 1 (x + 1)2
A
Bx + C
2x2 − 1
=
+ 2
; A = −14/17, B = 12/17, C = 3/17 so
(4x − 1)(x2 + 1)
4x − 1
x +1
Z
2x2 − 1
7
6
3
dx = − ln |4x − 1| +
ln(x2 + 1) +
tan−1 x + C
2
(4x − 1)(x + 1)
34
17
17
1
A Bx + C
1
1
= + 2
; A = , B = − , C = 0 so
+ 2)
x
x +2
2
2
x(x2
Z
x3
31.
32.
1
1
1
x2
1
dx = ln |x| − ln(x2 + 2) + C = ln 2
+C
+ 2x
2
4
4 x +2
x3 + 3x2 + x + 9
Ax + B
Cx + D
= 2
+ 2
; A = 0, B = 3, C = 1, D = 0 so
(x2 + 1)(x2 + 3)
x +1
x +3
Z 3
x + 3x2 + x + 9
1
dx = 3 tan−1 x + ln(x2 + 3) + C
2
2
(x + 1)(x + 3)
2
x3 + x2 + x + 2
Ax + B
Cx + D
= 2
+ 2
; A = D = 0, B = C = 1 so
(x2 + 1)(x2 + 2)
x +1
x +2
371
372
Chapter 7
Z
33.
34.
x3 + x2 + x + 2
1
dx = tan−1 x + ln(x2 + 2) + C
2
2
(x + 1)(x + 2)
2
x
x3 − 2x2 + 2x − 2
=x−2+ 2
,
x2 + 1
x +1
Z 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
Z
Z
Z
x
x
u−3
dx
=
dx
=
du,
x2 + 6x + 10
(x + 3)2 + 1
u2 + 1
=
Z
so
u=x+3
1
ln(u2 + 1) − 3 tan−1 u + C1
2
1
x4 + 6x3 + 10x2 + x
1
dx = x3 + ln(x2 + 6x + 10) − 3 tan−1 (x + 3) + C
2
x + 6x + 10
3
2
35. true
36. false; degree of numerator should be less than that of denominator
37. true
38. false; (x + 5)3 terms will not occur
Z
1
A
B
1
dx, and
=
+
; A = −1/6,
+ 4x − 5
(x + 5)(x − 1)
x+5 x−1
Z
Z
1
1 x − 1 1
1 − sin θ
1
1
1
dx +
dx = ln +
C
=
ln
+ C.
B = 1/6 so we get −
6
x+5
6
x−1
6
x + 5
6
5 + sin θ
39. Let x = sin θ to get
t
Z
40. Let x = e ; then
x2
et
dt =
e2t − 4
Z
1
dx,
x2 − 4
1
A
B
=
+
; A = −1/4, B = 1/4 so
(x + 2)(x − 2)
x+2 x−2
Z
Z
1
1
1 x − 2 1 et − 2 1
1
−
dx +
dx = ln + C = ln t
+ C.
4
x+2
4
x−2
4
x + 2
4
e + 2
41. u = ex , du = ex dx,
Z
e3x
dx =
2x
e +4
Z
u2
u
du = u − 2 tan−1 + C = ex − tan−1 (ex /2) + C
+4
2
u2
1
42. Set u = 1 + ln x, du = dx,
Z
Zx
5 + 2 ln x
3 + 2u
3
3
dx
=
du = − + 2 ln |u| + C = −
+ 2 ln |1 + ln x| + C
x(1 + ln x)2
u2
u
1 + ln x
Z
43. V = π
0
2
x4
x4
18x2 − 81
dx, 4
=1+ 4
,
2
2
2
(9 − x )
x − 18x + 81
x − 18x2 + 81
18x2 − 81
18x2 − 81
A
B
C
D
=
=
+
+
+
;
2
2
(9 − x )
(x + 3)2 (x − 3)2
x + 3 (x + 3)2
x − 3 (x − 3)2
Exercise Set 7.5
373
9
9
9
9
A = − , B = , C = , D = so
4
4
4
4
2
9
9/4
9
9/4
19 9
V = π x − ln |x + 3| −
+ ln |x − 3| −
− ln 5
=π
4
x+3 4
x−3 0
5
4
44. Let u = ex to get
Z
ln 5
− ln 5
dx
=
1 + ex
Z
ln 5
− ln 5
1
A
B
= +
; A = 1, B = −1;
u(1 + u)
u
1+u
45.
ex dx
=
x
e (1 + ex )
5
Z
1/5
Z
5
1/5
du
,
u(1 + u)
du
= (ln u − ln(1 + u))
u(1 + u)
5
= ln 5
1/5
Ax + B
Cx + D
x2 + 1
= 2
; A = 0, B = 1, C = D = −2 so
+ 2
2
2
(x + 2x + 3)
x + 2x + 3 (x + 2x + 3)2
Z
x2 + 1
dx =
2
(x + 2x + 3)2
Z
1
dx −
(x + 1)2 + 2
Z
2x + 2
dx
(x2 + 2x + 3)2
x+1
1
= √ tan−1 √ + 1/(x2 + 2x + 3) + C
2
2
46.
x5 + x4 + 4x3 + 4x2 + 4x + 4
Ax + B
Cx + D
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
Z
√
1
1
x+1
dx = ln(x2 + 2) + √ tan−1 (x/ 2) + C
x2 + 2
2
2
47. x4 − 3x3 − 7x2 + 27x − 18 = (x − 1)(x − 2)(x − 3)(x + 3),
A
B
C
D
1
=
+
+
+
;
(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
Z
1
1
1
1
dx
= ln |x − 1| − ln |x − 2| +
ln |x − 3| −
ln |x + 3| + C
x4 − 3x3 − 7x2 + 27x − 18
8
5
12
120
48. 16x3 − 4x2 + 4x − 1 = (4x − 1)(4x2 + 1),
1
A
Bx + C
=
+ 2
; A = 4/5, B = −4/5, C = −1/5 so
2
(4x − 1)(4x + 1)
4x − 1
4x + 1
Z
1
dx
1
1
= ln |4x − 1| −
ln(4x2 + 1) −
tan−1 (2x) + C
16x3 − 4x2 + 4x − 1
5
10
10
2
Z
49. Let u = x , du = 2x dx,
0
50.
1
x
1
dx =
4
x +1
2
Z
0
1
1
1
1
1π
π
−1
du = tan u =
= .
2
1+u
2
24
8
0
1
A
B
1
1
=
+
;A =
,B =
so
a2 − x2
a−x a+x
2a
2a
Z 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
51. If the polynomial has distinct roots r1 , r2 , r1 6= r2 , then the partial fraction decomposition will
A
B
,
, and they will give logarithms and no inverse tangents. If
contain terms of the form
x − r1 x − r2
374
Chapter 7
A
B
will appear, and neither
,
x − r (x − r)2
will give an inverse tangent term. The only other possibility is no real roots, and the integrand
1
can be written in the form
, which will yield an inverse tangent, specifically
b 2
b2
a
x
+
+
c
−
2a
4a
b
for some constant A.
of the form tan−1 A x +
2a
there are two roots not distinct, say x = r, then the terms
52. 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
decomposition form is one like
. Therefore the original quadratic had a multiple root.
(x − a)2
53. Yes, for instance the integrand
x2
1
, whose integral is precisely tan−1 x + C.
