Calculus Homework
Assignment 10
1. Evaluate
the integrals,
Z
a. y tan−1 (y 2 )dy.
Z
b. 5x sec2 3xdx.
[like §8.2 #7,10]
Sol.
Z
Z
1
−1 2
2
tan−1 xdx.
a. Let x = y , then dx = 2ydy and y tan (y )dy =
2
dx
Let u = tan−1 x, dv = dx, then du = 1+x
2 , v = x. So we
have that
Z
Z
x
−1
−1
tan xdx = x tan x −
dx
1 + x2
Z
1
dw
−1
, by letting w = 1 + x2
= x tan x −
2
w
1
1
= x tan−1 x − ln w + C0 = x tan−1 x − ln(1 + x2 ) + C0
2
2
Hence
Z
1
1
C0
y tan−1 (y 2 )dy =
x tan−1 x − ln(1 + x2 ) + C, where C =
2
4
2
1
1 2
y tan−1 (y 2 ) − ln(1 + y 4 ) + C
=
2
4
2
b. Let u = 5x, dv = sec 3x, then du = 5dx, v = 13 tan 3x. So
we have that
Z
Z
5
5
2
5x sec 3xdx =
x tan 3x −
tan 3xdx
3
3
Z
5
5
sin 3x
=
x tan 3x −
dx
3
3
cos 3x
Z
5
5
dw
=
x tan 3x +
, by letting w = cos 3x
3
9
w
5
5
=
x tan 3x + ln |w| + C
3
9
5
5
=
x tan 3x + ln | cos 3x| + C
3
9
2. Evaluate the integrals,
Z 1
3
a.
θ3 cos πθdθ.
Z0
b. e−2x cos 2xdx.
[like §8.2 #2,24]
Sol.
a. Let f (θ) = θ3 , g(θ) = cos πθ. Then
f (θ) and its derivatives
g(θ) and its integrals
θ3 \\\\\\\\\\\\\\\\\\\\\\(+)
\\
cos πθ
\\\\\\\\\\\\
\\\\\\. 1
2 \\\\\\\
sin πθ
\\\\\\\\\\\ (−)
3θ
π
\\\\\\\\\\\
\\\\\\\\\\\
− π12 cos πθ
6θ \\\\\\\\\\\\\\\\\\\\ (+)
\\\\\\\\\\\
\\\\\\\\\− π13 sin πθ
6 \\\\\\\\\\\\\\\\\\\\ (−)
\\\\\\\\\\\
\\\\\\\\\\\
- 1
0
π4
cos πθ
Hence
3
13
Z 1
3
θ
3θ2
6θ
6
3
θ cos πθdθ =
sin πθ + 2 cos πθ − 3 sin πθ − 4 cos πθ
π
π
π
π
0
0
√ 3
√
2
3π + 9π − 54 3π + 162
=
54π 4
b. Let f (x) = cos 2x, g(x) = e−2x . Then
f (x) and its derivatives
g(x) and its integrals
cos 2x \\\\\\\\\\\\\\\\\\\\(+)
e−2x
\\\\\\\\\\\\
\\\\\\\\. 1 −2x
−2e
−2 sin 2x \\\\\\\\\\\\\\\ (−)
\\\\\\\\\\\
\\\\\\\\\\\
\\\- 1 −2x
o
e
−4 cos 2x
(+)
4
So we have that
Z
Z
1 −2x
1 −2x
−2x
cos 2x+ e
sin 2x− e−2x cos 2xdx
e
cos 2xdx = − e
2
2
Z
1
Thus, 2 e−2x cos 2xdx = e−2x (sin 2x−cos 2x), and hence
2
Z
1
e−2x cos 2xdx = e−2x (sin 2x − cos 2x) + C
4
Z
3. Evaluate the integral,
x sin(2 ln x)dx.
[like §8.2 #29]
Sol. Let y = ln x, then x = ey and dy = dx
⇒ dx = xdy = ey dy.
x
Thus
Z
Z
x sin(2 ln x)dx = e2y sin 2ydy
Now let f (y) = e2y , g(y) = sin 2y, then
f (y) and its derivatives
g(y) and its integrals
e2y \\\\\\\\\\\\\\\\\\\\\\(+)
\
2e2y
4e2y
sin 2y
\\\\\\\\\\\\
\\\\\\. 1
\\\\\\\\\\\
− 2 cos 2y
\\\\\\\\\\(−)
\\\\\\\\\\\
\\\\\\\\- 1
o
(+)
− 4 sin 2y
So we have that
Z
Z
1 2y
1 2y
2y
e sin 2ydy = − e cos 2y + e sin 2y − e2y sin 2ydy
2
2
Z
1
Hence 2 e2y sin 2ydy = e2y (sin 2y − cos 2y) + C0 . Therefore,
2
Z
Z
1
x sin(2 ln x)dx =
e2y sin 2ydy = e2y (sin 2y − cos 2y) + C
4
1 2
C0
=
x [sin(2 ln x) − cos(2 ln x)] + C, C =
4
2
4. Find the volume of the solid generated by revolving the region
in the first quadrant bounded by the coordinate axes, the curve
y = ex , and the line x = ln 2 about the line x = 1.
