MATH 101 HOMEWORK 5 – SOLUTIONS
1. Find the area of the finite planar region that lies inside the circle x2 + y 2 = 25 and
above the line x + y = 7.
The intersection points are (3, 4) and (4, 3). The area is
Z
4
A=
p
Z
25 −
x2 dx
4
−
3
(7 − x)dx.
3
We evaluate the first integral using the substitution x = 5 sin θ, dx = 5 cos θ dθ:
Z p
Z
Z
25
25
25
2
(cos(2θ) + 1)dθ =
sin(2θ) + θ + C
25 − x2 dx = 25 cos θ dθ =
2
4
2
√
x
25
25
25 x 25 − x2
25
=
sin θ cos θ + θ + C =
+
sin−1
+C
2
2
2
25
2
5
√
x
25
x 25 − x2
+
sin−1
=
+ C,
2
2
5
Z
4
3
√
x 4 25 4
3 p
25
x 25 − x2
2
+
sin−1
25 − x dx =
sin−1
− sin−1
.
=
2
2
5 3
2
5
5
We also have
Z
4
(7 − x)dx = 7x −
3
Hence the area is
x2 4
7
= .
2 3
2
7
25 −1 4 −1 3
A=
sin
− sin
− .
2
5
5
2
2. Evaluate the integrals:
(a) Substitute x = sec θ, dx = sec θ tan θ dθ,
Z
x3
√
dx =
x2 − 1
Z
√
x2 − 1 = tan θ, then
sec3 θ
sec θ tan θ dθ =
tan θ
1
Z
Z
4
sec θdθ =
(1 + tan2 θ) sec2 θ dθ
Z
=
(1 + u2 )du = u +
tan3 θ
u3
+ C = tan θ +
+ C,
3
3
where at the last step we substituted u = tan θ, du = sec2 θ dθ. Converting this back to x,
we get
Z
p
(x2 − 1)3/2
x3
√
dx = x2 − 1 +
+ C.
3
x2 − 1
Z
x
dx =
2
2
(x − 1)(x − 4)(x2 − 9)
(b)
Z
x
dx
(x − 3)(x − 2)(x − 1)(x + 1)(x + 2)(x + 3)
Z A
B
C
D
E
F =
+
+
+
+
+
dx,
x−3 x−2 x−1 x+1 x+2 x+3
where
A=
3
1
2
1
1
1
=
, B=
=− , C=
=
,
1·2·4·5·6
80
(−1) · 1 · 3 · 4 · 5
30
(−2) · (−1) · 2 · 3 · 4
48
D=
−1
1
−2
1
=
, E=
=− ,
(−4) · (−3) · (−2) · 1 · 2
48
(−5) · (−4) · (−3) · (−1) · 1
30
F =
1
−3
=
.
(−6) · (−5) · (−4) · (−2) · (−1)
80
Hence the integral is
ln |x − 3| ln |x − 2| ln |x − 1| ln |x + 1| ln |x + 2| ln |x + 3|
−
+
+
−
+
.
80
30
48
48
30
80
Z
Z A
2
B
Cx + D (c)
dx
=
+
dx. To find A, B, C, D, we
+
(x − 1)2 (x2 + 4)
x − 1 (x − 1)2
x2 + 4
add the fractions on the right and compare the coefficients:
A(x − 1)(x2 + 4) + B(x2 + 4) + (Cx + D)(x − 1)2 = 2,
A(x3 − x2 + 4x − 4) + B(x2 + 4) + (Cx3 + Dx2 − 2Cx2 − 2Dx + Cx + D) = 2,
A+C =0
−A + B + D − 2C = 0
4A − 2D + C = 0
−4A + 4B + D = 2.
From the first equation, C = −A. Substitute this into third equation: 3A − 2D = 0,
D = 3A/2. Now substitute this into the remaining two equations to get 5A + 2B = 0
2
and −5A + 8B = 4. Add these two equations: 10B = 4, B = 2/5. Then A = −4/25,
D = −6/25, C = 4/25. We get
Z
4x
1
4
10
6 +
−
+
−
dx
25
x − 1 (x − 1)2
x2 + 4 x2 + 4
1
10
2
−1 x
=
− 4 ln |x − 1| −
+ 2 ln(x + 4) − 3 tan ( ) + C 0 .
25
x−1
2
Z
π/2
3. Evaluate the integral
0
dx
, or show that it diverges.
sin3 x
Using that sin x = cos(x − π/2), we get
Z
π/2
0
dx
=
sin3 x
Z
π/2
π
sec (x − ) dx =
2
Z
0
3
0
−π
2
sec3 u du.
The indefinite integral is evaluated in the textbook (Example 3, Section 6.1):
Z
1
1
sec3 u du = sec u tan u + ln | sec u + tan u| + C.
2
2
So, we need to find
limπ
1
t→− 2 +
=
limπ
1
t→− 2 +
2
2
sec u tan u +
0
1
ln | sec u + tan u| 2
t
1
1
1
ln | sec 0 + tan 0| − sec t tan t − ln | sec t + tan t|
2
2
2
1
1
= limπ
− sec t tan t − ln | sec t + tan t| .
t→− 2 +
2
2
sec 0 tan 0 +
As t → − π2 +, sec t → ∞, tan t → −∞, and
limπ (sec t + tan t) =
t→− 2 +
1 + sin t
cos t
= limπ
= 0,
t→− 2 +
t→− 2 + − sin t
cos t
limπ
using l’Hospital’s rule.
1
limπ − sec t tan t = −∞ · (−∞) = ∞,
t→− 2 +
2
1
1
limπ − ln | sec t + tan t| = − ln(0) = ∞,
t→− 2 +
2
2
and the integral is ∞ + ∞ = ∞, divergent.
Note: this can also be proved using the comparison test.)
3