+1
EXERCISE SET 7.6
1
4
+ ln |4 − 5x| + C
25 4 − 5x
1. Formula (60):
i
4h
3x + ln |−1 + 3x| + C
9
2. Formula (62):
3. Formula (65):
1 x ln +C
5
5 + 2x 4. Formula (66): −
1 − 5x 1
+C
− 5 ln x
x √
2
(−x − 4) 2 − x + C
3
5. Formula (102):
1
(x − 1)(2x + 3)3/2 + C
5
6. Formula (105):
7. Formula (108):
√
1 4 − 3x − 2 +C
ln √
2
4 − 3x + 2 8. Formula (108): tan−1
9. Formula (69):
11. Formula (73):
1 x + 4 ln +C
8
x − 4
10. Formula (70):
3x − 4
+C
2
1 x − 3 ln +C
6
x + 3
p
xp 2
3 x − 3 − ln x + x2 − 3 + C
2
2
√
12. Formula (94): −
13. Formula (95):
√
p
x2 − 5
+ ln(x + x2 − 5) + C
x
p
xp 2
x + 4 − 2 ln(x + x2 + 4) + C
2
√
14. Formula (90):
x2 − 2
+C
2x
√
15. Formula (74):
4 − x2
x
− sin−1 + C
x
2
2 + √ 4 − x2 √
17. Formula (79): 4 − x2 − 2 ln +C
x
16. Formula (80): −
xp
9
x
9 − x2 + sin−1 + C
2
2
3
Exercise Set 7.6
375
√
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
Z
1
1
u du
4
2x 2x
2x
=
+ ln 4 − 3e
+C
25. u = e , du = 2e dx, Formula (62):
2
(4 − 3u)2
18 4 − 3e2x
24. Formula (43):
Z
26. u = cos 2x, du = −2 sin 2xdx, Formula (65): −
√
3
2
27. u = 3 x, du = √ dx, Formula (68):
3
2 x
Z
√
du
1
−1 3 x
= tan
+C
u2 + 4
3
2
1
28. u = sin 4x, du = 4 cos 4xdx, Formula (68):
4
1
29. u = 2x, du = 2dx, Formula (76):
2
Z
√
du
1 cos 2x = − ln +C
2u(3 − u)
6
3 − cos 2x Z
1
du
sin 4x
=
tan−1
+C
2
9+u
12
3
p
du
1 = ln 2x + 4x2 − 9 + C
2
2
u −9
√
2x2 , du = 2 2xdx, Formula (72):
Z p
√
p
1
x2 p 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):
Z
√
u2 du
x2 p
1
1
√
=−
2 − 4x4 + sin−1 ( 2x2 ) + C
4
4
4
2 − u2
Z
du
1 p
√
32. u = 2x, du = 2dx, Formula (83): 2
=−
3 − 4x2 + C
3x
u2 3 − u2
Z
1
1
33. u = ln x, du = dx/x, Formula (26):
sin2 u du = ln x + sin(2 ln x) + C
2
4
34. u = e
−2x
, du = −2e
−2x
1
, Formula (27): −
2
1
35. u = −2x, du = −2dx, Formula (51):
4
Z
1
36. u = 3x+1, du = 3 dx, Formula (11):
3
Z
Z
1
1
cos2 u du = − e−2x − sin 2e−2x + C
4
8
ueu du =
1
(−2x − 1)e−2x + C
4
ln u du =
1
1
(u ln u−u)+C = (3x+1)[ln(3x+1)−1]+C
3
3
376
Chapter 7
1
37. u = sin 3x, du = 3 cos 3x dx, Formula (67):
3
38. u = ln x, du =
1
dx, Formula (105):
x
Z
1
39. u = 4x , du = 8xdx, Formula (70):
8
2
√
Z
1
40. u = 2ex , du = 2ex dx, Formula (69):
2
Z
sin 3x 1
du
1
+C
=
+ ln u(u + 1)2
3 1 + sin 3x
1 + sin 3x Z
√
1
u du
=
(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 1
du
√ +C
= √ ln x
3 − u2
4 3
2e − 3 41. u = 2ex , du = 2ex dx, Formula (74):
Z
√
√
1 p
1 p
3
1 p
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):
√
√
Z √
4 − u2 du
4 − u2
4 − 9x2
−1
=
−3
−
3
sin
(u/2)
+
C
=
−
− 3 sin−1 (3x/2) + C
3
u2
u
x
43. u = 3x, du = 3dx, Formula (112):
r
Z r
1
5
25
6u − 5
1
5
5
−1
2
2
u − u du =
u−
u−u +
sin
+C
3
3
6
6
3
216
5
18x − 5 p
18x − 5
25
=
sin−1
5x − 9x2 +
+C
36
216
5
√
5x, du =
√
5 dx, Formula (117):
q √
√
Z
(u/ 5) − u2
du
x − 5x2
q √
√
+C
=−
+ C = −2
x
u/(2 5)
u (u/ 5) − u2
44. u =
45. u = 2x, du = 2dx, Formula (44):
Z
u sin u du = (sin u − u cos u) + C = sin 2x − 2x cos 2x + C
46. u =
√
Z
2
x, u = x, 2udu = dx, Formula (45): 2
√
2
47. u = − x, u = x, 2udu = dx, Formula (51): 2
u cos u du = 2 cos
Z
√
√
√
x + 2 x sin x + C
√
√
ueu du = −2( x + 1)e− x + C
48. u = 2 + x2 , du = 2x dx, Formula (11):
Z
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):
Z
du
1 u − 4 1 x − 1 =
ln
+
C
=
ln
+C
u2 − 16
8 u + 4
8 x + 7
Exercise Set 7.6
377
50. x2 + 2x − 3 = (x + 1)2 − 4, u = x + 1, du = dx, Formula (74):
Z p
1 p
4 − u2 du = u 4 − u2 + 2 sin−1 (u/2) + C
2
p
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):
Z
Z
Z
p
u
u du
du
u+2
√
√
du =
+2 √
= − 9 − u2 + 2 sin−1 + C
3
9 − u2
9 − u2
9 − u2
p
x−2
= − 5 + 4x − x2 + 2 sin−1
+C
3
52. x2 + 6x + 13 = (x + 3)2 + 4, u = x + 3, du = dx, Formula (71):
Z
(u − 3) du
1
3
1
3
= 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;
Z
Z
2
4
2
4
2u2 (u2 + 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;
Z
√
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;
Z
Z
2
2
2 5 2 3
2 3
2
u2 (u2 − 1)du =
(u4 − u2 )du =
u − u +C =
(x + 1)5/2 − (x3 + 1)3/2 + C
3
3
15
9
15
9
√
56. u = x3 − 1, x3 = u2 + 1, 3x2 dx = 2u du;
Z
p
2
1
2
2
−1
−1
du
=
tan
u
+
C
=
tan
x3 − 1 + C
3
u2 + 1
3
3
57. u = x1/3 , x = u3 , dx = 3u2 du;
Z
Z Z
u
1
1
3u2
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;
Z
Z Z
6u5
u3
1
2
du
=
6
du
=
6
u
−
u
+
1
−
du
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
Z
1
du = 4
u(1 − u)
Z 1
1
x1/4
+
du = 4 ln
+C
u 1−u
|1 − x1/4 |
378
Chapter 7
60. u = x1/2 , x = u2 , dx = 2u du;
Z 2
Z
√
√
u +1−1
2u2
du
=
2
du = 2u − 2 tan−1 u + C = 2 x − 2 tan−1 x + C
2
2
u +1
u +1
61. u = x1/6 , x = u6 , dx = 6u5 du;
Z Z
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;
Z 2
Z √
√
u +u
2
−2
du = −2
du = −x − 4 x − 4 ln | x − 1| + C
u+2+
u−1
u−1
√
63. u = 1 + x2 , x2 = u2 − 1, 2x dx = 2u du, x dx = u du;
Z
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;
Z
5
15
5 (u8 − 3u3 )du = (x + 3)9/5 − (x + 3)4/5 + C
9
4
Z
Z
1
2
1
65.
du
=
du = ln | tan(x/2) + 1| + C
2 1 + u2
u+1
2u
1−u
+
1+
1 + u2
1 + u2
Z
Z
2
1
1
du
=
du
66.
2
2u 1 + u2
u +u+1
2+
1 + u2
Z
2
2 tan(x/2) + 1
1
−1
√
√
du =
=
tan
+C
(u + 1/2)2 + 3/4
3
3
Z
Z
dθ
1
1
67. u = tan(θ/2),
=
du = − + C = − cot(θ/2) + C
1 − cos θ
u2
u
68. u = tan(x/2),
Z
Z
Z
2
2
2
1
1
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 5
z + 5/3 5
tan(x/2) + 3 69. u = tan(x/2),
1
2
Z
Z
70. u = tan(x/2),
Z
71.
2
x
1 − u2
1
du =
u
2
Z
(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
t
1
dt = ln
(Formula (65), a = 4, b = −1)
t(4 − t)
4 4−t 2
1
x
1
x
1
x
x
=
ln
− ln 1 = ln
, ln
= 0.5, ln
= 2,
4
4−x
4 4−x 4 4−x
4−x
Exercise Set 7.6
379
x
= e2 , x = 4e2 − e2 x, x(1 + e2 ) = 4e2 , x = 4e2 /(1 + e2 ) ≈ 3.523188312
4−x
x
Z x
√
1
√
dt = 2 tan−1 2t − 1
72.
(Formula (108), a = −1, b = 2)
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),
x = [1 + tan2 (1/2 + π/4)]/2 ≈ 6.307993516
4
Z
73. A =
p
25 − x2 dx =
0
1 p
25
x
x 25 − x2 +
sin−1
2
2
5
=6+
2
Z
p
74. A =
4
(Formula (74), a = 5)
0
25
4
sin−1 ≈ 17.59119023
2
5
9x2 − 4 dx; u = 3x,
2/3
6
p
p
1 1 p 2
u2 − 4 du =
u u − 4 − 2 ln u + u2 − 4 (Formula (73), a2 = 4)
3 2
2
2
√
√
√
1 √
2
=
3 32 − 2 ln(6 + 32) + 2 ln 2 = 4 2 − ln(3 + 2 2) ≈ 4.481689467
3
3
Z 1
1
75. A =
dx; u = 4x,
2
0 25 − 16x
Z
u + 5 4
1
1
1 4
1
du =
ln
=
ln 9 ≈ 0.054930614 (Formula (69), a = 5)
A=
4 0 25 − u2
40 u − 5 0
40
A=
1
3
Z
6
Z
4
√
76. A =
x ln x dx =
1
=
Z
π/2
77. V = 2π
4 3/2
x
9
4
3
ln x − 1
2
1
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
0
Z
(Formula (45))
0
8
78. V = 2π
4
√
4π
x x − 4 dx =
(3x + 8)(x − 4)3/2
15
=
Z
(Formula (50), n = 1/2)
8
(Formula (102), a = −4, b = 1)
4
1024
π ≈ 214.4660585
15
3
xe−x dx; u = −x,
79. V = 2π
0
Z
−3
ueu du = 2πeu (u − 1)
V = 2π
−3
0
Z
80. V = 2π
1
= 2π(1 − 4e−3 ) ≈ 5.031899801
0
5
π
x ln x dx = x2 (2 ln x − 1)
2
= π(25 ln 5 − 12) ≈ 88.70584621
5
1
(Formula (50), n = 1)
(Formula (51))
380
Chapter 7
2
Z
p
1 + 16x2 dx; u = 4x,
81. L =
0
1
L=
4
Z
Z
3
8
0
82. L =
8
p
p
1 up
1 2
2
2
1 + u du =
1 + u + ln u + 1 + u
4 2
2
0
√
√
1
= 65 + ln(8 + 65) ≈ 8.409316783
8
(Formula (72), a2 = 1)
!#3
3 + √ x2 + 9 p
x2 + 9 − 3 ln x
1
1
√
√
√
3 + 10
√ ≈ 3.891581644 (Formula (89), a = 3)
= 3 2 − 10 + 3 ln
1+ 2
Z
p
2
1 + 9/x dx =
1
π
Z
83. S = 2π
3
√
x2 + 9
dx =
x
p
(sin x) 1 + cos2 x dx; u = cos x, a2 = 1,
0
−1p
Z
S = −2π
1+
u2
1
Z
4
1
1p
1 + 1/x4 dx = 2π
x
16
√
84. S = 2π
Z
p
1
p
u
1 2
2
du = 4π
1 + du = 4π
1 + u + ln u + 1 + u
2
2
0
0
h√
√ i
= 2π
2 + ln(1 + 2) ≈ 14.42359945 (Formula (72))
Z 1p
S=π
1
Z
4
u2
√
1
x4 + 1
dx; u = x2 ,
x3
√
!#
16
p
u2 + 1
u2 + 1
du = π −
+ ln u + u2 + 1
u2
u
1
!
