NAMES:
MATH 152
February 25, 2015
QUIZ 4
• Show all your work and indicate your final answer clearly. You will be graded not merely
on the final answer, but also on the work leading up to it.
1. (3 points) Evaluate the integral
Z
1
dx
− 9)3/2
Solution: Make the substitution x = 3 sec(θ) so that dx = 3 sec(θ) tan(θ) dθ. Then
Z
1
1
dx =
· 3 sec(θ) tan(θ) dθ
2
3/2
(x − 9)
27 tan3 (θ)
Z
1
sec(θ)
=
dθ
9
tan2 (θ)
Z
1
cos(θ)
=
dθ
9
sin2 (θ)
1
−1
·
+C
=
9 sin(θ)
√
−1
x2 − 9
=
·
+C .
9
x
(x2
2. (3 points) Evaluate the integral
Z
1
dx
16 − x2
Solution: Make the substitution x = 2 sin(θ). Thendx = 2 cos(θ) dθ. So
Z
Z
1
1
dx =
· 2 cos(θ) dθ
2
3/2
(4 − x )
8 cos3 (θ)
Z
1
=
sec2 (θ) dθ
4
1
= tan(θ) + C
4
1
x
= √
+C .
4 4 − x2
x2
√
NAMES:
MATH 152
February 25, 2015
3. (3 points) Evaluate the integral
Z √
e2t − 9 dt
Solution: Let et = 3 sec θ. Then et dt = 3 sec θ tan θ dθ ⇒ dt =
Substituting gives
Z √
Z √
2t
e − 9 dt =
9 sec2 θ − 9 tan θ dθ
Z √
= 3 tan2 θ tan θ dθ
Z
= 3 tan2 θ dθ
Z
= 3 sec2 θ − 1 dθ
3 sec θ tan θ
et
= 3(tan θ − θ)
√ 2t
e −9
3
−1
=3
− 3 cos
+ C.
3
et
√
3
−1
2t
= e − 9 − 3 cos
+C .
et
=
3 sec θ tan θ
3 sec θ
= tan θ.