Math 21B, Winter 2017 – Homework 7 Solutions
Section 8.4
Z
dx
√
1.
9 + x2
Solution. Let x = 3 tan(θ). Then dx = 3 sec2 (θ) dθ, so we have:
Z
Z
Z
Z
3 sec2 (θ) dθ
3 sec2 (θ) dθ
dx
√
p
=
= sec(θ) dθ = ln | sec(θ)+tan(θ)|+C
=
3 sec(θ)
9 + x2
9 + 9 tan2 (θ)
Since tan(θ) = x3 , we have that sec(θ) =
√
9+x2
.
3
As such:
√
9 + x2 + x ln | sec(θ) + tan(θ)| + C = ln +C
3
Z
4.
0
2
dx
8 + 2x2
Solution. Let x = 2 tan(θ). Then dx = 2 sec2 (θ) dθ. Furthermore, solving 0 = 2 tan(θ)
gives θ = 0, while solving 2 = 2 tan(θ) gives θ = π4 , so we have:
π
Z π
Z π
Z π
Z 2
4
4 2 sec2 (θ) dθ
4 1
θ 4
π
dx
2 sec2 (θ) dθ
=
=
=
dθ
=
=
2
2
8 sec2 (θ)
4 0
16
0 8 + 2x
0 8 + 8 tan (θ)
0
0 4
Z
6.
1
√
2 2
0
2 dx
√
1 − 4x2
Solution. Let 2x = sin(θ). Then 2 dx = cos(θ) dθ. Furthermore, solving 0 = sin(θ)
gives θ = 0 and solving √12 = sin(θ) gives θ = π4 . Then we have:
Z
0
Z
10.
√
1
√
2 2
2 dx
√
=
1 − 4x2
5 dx
,x>
25x2 − 9
Z
0
π
4
cos(θ) dθ
p
=
1 − sin2 (θ)
Z
0
π
4
π
4
cos(θ)
π
dθ = θ =
cos(θ)
4
0
3
5
Solution. Let 5x = 3 sec(θ). Then 5 dx = 3 sec(θ) tan(θ) dθ. This gives:
Z
Z
Z
Z
5 dx
3 sec(θ) tan(θ) dθ
3 sec(θ) tan(θ)
√
p
=
=
dθ = sec(θ) dθ
3 tan(θ)
25x2 − 9
9 sec2 (θ) − 9
√
5x + 25x2 − 9 +C
= ln | sec(θ) + tan(θ)| + C = ln 3
Z
23.
0
√
3
2
4x2 dx
(1 − x2 )3/2
Solution. Let x = √
sin(θ). Then dx = cos(θ) dθ. Furthermore, solving 0 = sin(θ) gives
θ = 0 and solving 23 = sin(θ) gives θ = π3 . Then:
√
Z
3
2
0
Z
=
4x2 dx
=
(1 − x2 )3/2
π
3
0
Z
2
4 tan (θ) dθ =
0
Z
57. Evaluate
π
3
Z
0
4 sin2 (θ)
cos(θ) dθ =
(1 − sin2 (θ))3/2
π
3
π
3
4 sin2 (θ) cos(θ)
dθ
cos3 (θ)
0
π
√
3
4π
2
4(sec (θ) − 1) dθ = 4 tan(θ) − 4θ = 4 3 −
3
0
Z
√
x3 1 − x2 dx using:
(a) Integration by parts
√
Solution. Let u = 12 x2 and dv = 2x 1 − x2 dx. Then du = x dx and (using the
substitution w = 1 − x2 ) v = − 23 (1 − x2 )3/2 . This gives:
Z
Z
√
1
2
3
2
2
3/2
x 1 − x2 dx = − x (1 − x ) +
x(1 − x2 )3/2
3
3
Again using the substitution w = 1 − x2 , we obtain:
Z
2
2
x(1 − x2 )3/2 dx = − (1 − x2 )5/2 + C
3
15
Therefore:
Z
√
1
2
x3 1 − x2 dx = − x2 (1 − x2 )3/2 − (1 − x2 )5/2 + C
3
15
(b) Substitution
Solution. Let u = 1 − x2 . Then du = −2x dx and x2 = 1 − u, so we have:
Z
Z
Z
√
√
√
1
1
3
2
u3/2 − u du
x 1 − x dx = − (1 − u) u du =
2
2
1
1
1
1
= u5/2 − u3/2 + C = (1 − x2 )5/2 − (1 − x2 )3/2 + C
5
3
5
3
(c) Trigonometric substitution
Solution. Let x = sin(θ). Then dx = cos(θ) dθ, so we have:
Z
Z
Z
q
√
3
3
2
x 1 − x2 dx = sin (θ) 1 − sin (θ) cos(θ) dθ = sin3 (θ) cos2 (θ) dθ
Let u = cos(θ). Then du = − sin(θ) dθ, so since sin2 (θ) = 1 − cos2 (θ), we have:
Z
Z
Z
u5 u3
3
2
2 2
4
2
−
+C
sin (θ) cos (θ) dθ = −(1 − u )u du = u − u du =
5
3
√
√
cos5 (θ) cos3 (θ)
( 1 − x2 )5 ( 1 − x2 )3
=
−
+C =
−
+C
5
3
5
3
Section 8.5
2x + 2
4. Expand 2
in partial fractions.