√
√
√
16 + 257
257
√
=π
2−
+ ln
≈ 9.417237485 (Formula (93), a2 = 1)
16
1+ 2
Z
t
20 cos6 u sin3 u du
85. (a) s(t) = 2 +
=−
s(t)
(b)
4
0
20
40
166
sin2 t cos7 t −
cos7 t +
9
63
63
3
2
1
t
3
t
Z
a(u) du = −
86. (a) v(t) =
0
Z
s(t) = 10 +
6
9
12
15
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
t
v(u) du
0
=−
2
35 −t
6 −t
16
343866
3 −t
e cos 2t − e−t sin 2t +
e cos 6t +
e sin 6t +
t+
50
25
2738
1369
185
34225
Exercise Set 7.6
(b)
12
10
8
6
4
2
381
s(t)
t
2
Z
87. (a)
6
10
14
18
Z
1 + u
1 + tan(x/2) 1
2
+C
du = ln + C = ln sec x dx =
dx =
cos x
1 − u2
1 − u
1 − tan(x/2) cos(x/2) + sin(x/2) cos(x/2) + sin(x/2) 1 + sin x +C
= ln + C = ln cos(x/2) − sin(x/2) cos(x/2) + sin(x/2) cos x Z
= ln |sec x + tan x| + C
x
x
π
1 + tan
tan + tan
x
4
2
2
(b) tan
=
+
π
x =
x
4
2
1 − tan tan
1 − tan
4
2
2
Z
Z
Z
1
dx = 1/u du = ln | tan(x/2)| + C but
88.
csc x dx =
sin x
π
ln | tan(x/2)| =
1 (1 − cos x)/2
1 1 − cos x
1 sin2 (x/2)
ln
= ln
= ln
; also,
2
2 cos (x/2)
2 (1 + cos x)/2
2 1 + cos x
1 − cos x
1 − cos2 x
1
1 1 − cos x
=
=
so ln
= − ln | csc x + cot x|
1 + cos x
(1 + cos x)2
(csc x + cot x)2
2 1 + cos x
q
√
89. Let u = tanh(x/2) then cosh(x/2) = 1/ sech(x/2) = 1/ 1 − tanh2 (x/2) = 1/ 1 − u2 ,
√
sinh(x/2) = tanh(x/2) cosh(x/2) = u/ 1 − u2 , so sinh x = 2 sinh(x/2) cosh(x/2) = 2u/(1 − u2 ),
cosh x = cosh2 (x/2) + sinh2 (x/2) = (1 + u2 )/(1 − u2 ), x = 2 tanh−1 u, dx = [2/(1 − u2 )]du;
Z
Z
2
2
1
2u + 1
2 tanh(x/2) + 1
dx
√
=
du = √ tan−1 √
+ C = √ tan−1
+ C.
2
2 cosh x + sinh x
u +u+1
3
3
3
3
1
4
Z
1
1
1
du = sin−1 u + C = sin−1 (x4 ) + C.
2
4
4
1−u
Z
Z
91.
(cos32 x sin30 x − cos30 x sin32 x)dx = cos30 x sin30 x(cos2 x − sin2 x)dx
90. Let u = x4 to get
√
1
= 30
2
Z
sin30 2x cos 2x dx =
sin31 2x
+C
31(231 )
√
Z
√
√
2
1
92.
x−
− 4dx = √
( x + 2 − x − 2)dx =
[(x + 2)3/2 − (x − 2)3/2 ] + C
3
2
Z
Z
1
1
1
1
1
93.
dx
=
−
du = − ln |u| + C = − ln |1 + x−9 | + C
10
−9
x (1 + x )
9
u
9
9
Z q
p
x2
94. (a) (x + 4)(x − 5)(x2 + 1)2 ;
A
B
Cx + D
Ex + F
+
+ 2
+ 2
x+4 x−5
x +1
(x + 1)2
382
Chapter 7
(b) −
3
2
x−2
3
+
−
−
x + 4 x − 5 x2 + 1 (x2 + 1)2
(c) −3 ln |x + 4| + 2 ln |x − 5| + 2 tan−1 x −
1
3
ln(x2 + 1) −
2
2
x
−1
+
tan
x
+C
x2 + 1
EXERCISE SET 7.7
1. exact value = 14/3 ≈ 4.666666667
(a) 4.667600662, |EM | ≈ 0.000933995
(b) 4.664795676, |ET | ≈ 0.001870991
(c) 4.666666602, |ES | ≈ 9.9 · 10−7
3. exact value = 1
(a) 1.001028824, |EM | ≈ 0.001028824
(b) 0.9979429864, |ET | ≈ 0.0020570136
(c) 1.000000013, |ES | ≈ 2.12 · 10−7
5. exact value = 12 (e−2 − e−6 ) ≈ 0.06642826551
(a) 0.06598746840, |EM | ≈ 0.00044079711
(b) 0.06731162281, |ET | ≈ 0.00088335730
(c) 0.06642830240, |ES | ≈ 5.88 · 10−7
7. f (x) =
√
2. exact value = 2
(a) 1.999542900, |EM | ≈ 0.000457100
(b) 2.000915091, |ET | ≈ 0.000915091
(c) 2.000000019, |ES | ≈ 2.97 · 10−7
4. exact value = 1 − cos(2) ≈ 1.416146836
(a) 1.418509838, |EM | ≈ 0.002363002
(b) 1.411423197, |ET | ≈ 0.004723639
(c) 1.416146888, |ES | ≈ 7.88 · 10−7
1
ln 10 ≈ 0.7675283641
3
(a) 0.7575800751, |EM | ≈ 0.0099482890
(b) 0.7884048101, |ET | ≈ 0.0208764460
(c) 0.767855, |ES | ≈ 0.000326622
6. exact value =
15
1
x + 1, f 00 (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
81
(c) |ES | ≤
≈ 0.000007910156250
10240000
(a) |EM | ≤
(b) |ET | ≤
27
(1/4) = 0.00562500
1200
√
3
105 −9/2
8. f (x) = 1/ x, f 00 (x) = x−5/2 , f (4) (x) =
x
; K2 = 3/128, K4 = 105/8192
4
16
125
(3/128) = 0.001220703125
2400
35
(c) |ES | ≤
≈ 0.000001390775045
25165824
(a) |EM | ≤
(b) |ET | ≤
125
(3/128) = 0.002441406250
1200
9. f (x) = cos x, f 00 (x) = − cos x, f (4) (x) = cos x; K2 = K4 = 1
π 3 /8
(1) ≈ 0.001028824
2400
π5
(c) |ES | ≤
(1) ≈ 3.320526095 · 10−7
180 × 104
(a) |EM | ≤
(b) |ET | ≤
π 3 /8
(1) ≈ 0.003229820488
1200
10. f (x) = sin x, f 00 (x) = − sin x, f (4) (x) = sin x; K2 = K4 = 1
8
(1) ≈ 0.003333333
2400
32
(c) |ES | ≤
(1) ≈ 0.000017778
180 × 104
(a) |EM | ≤
(b) |ET | ≤
8
(1) ≈ 0.006666667
1200
Exercise Set 7.7
383
11. f (x) = e−2x , f 00 (x) = 4e−2x ; f (4) (x) = 16e−2x ; K2 = 4e−2 ; K4 = 16e−2
8
(4e−2 ) ≈ 0.0018044704
2400
32
(16e−2 ) ≈ 0.000017778
(c) |ES | ≤
180 × 105
(a) |EM | ≤
(b) |ET | ≤
8
(4e−2 ) ≈ 0.0036089409
1200
12. f (x) = 1/(3x + 1), f 00 (x) = 18(3x + 1)−3 , f (4) (x) = 1944(3x + 1)−5 ; K2 = 18, K4 = 1944
27
(18) ≈ 0.202500
2400
243
(c) |ES | ≤
(1944) ≈ 0.2624400000
180 × 104
(a) |EM | ≤
1/2
(27)(1/4)
13. (a) n >
≈ 23.7; n = 24
(24)(5 × 10−4 )
1/4
(243)(15/16)
≈ 7.1; n = 8
(c) n >
(180)(5 × 10−4 )
(b) |ET | ≤
27
(18) ≈ 0.4050000000
1200
(27)(1/4)
(b) n >
(12)(5 × 10−4 )
1/2
≈ 33.5; n = 34
1/2
(125)(3/128)
14. (a) n >
≈ 15.6; n = 16
(24)(5 × 10−4 )
1/4
(243)(105/16)
(c) n >
≈ 11.6; n = 12
(180)(5 × 10−4 )
(125)(3/128)
(b) n >
(12)(5 × 10−4 )
1/2
(π 3 /8)(1)
≈; n = 13
15. (a) n >
(24)(10−3 )
5
1/4
(π /32)(1)
(c) n >
≈ 2.7; n = 4
(180)(10−3 )
(π 3 /8)(1)
(b) n >
(12)(10−3 )
1/2
(8)(1)
(12)(10−3 )
1/2
1/2
(8)(4e−2 )
≈ 42.5; n = 43
17. (a) n >
(24)(10−6 )
1/4
(32)(16e−2 )
(c) n >
≈ 7.9; n = 8
(180)(10−6 )
(8)(4e−2 )
(b) n >
(12)(10−6 )
1/2
1/2
(27)(18)
18. (a) n >
= 450; n = 451
(24)(10−6 )
1/4
(243)(1944)
(c) n >
≈ 71.6; n = 72
(180)(10−6 )
1/2
16. (a) n >
(8)(1)
(24)(10−3 )
1/2
(32)(1)
(c) n >
(180)(10−3 )
≈ 18.25; n = 19
(b) n >
1/2
≈ 22.1; n = 23
≈ 17.9; n = 18
≈ 25.8; n = 26
1/4
≈ 3.7; n = 4
19. false; Tn is the average of Ln and Rn
20. true, see Theorem 7.7.1(b)
(27)(18)
(b) n >
(12)(10−6 )
≈ 60.2; n = 61
≈ 636.4; n = 637
384
Chapter 7
21. false, it is the weighted average of M25 and T25
22. true
23. 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.