[like §8.2 #33]
Sol. It is clear that the curve y = ex and the line x = ln 2
intersect at the point (ln 2, 2). The region is sketched as follows
Observe that
shell radius is 1 − x and the shell height is ex .
Z the
ln 2
2πex (1 − x)dx. Note that if we let u = x, dv =
Hence V =
0
ex dx ⇒ du = dx, v = ex , and then
Z
Z
Z
Z
Z
x
x
x
x
x
e (1 − x)dx =
e dx − xe dx = e dx − xe + ex dx
Z
x
= −xe + 2 ex dx = ex (2 − x) + C
This implies that
Z ln 2
h
iln 2
2πex (1 − x)dx = 2π ex (2 − x)
= 4π(1 − ln 2)
V =
0
0
Z
5. Evaluate the integral,
e4t + 16
dt.
(e2t + 4)2
[like §8.3 #26]
Sol. Observe that
Z
Z Z 4t
(e2t + 4)2 − 8e2t
8e2t e + 16
dt
=
dt
=
1
−
dt
(e2t + 4)2
(e2t + 4)2
(e2t + 4)2
Z
e2t
= t−8
dt
(e2t + 4)2
Let x = e2t , then dx = 2e2t dt, so
Z 4t
Z
e + 16
dx
4
4
dt = t−4
= t+
+C = t+ 2t
+C
2t
2
2
(e + 4)
(x + 4)
x+4
e +4
6. Evaluate the integral,
Z
(x − 1)2 tan−1 (3x) − 18x3 − 2x
dx.
(9x2 + 1)(x − 1)2
[like §8.3 #40]
Sol. Observe that
Z
Z (x − 1)2 tan−1 (3x) − 18x3 − 2x
tan−1 (3x)
2x dx
=
−
dx
(9x2 + 1)(x − 1)2
9x2 + 1
(x − 1)2
Z
Z
tan−1 (3x)
x
=
dx − 2
dx
2
9x + 1
(x − 1)2
B
x
A
A(x − 1) + B
+
=
=
.
2
2
(x − 1)
x − 1 (x − 1)
(x − 1)2
3dx
and
Then we have that dy = 2
9x + 1
Let y = tan−1 (3x) and
A
=1
⇒
−A + B = 0
A=1
B=1
Thus,
Z
Z (x − 1)2 tan−1 (3x) − 18x3 − 2x
1
1
1
dx =
ydy − 2
+
dx
(9x2 + 1)(x − 1)2
3
x − 1 (x − 1)2
Z
Z
y2
dx
dx
−2
−2
=
6
x−1
(x − 1)2
1
2
= [tan−1 (3x)]2 − 2 ln |x − 1| +
+C
6
x−1
1
2
= [tan−1 (3x)]2 − ln(x − 1)2 +
+C
6
x−1
Z
7. Evaluate
the integrals,
Z π √
6
sec2 x − 1dx.
a.
π
−
Z π6
√
4
b.
θ 1 − cos 4θdθ.
0
[like §8.4 #20,21]
Sol.
a.
Z
π
6
√
Z
sec2
x − 1dx =
− π6
π
6
Z
Z
| tan x|dx = −
tan xdx +
− π6
− π6
π
6
Z
0
= 2
Z
tan xdx = 2
0
Z
√
3
2
0
π
6
sin x
dx
cos x
du
by letting u = cos x
u
1
√3
4
2
= −2 ln |u| = ln
3
1
= −2
π
6
tan xdx
0
b.
Z
π
4
√ Z
θ 1 − cos 4θdθ =
2
√
0
π
4
θ
0
√ Z
=
2
r
√
1 − cos 4θ
dθ = 2
2
π
4
Z
θ| sin 2θ|dθ
0
π
4
θ sin 2θdθ
0
Let u = θ, v = sin 2θdθ, then du = dθ, v = − 21 cos 2θ and
hence
#
"
π
Z π
Z π
√
√
4 1 4
4
1
θ 1 − cos 4θdθ =
cos 2θdθ
2 − θ cos 2θ +
2
2 0
0
0
√
√
π
4
2
2
=
sin 2θ =
4
4
0
8. Find the length of the curve
y = ln(sin x),
π
π
≤x≤
3
2
[like §8.4 #40]
dy
cos x
Sol. Since y = ln(sin x), so
=
= cot x. Hence the
dx
sin x
length of the curve is
Z πp
Z πr
dy 2
2
2
1+
dx =
1 + cot2 xdx
L =
π
π
dx
3
3
Z π
Z π
Z π
2
2
2 csc x(csc x + cot x)
=
dx
| csc x|dx =
csc xdx =
π
π
π
cot
x
+
csc
x
3
3
3
Z π
2 − csc2 x − csc x cot x
= −
dx
π
cot x + csc x
3
Z 1
du
= − √
be letting u = cot x + csc x
3 u
1
1
= − ln |u|√ = ln 3
2
3