x − 2x + 1
A
B
= x−1
+ (x−1)
Solution. By factoring x2 − 2x + 1 = (x − 1)2 , we can write x22x+2
2.
−2x+1
Clearing denominators gives 2x + 2 = A(x − 1) + B. Then taking x = 1 gives B = 4
and taking x = 0 gives A = 2. As such, the partial fractions decomposition is
x2
6. Expand
z3
2
4
2x + 2
=
+
− 2x + 1
x − 1 (x − 1)2
z
in partial fractions.
− z 2 − 6z
Solution. By factoring z 3 − z 2 − 6z = z(z 2 − z − 6) = z(z − 3)(z + 2), we can write
B
C
z
= Az + z−3
+ z+2
. Clearing denominators gives z = A(z − 3)(z + 2) + Bz(z +
z 3 −z 2 −6z
2) + Cz(z − 3). Then we can take z = 0 to get A = 0, z = 3 to get B = 15 , and z = −2
to get C = − 51 . As such, the partial fractions decomposition is
z3
Z
10.
1
1
z
=
−
2
− z − 6z
5(z − 3) 5(z + 2)
dx
x2 + 2x
1
B
Solution. We can write x2 +2x
= Ax + x+2
. Then 1 = A(x + 2) + Bx, so taking x = 0
and x = −2 gives A = 21 and B = − 21 . As such:
Z
Z
dx
1
1
1
1 x =
−
dx = (ln |x| − ln |x + 2|) + C = ln +C
x2 + 2x
2x 2(x + 2)
2
2
x + 2
Z
17.
0
1
x3
dx
x2 + 2x + 1
Solution. By polynomial divison, we have that
x3
3x + 2
=x−2+ 2
2
x + 2x + 1
x + 2x + 1
A
B
We can then write x23x+2
= x+1
+ (x+1)
2 which gives 3x + 2 = A(x + 1) + B. Then
+2x+1
A = 3, and taking x = −1 gives B = −1. As such:
Z 1
Z 1
x3
3
1
x
−
2
+
dx
=
−
dx
2
x + 1 (x + 1)2
0 x + 2x + 1
0
1
1 x2
− 2x + 3 ln |x + 1| +
=
= 3 ln(2) − 2
2
x + 1 0
Z
39.
et
dt
e2t + 3et + 2
Solution. Let x = et . Then dx = et dt, so we have:
Z
Z
et
1
dt =
dx
2t
t
2
e + 3e + 2
x + 3x + 2
1
A
B
We can write x2 +3x+2
= x+1
+ x+2
. Then 1 = A(x + 2) + B(x + 1), so taking x = −1
and x = −2 gives A = 1 and B = −1. As such:
Z
Z
1
1
1
dx
=
−
dx = ln |x + 1| − ln |x + 2| + C
x2 + 3x + 2
x+1 x+2
t
x + 1
e + 1
+ C = ln = ln et + 2 + C
x + 2
Z √
47.
x+1
dx
x
Solution. Let x + 1 = u2 . Then dx = 2u du and x = u2 − 1, so we have:
Z √
Z
Z
Z
1
x+1
u
u2
dx =
2u
du
=
2
du
=
2
1
+
du
x
u2 − 1
u2 − 1
u2 − 1
A
B
We can write u21−1 = u−1
+ u+1
. Then 1 = A(u + 1) + B(u − 1), so taking u = 1 and
1
u = −1 gives A = 2 and B = − 12 . As such:
Z
Z
1
1
1
du = 2 1 +
−
du = 2u + ln |u − 1| − ln |u + 1| + C
2 1+ 2
u −1
2(u − 1) 2(u + 1)
√
u − 1
x + 1 − 1
√
+ C = 2 x + 1 + ln √
= 2u + ln x + 1 + 1 + C
u + 1