Z 4
Z 4 2
x
3x 13
25
g(x) dx =
−
+
dx =
24
8
12
36
2
2
∆x
1 1 4 1
25
[f (X0 ) + 4f (X1 ) + f (X2 )] =
+ +
=
3
3 2 3 4
36
24. 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 + Y2 ) = (1/18)(0 + 4(1/4) + 3/4) = 7/72, and
1/3
Z 1/3
3
2
= 1/18 + 1/24 = 7/72
g(x) dx = (3/2)x + (3/8)x
0
0
25. 1.493648266,
1.493649897
26. 1.367479484,
1.367479548
27. 3.805639712,
3.805537256
28. 1.899407630,
1.899406253
29. 0.9045242448,
0.9045242380
30. 0.2652884045,
0.265280129
−1
31. exact value = 4 tan
2
=π
(x/2)
0
32. exact value
√
3
= 29 x 9 − x2 + 2 sin−1 x3
=π
0
(a) 3.142425985, |EM | ≈ 0.000833331
(b) 3.139925989, |ET | ≈ 0.001666665
(c) 3.141592654, |ES | ≈ 0
(a) 3.152411433, |EM | ≈ 0.010818779
(b) 3.104518327, |ET | ≈ 0.037074327
(c) 3.139775352, |ES | ≈ 0.001817302
33. 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.
34. (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)
35. f (x) = x sin x, f 00 (x) = 2 cos x − x sin x, |f 00 (x)| ≤ 2| cos x| + |x| | sin x| ≤ 2 + 2 = 4 so K2 ≤ 4,
n>
(8)(4)
(24)(10−4 )
1/2
≈ 115.5; n = 116 (a smaller n might suffice)
36. f (x) = ecos x , f 00 (x) = (sin2 x)ecos x − (cos x)ecos x , |f 00 (x)| ≤ ecos x (sin2 x + | cos x|) ≤ 2e so
K2 ≤ 2e, n >
(1)(2e)
(24)(10−4 )
1/2
≈ 47.6; n = 48 (a smaller n might suffice)
Exercise Set 7.7
385
√
3
37. f (x) = x x, f 00 (x) = √ , lim+ |f 00 (x)| = +∞
4 x x→0
√
√
√
√
x sin x + cos x
, lim |f 00 (x)| = +∞
38. f (x) = sin x, f 00 (x) = −
x→0+
4x3/2
Z xp
Z πp
Z xp
1 + (y 0 (t))2 dt =
1 + cos2 t dt, ` =
1 + cos2 t dt ≈ 3.820197788
39. s(x) =
0
0
Z
x
p
40. s(x) =
1 + (y 0 (t))2 dt =
1
Z
v dt ≈
0
42.
t
a
Z
0
0
8
x
Z
p
1 + 4/t6 dt, ` =
1
15
41.
Z
0
2
p
1 + 4/t6 dt ≈ 1.296766956
1
15
[0 + 4(37.1) + 2(60.1) + 4(90.9) + 2(98.3) + 4(104.1) + 114.4] ≈ 772.5 ft
(3)(6)
7
1
4
6
3
2
5
0.02 0.08 0.20 0.40 0.60 0.70 0.60
8
0
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)
≈ 2.7 cm/s
a dt ≈
0
Z
180
v dt ≈
43.
0
Z
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 ≈
44.
0
1
4
2
4
2
4
1
1800
+
+
+
+
+
+
≈ 0.71 s
(3)(6) 3100 2908 2725 2549 2379 2216 2059
16
Z
πr2 dy = π
45. V =
0
Z
0
16
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
Z
600
600
[0 + 4(7) + 2(16) + 4(24) + 2(25) + 4(16) + 0] = 9000 ft2 ,
(3)(6)
0
V = 75A ≈ 75(9000) = 675, 000 ft3
46. A =
h dx ≈
47. (a) The maximum value of |f 00 (x)| is approximately 3.8442
(b) n = 18
(c) 0.9047406684
48. (a) The maximum value of |f 00 (x)| is approximately 1.46789
(b) n = 12
(c) 1.112830350
49. (a) K4 = max |f (4) (x)| ≈ 12.40703740 (calculator)
0≤x≤1
104 K4
(b − a)5 K4
(b)
< 10−4 provided n4 >
, n > 5.12, so n ≥ 6
4
180n
180
K4 4
(c)
· 6 ≈ 0.0000531 with S12 = 0.9833961318
180
386
Chapter 7
50. (a) K4 = max |f (4) (x)| ≈ 3.94136281877
0≤x≤1
(b − a)5 K4
106 K4
−6
4
(b)
<
10
provided
n
>
≈ 21896.5, n > 12.2 and even, so n ≥ 14
180n4
180
(b − a)5 K4
≈ 0.00000057
(c) With n = 14, S14 ≈ 1.497185354, with possible error |ES | ≤
180n4
b−a
[y0 + y1 + . . . + yn−2 + yn−1 ]
n
b−a
Right endpoint approximation ≈
[y1 + y2 + . . . + yn−1 + yn ]
n
b−a1
Average of the two =
[y0 + 2y1 + 2y2 + . . . + 2yn−2 + 2yn−1 + yn ]
n 2
yk + yk+1
y
(b) Area of trapezoid = (xk+1 − xk )
. If we sum from k = 0
3
2
to k = n − 1 then we get the right hand side of (2).
51. (a) Left endpoint approximation ≈
2
1
yk
xk
yk + 1
xk + 1
t
52. right endpoint, trapezoidal, midpoint, left endpoint
53. 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 = 21 (Y2 − Y0 ), and A = 12 (Y0 + Y2 ) − Y1 .
Z 1
Z 1
1
2
1
2
Then
g(x) dx = 2
(Ax2 + C) dx = A + 2C = (Y0 + Y2 ) − Y1 + 2Y1 = (Y0 + 4Y1 + Y2 ),
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
Z
Z
m+∆x
∆x
1
g(x) dx. Finally, Y0 = g(m − ∆x) = h(−1),
h(z) dz =
−1
m−∆x
Y1 = g(m) = h(0), Y2 = g(m + ∆x) = h(1).
54. From Exercise 53 we know, for i = 0, 1, . . . , n − 1 that
Z x2i+2
1b−a
gi (x) dx =
[y2i + 4y2i+1 + y2i+2 ],
3 2n
x2i
b−a
because
is the width of the partition and acts as ∆x in Exercise 53.
2n
Summing over all the subintervals note that y0 and y2n are only listed once; so
Z b
n−1
X Z x2i+2
f (x) dx =
gi (x) dx
a
i=0
x2i
1b−a
=
[y0 + 4y1 + 2y2 + 4y3 + . . . + 4y2n−2 + 2y2n−1 + y2n ]
3 2n
EXERCISE SET 7.8
1. (a) improper; infinite discontinuity at x = 3
Exercise Set 7.8
(b)
(c)
(d)
(e)
(f )
387
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
3.
`
1
1 −2x
1
lim (−e−2` + 1) =
lim − e
=
`→+∞
2
2 `→+∞
2
0
4.
`
1
1
ln(1 + x2 )
= lim [ln(1 + `2 ) − ln 2] = +∞, divergent
`→+∞ 2
`→+∞ 2
−1
5.
6.
7.
8.
9.
10.
11.
12.
lim
`
−1
lim −2 coth
x
`→+∞
`→+∞
3
`
2
1
lim − e−x
`→+∞
2
0
`
`→−∞
`→+∞
`→+∞
1
4(2x − 1)2
1
x
tan−1
`→−∞ 3
3
1 3x
e
`→−∞ 3
0
= lim
1 π
`
1
− tan−1
= [π/4 − (−π/2)] = π/4
`→−∞ 3 4
3
3
`
0
= lim
`→−∞
`
1 1 3`
1
− e
=
3 3
3
0
= lim
`→−∞
`
+∞
13.
Z
−∞
Z
+∞
+∞
√
0
x
x2 + 2
∞
so
−∞
1 2
x
`→+∞ 2
√
dx = lim
+∞
`→+∞
p
x dx both converge; it diverges if either (or both)
0
`
1 2
` = +∞ so
`→+∞ 2
0
`
x2 + 2
x
dx is divergent.
x2 + 2
Z
+∞
= lim
x dx = lim
0
Z
Z
x dx and
−∞
diverges.
Z
1
1
ln(3 − 2e` ) = ln 3
2
2
0
x dx converges if
14.
1
[−1 + 1/(2` − 1)2 ] = −1/4
4
= lim
1
lim − ln(3 − 2ex )
`→−∞
2
Z
`→−∞
`
3
lim
lim
1
1
1
2 + 2 = 2
2 ln `
√
√
= lim (2 ln ` − 2 ln 2) = +∞, divergent
2
lim −
= lim −
e
`
`→+∞
1 −`2
−e
+ 1 = 1/2
`→+∞ 2
= lim
1
lim −
`→+∞
2 ln2 x
√
lim 2 ln x
2 coth−1 3 − 2 coth−1 ` = 2 coth−1 3
= lim
0
x dx is divergent.
−∞
p
√
= lim ( `2 + 2 − 2) = +∞
`→+∞
388
Chapter 7
`
x
1
1
1
15.
dx = lim −
= lim [−1/(`2 + 3) + 1/3] = ,
2 + 3)2
2 + 3)
`→+∞
`→+∞
(x
2(x
2
6
0
0
Z ∞
Z 0
x
x
dx = −1/6 so
dx = 1/6 + (−1/6) = 0
similarly
2
2
2
2
−∞ (x + 3)
−∞ (x + 3)
Z
+∞
Z
+∞
16.
0
0
Z
−∞
`
h
e−t
πi π
−1 −t
dt = lim − tan (e ) = lim − tan−1 (e−` ) +
= ,
−2t
`→+∞
`→+∞
1+e
4
4
0
0
h π
i π
e−t
−1 −t
−1 −`
dt
=
lim
−
tan
(e
)
=
lim
−
+
tan
(e
)
= ,
`→−∞
`→−∞
1 + e−2t
4
4
`
+∞
Z
−∞
17.
18.
e−t
π π
π
dt = + =
−2t
1+e
4
4
2
1
lim −
x−4
`→4−
3
lim x2/3
`→0+ 2
`
8
`
1
1
= +∞, divergent
= lim −
−
`−4 4
`→4−
0
3
= lim+ (4 − `2/3 ) = 6
`→0 2
`
lim − − ln(cos x) =
19.
`→π/2
20.
0
√
lim− −2 4 − x
`→4
21.
lim− sin−1 x
`→1
22.
`
0
`
0
lim − ln(cos `) = +∞, divergent
`→π/2−
√
= lim− 2(− 4 − ` + 2) = 4
`→4
= lim− sin−1 ` = π/2
`→1
1
p
p
√
√
lim − 9 − x2 = lim (− 8 + 9 − `2 ) = − 8
`→−3+
23.
`→−3+
`
lim +
√
π/2
1 − 2 cos x
`→π/3
lim + (1 −
=
√
1 − 2 cos `) = 1
`→π/3
`
`
24.
lim − ln(1 − tan x)
`→π/4−
Z
2
25.
0
Z
2
26.
0
Z
dx
= lim− ln |x − 2|
x−2
`→2
dx
= lim+ −1/x
x2
`→0
8
27.
x
−1/3
0
Z
=
0
x−1/3 dx = lim
`→0−
−1
Z
`
= lim− (ln |` − 2| − ln 2) = −∞, divergent
`→2
0
2
`
Z
= lim+ (−1/2 + 1/`) = +∞ so
`→0
3
dx = lim x2/3
`→0+ 2
0
lim − ln(1 − tan `) = +∞, divergent
`→π/4−
8
3 2/3
x
2
= lim
`
`→0+
= lim
−1
so
`→0−
−1
x−1/3 dx = 6 + (−3/2) = 9/2
−2
3
(4 − `2/3 ) = 6,
2
`
8
2
3 2/3
(` − 1) = −3/2
2
dx
is divergent
x2
Exercise Set 7.8
Z
1
28.
0
389
dx
= lim− 3(x − 1)1/3
`→1
(x − 1)2/3
Z
+∞
1
dx =
x2
Z
+∞
dx
=
x x2 − 1
`
0
= lim− 3[(` − 1)1/3 − (−1)1/3 ] = 3
`→1
a
Z +∞
1
1
29. Define
dx +
dx where a > 0; take a = 1 for convenience,
2
2
x
x
0
0
a
1
Z 1
Z +∞
1
1
dx = lim+ (−1/x) = lim+ (1/` − 1) = +∞ so
dx is divergent.
2
2
x
x
`→0
`→0
0
0
`
Z
1
a
Z
√
30. Define
dx
+
x x2 − 1
√
1
+∞
Z
a
dx
√
where a > 1,
x x2 − 1
take a = 2 for convenience to get
2
Z 2
dx
√
= lim+ sec−1 x = lim+ (π/3 − sec−1 `) = π/3,
2
`→1
`→1
1 x x −1
`
`
Z +∞
Z +∞
dx
dx
−1
√
√
= lim sec x = π/2 − π/3 so
= π/2.
2
`→+∞
x x −1
x x2 − 1
2
1
2
√
31. Z
Let u = x, x = Zu2 , dx = 2u du. Then
√
du
dx
√
= 2 2
= 2 tan−1 u + C = 2 tan−1 x + C and
u +1
x(x + 1)
Z 1
√ 1
√
dx
−1
√
= 2 lim+ tan
x = 2 lim+ (π/4 − tan−1 ) = π/2
x(x
+
1)
→0
→0
0
Z
+∞
32.
Z
=
1
Z
+∞
+
0
0
. The first integral was studied in Exercise 31 through the transformation
1
x = u2 . We apply the same transformation to the second integral and obtain
Z +∞
Z A
2du
π
π
π
dx
√
= lim
= lim [2 tan−1 A − 2 tan−1 1] = 2 − 2 = , and thus
2+1
A→+∞
A→+∞
u
2
4
2
x(x
+
1)
1
Z1 +∞
dx
√
=π
x(x + 1)
0
33. true, Theorem 7.8.2
34. false; consider f (x) = 1
35. false, neither 0 nor 3 lies in [1, 2], so the integrand is continuous
36. false, the integral does not converge at x = 0
Z
+∞
37.
√
e−
√
0
Z
+∞
√
38.
12
Z
+∞
√
39.
0
Z
40.
0
+∞
√
x
x
Z
+∞
e−u du = 2 lim
dx = 2
`→+∞
0
+∞
dx
=2
x(x + 4)
Z
e−x
dx =
1 − e−x
Z
√
2 3
1
0
e−x
dx = −
1 − e−2x
−e−u
`
= 2 lim
0
du
1
u
= 2 lim
tan−1
2
`→+∞ 2
u +4
2
`→+∞
`
√
2 3
1 − e−` = 2
= lim tan−1
`→+∞
√
`
π
− tan−1 3 =
2
6
√
√ 1
du
√ = lim 2 u = lim 2(1 − `) = 2
u
`→0+
`→0+
`
Z
1
0
√
du
=
1 − u2
Z
0
1
√
`
du
π
= lim sin−1 u = lim sin−1 ` =
`→1
`→1
2
1 − u2
0
390
Chapter 7
Z
41.
lim
e
`→+∞
`
1 −x
cos x dx = lim e (sin x − cos x) = 1/2
`→+∞ 2
0
`
−x
0
Z
+∞
42. A =
0
1
xe−3x dx = lim − (3x + 1)e−3x
`→+∞
9
43. (a) 2.726585
45. 1 +
dy
dx
2
`
= 1/9
0
(b) 2.804364
(c) 0.219384
4 − x2/3
4
=1+
= 2/3 ; the arc length is
2/3
x
x
Z
0
8
2
x1/3
dx = 3x
(d) 0.504067
8
= 12
2/3 0
2
dy
4
x2
46. 1 +
=
; the arc length is
=1+
dx
4 − x2
4 − x2
r
`
Z `
Z 2
2
4
−1 x
√
dx
=
lim
dx
=
lim
2
sin
= 2 sin−1 1 = π
4 − x2
2 0
`→2− 0
`→2−
4 − x2
0
Z
ln x dx = x ln x − x + C,
47.
Z
1
Z
`→0+
`→0+
`
but lim+ ` ln ` = lim+
`→0
= lim (−1 − ` ln ` + `),
ln x dx = lim (x ln x − x)
ln x dx = lim
0
1
1
`→0
ln `
= lim (−`) = 0 so
1/` `→0+
Z
`
1
`→0+
ln x dx = −1
0
Z
48.
ln x
ln x 1
dx = −
− + C,
2
x
x
x
`
Z +∞
Z `
ln x 1
ln ` 1
ln x
ln x
dx = lim
dx = lim −
−
= lim −
− +1 ,
`→+∞ 1 x2
`→+∞
x2
x
x 1 `→+∞
`
`
1
Z +∞
ln `
1
ln x
but lim
= lim
= 0 so
=1
`→+∞ `
`→+∞ `
x2
1
49. (a) From the graph of f (4) (x) it is evident that |f (4) (x)| ≤ 12.5 on [0, 1]. We take K4 = 12.5.
1/4
12.5
12.5
−4
(b) Set n = 2k, then
≤
10
,
or
k
≥
= 2.57, so
180 · 24 k 4
180 · 24 · 10−4
k ≥ 2.57, k ≥ 3, n ≥ 6.
(c) S6 ≈ 0.983347416
50. (a) From its graph it is evident that |f (4) (x)| ≤ 3 on [0, 1], set K4 = 3.94136.
1/4
3.94136
3.94136
−6
≤
10
,
or
k
≥
≈ 6.08, k ≥ 7, n ≥ 14
(b) Set n = 2k, then
180 · 24 · k 4
180 · 24 · 10−6
(c) S14 ≈ 1.49718504
+∞
Z
e−2x dx = −
51. (a) V = π
0
Z
+∞
e−x
π
2
lim e−2x
`→+∞
p
`
= π/2
0
1 + e−2x dx, let u = e−x to get
p
Z 0p
1
h√
p
√ i
u
1 S = −2π
1 + u2 du = 2π
1 + u2 + ln u + 1 + u2 = π
2 + ln(1 + 2)
2
2
1
0
(b) S = 2π
0
2
53. (a) For x ≥ 1, x2 ≥ x, e−x ≤ e−x
Exercise Set 7.8
Z
391
+∞
Z
`
i`
e−x dx = lim −e−x = lim (e−1 − e−` ) = 1/e
`→+∞
`→+∞ 1
`→+∞
1
1
Z +∞
2
(c) By Parts (a) and (b) and Exercise 52(b),
e−x dx is convergent and is ≤ 1/e.
e−x dx = lim
(b)
1
1
ex
54. (a) If x ≥ 0 then ex ≥ 1,
≤
2x + 1
2x + 1
`
Z `
dx
1
(b) lim
= lim
ln(2x + 1) = +∞
`→+∞ 0 2x + 1
`→+∞ 2
0
Z +∞
(c) By Parts (a) and (b) and Exercise 52(a),
0
Z
`
55. V = lim
`→+∞
1
Z `
A = lim
`→+∞
(π/x2 ) dx = lim −(π/x)
i`
`→+∞
1
ex
dx is divergent.
2x + 1
= lim (π − π/`) = π
`→+∞
p
p
2π(1/x) 1 + 1/x4 dx; use Exercise 52(a) with f (x) = 2π/x, g(x) = (2π/x) 1 + 1/x4
1
and a = 1 to see that the area is infinite.
√
Z +∞
x3 + 1
for x ≥ 2,
1dx = +∞
x
2
`
Z +∞
Z +∞
dx
1
x
dx ≤
= lim − 3 = 1/24
(b)
`→+∞
x5 + 1
x4
3x 2
2
2
Z
Z
Z ∞
+∞
+∞
xex
dx
xex
dx ≥
≥
= +∞
(c)
2x
+
1
2x
+
1
2x
+1
1
1
0
56. (a) 1 ≤
1
, above the x-axis, and to the
1 + x2
Z ∞
1
right of the y-axis is given by
. Solving for
1
+
x2
0
r
1−y
x=
, the area is also given by the improper integral
y
Z 1r
1−y
dy.
y
0
y
57. The area under the curve y =
58. (b) u =
√
Z
x,
0
+∞
1
0.5
x
1
2
3
4
√
Z +∞
Z +∞
cos x
√ dx = 2
cos u du;
cos u du diverges by Part (a).
x
0
0
Z
Z
dx
1
1
x
59. Let x = r tan θ to get
= 2
cos θ dθ = 2 sin θ + C = √
+C
2
r
r
(r2 + x2 )3/2
r r 2 + x2
`
p
p
2πN Ir
x
2πN I
√
so u =
lim
=
lim (`/ r2 + `2 − a/ r2 + a2 )
`→+∞ r 2 r 2 + x2 a
k
kr `→+∞
p
2πN I
=
(1 − a/ r2 + a2 ).
kr
60. Let a2 =
M
to get
2RT
5
392
Chapter 7
r
r
3/2 −2
2
1
M
2RT
8RT
M
=√
=
2RT
2 2RT
M
πM
π
r
3/2 √ −5/2
4
3RT
3RT
M
3 π
M
=√
=
so vrms =
2RT
8
2RT
M
M
π
4
(a) v̄ = √
π
2
(b) vrms
61. (a) Satellite’s weight = w(x) = k/x2 lb when x = distance from center of Earth; w(4000) = 6000
Z 4000+b
9.6 × 1010 x−2 dx mi·lb.
so k = 9.6 × 1010 and W =
4000
Z
+∞
9.6 × 1010 x−2 dx = lim −9.6 × 1010 /x
(b)
`
`→+∞
4000
= 2.4 × 107 mi·lb
4000
`
1 −st
1
62. (a) L{1} =
e dt = lim − e
=
`→+∞
s
s
0
0
`
Z +∞
Z +∞
1 −(s−2)t
1
(b) L{e2t } =
e−st e2t dt =
e−(s−2)t dt = lim −
e
=
`→+∞
s−2
s−2
0
0
0
`
Z +∞
−st
e
1
(c) L{sin t} =
e−st sin t dt = lim 2
(−s sin t − cos t) = 2
`→+∞
s
+
1
s
+1
0
0
Z +∞
`
s
e−st
(−s cos t + sin t) = 2
(d) L{cos t} =
e−st cos t dt = lim 2
`→+∞
s
+
1
s
+1
0
0
Z
+∞
−st
Z
+∞
−st
63. (a) L{f (t)} =
te
2
dt = lim −(t/s + 1/s )e
`→+∞
0
Z
0
`→+∞
+∞
(c) L{f (t)} =
3
1
s2
100
1000
10,000
0.8862269
0.8862269
0.8862269
0.8862269
√
Z
+∞
`
=
0
1
e−st dt = lim − e−st
`→+∞
s
10
√
=
t2 e−st dt = lim −(t2 /s + 2t/s2 + 2/s3 )e−st
0
64.
`
+∞
(b) L{f (t)} =
Z
−st
`
3
2
s3
e−3s
=
s
2
dx = √
a
Z
+∞
p
2
e−u du = π/a
0
0
Z +∞
Z +∞
√
√
2
2
2
2
2
−x /2σ
(b) x = 2σu, dx = 2σ du, √
e
dx = √
e−u du = 1
π 0
2πσ 0
65. (a) u =
Z
3
Z
4
ax, du =
a dx, 2
e
−ax2
√
2
e−x dx ≈ 0.8862; π/2 ≈ 0.8862
0
Z +∞
Z 3
Z +∞
2
2
2
(b)
e−x dx = e−x dx +
e−x dx so
0
0
3
∞
Z +∞
Z
2
1 +∞ −x2
1
1
−x2
E=
e
dx <
xe
dx = − e−x
= e−9 < 2.1 × 10−5
3
6
6
3
3
3
66. (a)
67. (a)
0
1
dx ≈ 1.047; π/3 ≈ 1.047
x6 + 1
Review Exercises, Chapter 7
+∞
Z
393
Z +∞
1
1
dx
+
dx so
6+1
6+1
x
x
0
4
Z +∞
1
1
1
dx =
< 2 × 10−4
dx <
6
5
x6 + 1
x
5(4)
4
1
dx =
x6 + 1
(b)
0
+∞
Z
E=
4
+∞
Z
68. If p = 0, then
Z
4
`
(1)dx = lim x = +∞,
`→+∞
0
+∞
Z
0
1 px
e
`→+∞ p
epx dx = lim
if p 6= 0, then
0
1
Z
69. If p = 1, then
0
Z
if p 6= 1, then
0
1
dx
= lim ln x
x
`→0+
`
1 p`
(e − 1) =
`→+∞ p
= lim
0
−1/p, p < 0
.
+∞,
p>0
1
= +∞;
`
x1−p
dx
=
lim
xp
`→0+ 1 − p
1
1−p
= lim [(1 − `
`
`→0+
)/(1 − p)] =
1/(1 − p), p < 1
.
+∞,
p>1
√
70. u = 1 − x, u2 = 1 − x, 2u du = −dx;
Z 0p
Z 1p
h p
√ i1
2
−2
2 − u du = 2
2 − u2 du = u 2 − u2 + 2 sin−1 (u/ 2) = 1 + π/2
1
0
0
1
Z
cos(u2 )du ≈ 1.809
71. 2
0
Z
72. −2
0
sin(1 − u2 )du = 2
1
Z
1
sin(1 − u2 )du ≈ 1.187
0
REVIEW EXERCISES, CHAPTER 7
1. u = 4 + 9x, du = 9 dx,
2. u = πx, du = π dx,
1
π
1
9
Z
u1/2 du =
Z
cos u du =
2
(4 + 9x)3/2 + C
27
1
1
sin u + C = sin πx + C
π
π
Z
2
u1/2 du = − cos3/2 θ + C
3
Z
dx
du
4. u = ln x, du =
,
= ln |u| + C = ln |ln x| + C
x
u
Z
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,
Z 3
Z 3
3
u2
9
3
−1 u
2
du = 2
1− 2
du = 2u − 6 tan
=6− π
2+9
u
u
+
9
3
2
0
0
0
√
7. (a) Z
With u = x:
Z
p
√
1
1
dx = 2 √
du = 2 sin−1 (u/ 2) + C = 2 sin−1 ( x/2) + C;
√ √
x 2√
−x
2 − u2
with u = 2 − x :
394
Chapter 7
Z
√
√
√
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:
Z
1
p
dx = sin−1 (x − 1) + C.
1 − (x − 1)2
Z
(b) In the three
differ by a constant, in particular
p 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 = √
Z
0
1
√
x3
x2 + 1
x
dx, du = 2x dx, v =
x2 + 1
Z
i1
p
2
2
dx = x x + 1 − 2
0
=
√
2
2 − (x2 + 1)3/2
3
√
x2 + 1;
1
x(x2 + 1)1/2 dx
0
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;
Z 1
Z 1
Z √2 2
u −1
x3
x2
√
√
u du
dx =
x dx =
2
2
u
x +1
x +1
1
0
0
√2
Z √2
√
1 3
2
=
u −u
= (2 − 2)/3
(u − 1)du =
3
1
1
9. Z
u = x, dv = e−x dx, du Z= 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
Z
Z
1
1
1
1
x sin 2x dx = − x cos 2x +
cos 2x dx = − x cos 2x + sin 2x + C
2
2
2
4
Z
Z
2
2x
11. u = ln(2x + 3), dv = dx, du =
dx, v = x;
ln(2x + 3)dx = x ln(2x + 3) −
dx
2x + 3
2x + 3
Z
Z 2x
3
3
but
dx =
1−
dx = x − ln(2x + 3) + C1 so
2x + 3
2x + 3
2
Z
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
Z
Z
2x
1
tan−1 (2x)dx = x tan−1 (2x) −
dx = x tan−1 (2x) − ln(1 + 4x2 ) + C, thus
2
1 + 4x
4
Z 1/2
1
1
−1
tan (2x) dx = (1/2)(π/4) − ln 2 = π/8 − ln 2
4
4
0
Review Exercises, Chapter 7
13.
395
Repeated
Differentiation
Repeated
Antidifferentiation
8x4
cos 2x
PP +
PP
PP 1
32x3
sin 2x
2
PP −
PP
PP 1
96x2
− cos 2x
4
PP +
PP
PP 1
192x
− sin 2x
8
PP −
PP
PP 1
cos 2x
192
16
PP +
PP
PP 1
0
sin 2x
32
Z
8x4 cos 2x dx = (4x4 − 12x2 + 6) sin 2x + (8x3 − 12x) cos 2x + C
5
Z
t2 e−t dt; u = t2 , dv = e−t dt, du = 2tdt, v = −e−t ,
14. distance =
0
distance = −t2 e−t
i5
0
−5
distance = −25e
5
Z
te−t dt; u = 2t, dv = e−t dt, du = 2dt, v = −e−t ,
+2
0
− 2te
−t
i5
Z
5
+2
0
e−t dt = −25e−5 − 10e−5 − 2e−t
i5
0
0
= −25e−5 − 10e−5 − 2e−5 + 2 = −37e−5 + 2
Z
15.
sin2 5θ dθ =
Z
16.
3
1
2
Z
(1 − cos 10θ)dθ =
Z
2
sin 2x cos 2x dx =
Z
=
Z
1
sin x cos 2x dx =
2
Z
π/6
17.
18.
0
Z
1
1
θ−
sin 10θ + C
2
20
(1 − cos2 2x) cos2 2x sin 2x dx
1
1
cos5 2x + C
(cos2 2x − cos4 2x) sin 2x dx = − cos3 2x +
6
10
1
1
(sin 3x − sin x)dx = − cos 3x + cos x + C
6
2
1
sin 2x cos 4x dx =
2
Z
0
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,
396
Chapter 7
Z
1
1
− sin3 u cos u +
2
4
1
3
1
= − sin3 u cos u +
− sin u cos u +
8
8
2
sin4 2x dx =
1
2
Z
sin4 u du =
3
4
Z
sin2 u du
1
2
Z
du
3
3
1
sin u cos u + u + C
= − sin3 u cos u −
8
16
16
1
3
3
= − sin3 2x cos 2x −
sin 2x cos 2x + x + C
8
16
8
20. u = x2 ,
Z
Z
Z
1
1
cos5 u du =
(cos u)(1 − sin2 u)2 du
x cos5 (x2 )dx =
2
2
Z
Z
Z
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,
Z
Z
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
p
9
1
= sin−1 (x/3) − x 9 − x2 + C
2
2
22. x = 4 sin θ, dx = 4 cos θ dθ,
√
Z
Z
1
1
1
16 − x2
1
2
csc θ dθ = − cot θ + C = −
+C
dθ =
2
16
16
16
16x
sin θ
Z
23. x = sec θ, dx = sec θ tan θ dθ,
p
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,
Z
25
25
25 sec3 θ dθ =
sec θ tan θ +
ln | sec θ + tan θ| + C1
2
2
p
1 p
25
= x x2 − 25 +
ln |x + x2 − 25| + C
2
2
25. x Z= 3 tan θ, dx = 3 secZ2 θ dθ,
Z
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 p
9
1p
1
= x 9 + x2 − ln |
9 + x2 + x| + C
2
2
3
3
=
26. 2x = tan θ, 2 dx = sec2 θ dθ,
Review Exercises, Chapter 7
Z
2
397
Z
sec θ csc θ dθ =
(sec θ tan θ + csc θ) dθ
= sec θ − ln | csc θ+ cot θ| + C
√1 + 4x2
p
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
Z
Z
1
1
1
1
1
1
1 x − 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
Z
Z
1
1
1
1
1
1
1 x + 1 +C
dx −
dx = ln |x + 1| − ln |x + 7| + C = ln 6
x+1
6
x+7
6
6
6
x + 7
x2 + 2
6
=x−2+
,
x+2
x+2
Z 6
x−2+
x+2
32.
dx =
1 2
x − 2x + 6 ln |x + 2| + C
2
A
B
C
x2 + x − 16
=
+
+
; A = −1, B = 2, C = −1 so
(x + 1)(x − 3)2
x + 1 x − 3 (x − 3)2
Z
Z
Z
1
1
1
−
dx + 2
dx −
dx
x+1
x−3
(x − 3)2
= − ln |x + 1| + 2 ln |x − 3| +
31.
(x − 3)2
1
1
+ C = ln
+
+C
x−3
|x + 1|
x−3
x2
B
A
C
+
=
+
; A = 1, B = −4, C = 4 so
(x + 2)3
x + 2 (x + 2)2
(x + 2)3
Z
Z
Z
1
1
4
2
1
dx − 4
dx + 4
dx = ln |x + 2| +
−
+C
x+2
(x + 2)2
(x + 2)3
x + 2 (x + 2)2
1
A Bx + C
= + 2
; A = 1, B = −1, C = 0 so
+ 1)
x
x +1
x(x2
Z
x3
1
1
x2
1
dx = ln |x| − ln(x2 + 1) + C = ln 2
+C
+x
2
2 x +1
33. (a) With x = sec θ:
√
Z
Z
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:
Z
Z
Z
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|
398
Chapter 7
1
A
B
C
1
1
1
= +
+
=− +
+
x
x−1 x+1
x 2(x − 1) 2(x + 1)
Z −x
1
1
1
dx = − ln |x| + ln |x − 1| + ln |x + 1| + C, valid on any interval not containing
3
x −x
2
2
the numbers x = 0, ±1
(c)
x3
2
3−x
C
A
B
3−x
dx, 2
= + 2+
; A = −4, B = 3, C = 4
3 + x2
x
x
(x
+
1)
x
x
x
+1
1
2
3
A = −4 ln |x| − + 4 ln |x + 1|
x
1
Z
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 = 4 − 2 2 ≈ 1.17157
(a) 1.17138, |EM | ≈ 0.000190169
(b) 1.17195, |ET | ≈ 0.000380588
(c) 1.17157, |ES | ≈ 8.35 × 10−8
π
≈ 1.57080
2
(a) 1.57246, |EM | ≈ 0.00166661
(b) 1.56746, |ET | ≈ 0.00333327
(c) 1.57080, |ES | ≈ 2.0 × 10−8
42. exact value =
43. f (x) = √
1
105
3
105
3
, f (4) (x) =
(x+1)−7/2 ; K2 = √ , K4 = √
, f 00 (x) =
5/2
9/2
4
4(x + 1)
16(x + 1)
x+1
2 2
28 2
23
3
1
√ =
√ ≈ 4.419417 × 10−4
2400 24 2
102 24 2
3
23
√ = 8.838834 × 10−4
(b) |ET | ≤
4
1200 2 2
25
7
105
√ =
√ ≈ 3.2224918 × 10−7
(c) |ES | ≤
4
9
180 × 204 28 2
3 · 10 · 2 2
(a) |EM | ≤
1
and use a CAS to compute and then plot f 00 (x) and f (4) (x). One sees that the
1 + x2
absolute maxima |f 00 (x)| and |f (4) (x)| both occur at x = 0, and hence K2 = 2, K4 = 24.
44. Let f (x) =
(a) |EM | ≤
(c) |ES | ≤
8
(2) ≈ 0.006666667
2400
25 · 24
≈ 0.000026666667
180 · 24 · 104
(b) |ET | ≤
8
(2) ≈ 0.013333333
1200
Review Exercises, Chapter 7
399
102
8·3
√ , so n ≥ 2 1/4 ≈ 21.02, n ≥ 22
4
2 2
24 × 2 2
2
4
10
10
(b) n2 ≥ √ , so n ≥
≈ 29.73, n ≥ 30
2 · 23/4
23 2
5
25 K4
7
105
−4
4
4 2
√ = 104
√ , so
(c) Let n = 2k, then want
≤
10
,
or
k
≥
10
9
180(2k)4
180 24 · 28 2
2 ·3 2
1/4
7
√
k ≥ 10
≈ 2.38; so k ≥ 3, n ≥ 6
3 · 29 2
45. (a) n2 ≥ 104
46. Recall from Exercise 44 that K2 = 2, K4 = 24.
1/2
(8)(2)
(a) n >
≈ 81.6; n ≥ 82
(24)(10−4 )
(b) n >
(8)(2)
(12)(10−4 )
1/2
≈ 115.47; n ≥ 116
25 K4
(c) Let n = 2k, then want
≤ 10−4 , or
180(2k)4
1/4
5
4 2 24
≈ 7.19, k ≥ 8, n ≥ 16
k ≥ 10
180 24
`
47.
lim (−e−x ) = lim (−e−` + 1) = 1
`→+∞
48.
49.
`→+∞
0
1
x
lim
tan−1
`→−∞ 2
2
√
lim− −2 9 − x
Z
1
0
`
`
`→9
50.
2
√
= lim− 2(− 9 − ` + 3) = 6
`→9
0
Z
1
dx =
2x − 1
1 π
1
−1 `
= lim
− tan
= [π/4 − (−π/2)] = 3π/8
`→−∞ 2 4
2
2
0
1/2
1
dx +
2x − 1
Z
1
1
1
1
dx = lim − ln(1 − 2`) + lim + ln(2` − 1) + C
2x − 1
`→1/2 2
`→1/2 2
1/2
neither limit exists hence the integral diverges
+∞
Z
51. A =
e
Z
52. V = 2π
ln x − 1
ln x
dx = lim c −
2
`→+∞
x
x
= 1/e
e
`
+∞
−x
xe
0
`
dx = 2π lim −e
`→+∞
−x
(x + 1)
= 2π lim
0
`→+∞
1 − e−` (` + 1)
1
`+1
but lim e−` (` + 1) = lim
= lim ` = 0 so V = 2π
`→+∞
`→+∞ e`
`→+∞ e
Z
+∞
53.
0
54. (a)
(c)
(e)
(g)
(i)
`
1
dx
1
π
−1
tan
(x/a)
= lim
tan−1 (`/a) =
= 1, a = π/2
=
lim
2
2
`→+∞ a
`→+∞ a
x +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,
Z
Z
1
1
1
1
x
dθ =
cos θ dθ = sin θ + C = √
+C
3
sec θ
3
3
3 3 + x2
400
Chapter 7
1
56. u = x, dv = cos 3x dx, du = dx, v = sin 3x;
3
Z
Z
1
1
1
1
x cos 3x dx = x sin 3x −
sin 3x dx = x sin 3x + cos 3x + C
3
3
3
9
57. Use Endpaper Formula (31) to get
π/4
π/4
π/4
π/4
Z π/4
1
1
1
7
6
4
2
tan θ dθ = tan θ
− tan θ
+ tan θ
+ ln | cos θ|
6
4
2
0
0
0
0
0
√
√
1 1 1
5
= − + − ln 2 =
− ln 2.
6 4 2
12
Z
cos θ
dθ, let u = sin θ − 3,
(sin θ − 3)2 + 3
Z
sin2 2x cos3 2x dx =
58.
59.
=
Z
60.
3
Z
Z
sin2 2x(1 − sin2 2x) cos 2x dx =
`
Z
0
1
1
dx = lim− −
(x − 3)2
x−3
`→3
61. u = e2x , dv = cos 3x dx, du = 2e2x dx, v =
e
Z
Z
13
9
2x
1
2
cos 3x dx = e2x sin 3x −
3
3
Z
1
2
e2x sin 3x dx = − e2x cos 3x +
3
3
e2x cos 3x dx =
Z
(sin2 2x − sin4 2x) cos 2x dx
`
which is clearly divergent, so the integral
0
1
sin 3x;
3
e2x sin 3x dx. Use u = e2x , dv = sin 3x dx to get
Z
e2x cos 3x dx so
1 2x
2
4
e sin 3x + e2x cos 3x −
3
9
9
e2x cos 3x dx =
Z
1
1
sin3 2x −
sin5 2x + C
6
10
1
dx = lim−
(x
−
3)2
`→3
0
diverges.
Z
√
1
1
du = √ tan−1 [(sin θ − 3)/ 3] + C
u2 + 3
3
Z
e2x cos 3x dx,
1 2x
e (3 sin 3x + 2 cos 3x) + C1 ,
9
Z
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,
1
√
2
63.
Z
π/2
)
π/2
Z π/2
1
3
cos3 θ sin θ
+
cos2 θ dθ
4
4 −π/2
−π/2
(
)
π/2
Z
3
1
1 π/2
3 1
3π
= √
cos θ sin θ
+
dθ = √ π = √
2 −π/2
4 2 2
4 22
8 2
−π/2
1
cos4 θ dθ = √
2
−π/2
(
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
Z
Z
Z
1
1
1
1
1
1
−
dx +
dx +
dx
6
x−1
15
x+2
10
x−3
1
1
1
= − ln |x − 1| +
ln |x + 2| +
ln |x − 3| + C
6
15
10
Review Exercises, Chapter 7
401
2
2
sin θ, dx = cos θ dθ,
3
3
π/6
Z π/6
Z 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
48 3 2
64. x =
65. u =
Z 2
0
x − 4, x = u2 + 4, dx = 2u du,
2
Z 2
2u2
4
−1
du
=
2
1
−
du
=
2u
−
4
tan
(u/2)
=4−π
u2 + 4
u2 + 4
0
0
66. u =
Z
1
0
67. u =
Z
68.
√
√
2u
du,
u2 + 1
1
Z 1
1
2u2
π
−1
1
−
du
=
2
du
=
(2u
−
2
tan
u)
=2−
u2 + 1
u2 + 1
2
0
0
√
ex − 1, ex = u2 + 1, x = ln(u2 + 1), dx =
ex + 1, ex = u2 − 1, x = ln(u2 − 1), dx =
2
du =
2
u −1
Z 2u
du,
u2 − 1
√ x
1
1
e +1−1
−
du = ln |u − 1| − ln |u + 1| + C = ln √ x
+C
u−1 u+1
e +1+1
A
Bx + C
1
= + 2
; A = 1, B = C = −1 so
x(x2 + x + 1)
x
x +x+1
Z
−x − 1
dx = −
x2 + x + 1
Z
x+1
dx = −
(x + 1/2)2 + 3/4
Z
u + 1/2
du,
u2 + 3/4
u = x + 1/2
√
1
1
= − ln(u2 + 3/4) − √ tan−1 (2u/ 3) + C1
2
3
Z
so
x(x2
1
dx
1
2x + 1
= ln |x| − ln(x2 + x + 1) − √ tan−1 √
+C
+ x + 1)
2
3
3
1
dx, v = x;
1 − x2
1/2 Z 1/2
1/2
Z 1/2
1
1 p
x
√
dx = sin−1 + 1 − x2
sin−1 x dx = x sin−1 x
−
2
2
1 − x2
0
0
0
0
r
√
1 π
3
π
3
=
+
−1=
+
−1
2 6
4
12
2
69. u = sin−1 x, dv = dx, du = √
Z
70.
2
2
tan 4x(1 + tan 4x) sec 4x dx =
Z
71.
3
Z
(tan3 4x + tan5 4x) sec2 4x dx =
1
1
tan4 4x +
tan6 4x + C
16
24
x+3
p
dx, let u = x + 1,
(x + 1)2 + 1
Z
Z p
u+2
2
2
−1/2
√
du =
u(u + 1)
+√
du = u2 + 1 + 2 sinh−1 u + C
u2 + 1
u2 + 1
p
= x2 + 2x + 2 + 2 sinh−1 (x + 1) + C
402
Chapter 7
Alternate solution: let x + 1 = tan θ,
Z
Z
Z
(tan θ + 2) sec θ dθ = sec θ tan θ dθ + 2 sec θ dθ = sec θ + 2 ln | sec θ + tan θ| + C
=
p
x2 + 2x + 2 + 2 ln(
Z
72. Let x = tan θ to get
x3
p
x2 + 2x + 2 + x + 1) + C.
1
dx.
− x2
1
C
A
B
= + 2+
; A = −1, B = −1, C = 1 so
x2 (x − 1)
x
x
x−1
Z
Z
Z
1
1
1
1
dx +
dx −
dx = − ln |x| + + ln |x − 1| + C
−
x
x2
x−1
x
tan θ − 1 x − 1
1
+ C = cot θ + ln |1 − cot θ| + C
+ C = cot θ + ln = + ln x
x tan θ 73.
74.
lim −
`→+∞
1
2(x2 + 1)
`
bx
1
lim
tan−1
`→+∞ ab
a
= lim
a
`
`→+∞
−
1
1
1
+
=
2(`2 + 1) 2(a2 + 1)
2(a2 + 1)
b`
π
1
tan−1
=
`→+∞ ab
a
2ab
= lim
0
MAKING CONNECTIONS, CHAPTER 7
1. (a) u = f (x), dv = dx, du = f 0 (x), v = x;
b Z b
Z b
Z
f (x) dx = xf (x) −
xf 0 (x) dx = bf (b) − af (a) −
a
a
a
b
xf 0 (x) dx
a
(b) Substitute y = f (x), dy = f 0 (x) dx, x = a when y = f (a), x = b when y = f (b),
Z b
Z f (b)
Z f (b)
0
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
Z β
Z β
Z f (b)
f −1 (x) dx =
f −1 (y) dy =
f −1 (y) dy,
α
α
A1
A2
a
f (a)
which, by Part (b), yields
Z β
Z
−1
f (x) dx = bf (b) − af (a) −
α
x
a=f
–1(a)
b
f (x) dx
a
= βf −1 (β) − αf −1 (α) −
Z
f −1 (β)
f (x) dx
f −1 (α)
Z
β
Note from the figure that A1 =
f −1 (x) dx, A2 =
α
A1 + A2 = βf −1 (β) − αf −1 (α), a ”picture proof”.
Z
f −1 (β)
f (x) dx, and
f −1 (α)
b=f
–1(b)
Making Connections, Chapter 7
403
2. (a) Use Exercise 1(c);
Z π/6
Z 1/2
Z sin−1 (1/2)
1
1
1
1
sin−1 x dx = sin−1
−0·sin−1 0−
sin x dx = sin−1
−
sin x dx
2
2
2
2
0
sin−1 (0)
0
(b) Use Exercise 1(b);
Z e2
Z
ln x dx = e2 ln e2 − e ln e −
e
ln e2
f −1 (y) dy = 2e2 − e −
ln e
Z
+∞
−t
e dt = lim −e
3. (a) Γ(1) =
−t
`
`→+∞
0
Z
0
Z
2
ey dy = 2e2 − e −
Z
1
= lim (−e−` + 1) = 1
`→+∞
+∞
tx e−t dt; let u = tx , dv = e−t dt to get
(b) Γ(x + 1) =
0
x −t
+∞
Γ(x + 1) = −t e
Z
+x
+∞
t
x−1 −t
x −t
e dt = −t e
0
0
+∞
+ xΓ(x)
0
tx
= 0 (by multiple applications of L’Hôpital’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
Thus Γ(n) = (n − 1)! if n is a positive integer.
Z +∞
Z +∞
√
√
√
2
1
(d) Γ
=
t−1/2 e−t dt = 2
e−u du (with u = t) = 2( π/2) = π
2
0
0
3
1
1
1√
3
3
3√
5
(e) Γ
= Γ
=
π, Γ
= Γ
=
π
2
2
2
2
2
2
2
4
4. (a) t = − ln x, x = e−t , dx = −e−t dt,
Z 1
Z 0
Z
(ln x)n dx = −
(−t)n e−t dt = (−1)n
0
+∞
+∞
tn e−t dt = (−1)n Γ(n + 1)
0
(b) t = xn , x = t1/n , dx = (1/n)t1/n−1 dt,
Z +∞
Z +∞
n
e−x dx = (1/n)
t1/n−1 e−t dt = (1/n)Γ(1/n) = Γ(1/n + 1)
0
0
q p
q
2
2
cos θ − cos θ0 = 2 sin (θ0 /2) − sin (θ/2) = 2(k 2 − k 2 sin2 φ) = 2k 2 cos2 φ
5. (a)
q
√
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
q
1
2k cos φ
=
1 − k 2 sin2 φ dθ, thus dθ = p
dφ and hence
2
1 − k 2 sin2 φ
s
s Z
Z
8L π/2
1
2k cos φ
L π/2
1
√
p
T =
·p
dφ
=
4
dφ
2
2
g 0
g
2k cos φ
0
1 − k sin φ
1 − k 2 sin2 φ
p
(b) If L = 1.5 ft and θ0 = (π/180)(20) = π/9, then
√ Z π/2
3
dφ
q
T =
≈ 1.37 s.
2 0
1 − sin2 (π/18) sin2 φ
1
2
ex dx