Solutions to quizzes
1. Quiz 1
Evaluate the integral
Z
sin x ln (cos x) dx
Solution
Recall the formula for integration by parts:
Z
Z
0
f (x)g (x) dx = f (x)g(x) − f 0 (x)g(x) dx
• Since we don’t know how to find the antiderivative of ln cos x we cannot choose ln cos x for g 0 . So we
choose f = ln cos x and g 0 = sin x.
• We find that g = − cos x by antidifferentiating g 0 = sin x and f 0 = cos1 x (− sin x) by differentiating
f = ln cos x (and using the Chain Rule).
• Applying the formula for integration by parts we get:
Z
Z
1
(−sinx)(− cos x) dx =
sin x ln (cos x) dx = ln (cos x) · (− cos x) −
cos
x
Z
= − cos x ln (cos x) − sin x dx = − cos x ln (cos x) − (− cos x) + C = − cos x ln (cos x) + cos x + C
2
2. Quiz 2
Evaluate the integral
Z
cos5 x tan2 x dx
Solution
• We can rewrite the integrand as the product of a power of sine and a power of cosine using the definition
sin x
of tan x, that is tan x = cos
x
Z
Z
Z
sin2 x
5
2
5
= cos3 x sin2 x dx
cos x tan x dx = cos x ·
cos2 x
• Since the power of cosine is odd, we save a factor of cos x to prepare for the substitution u = sin x
Z
Z
Z
3
2
2
2
cos x sin x dx = cos x sin x cos x dx = (1 − sin2 x) sin2 x cos x dx
• Where we used the trigonometric identity cos2 x = 1 − sin2 x in the last step. If we now set u = sin x,
then du = cos x dx, so using the substitution rule:
Z
(1 − sin2 x) sin2 x cos x dx =
Z
(1 − u2 )u2 du =
Z
• It turns out that
Z
cos5 x tan2 x dx =
(u2 − u4 ) du =
u3 u5
sin3 x sin5 x
−
+C =
−
+C
3
5
3
5
sin3 x sin5 x
−
+C
3
5
3
3. Quiz 3
Evaluate the integral
Z
√
3
x
dx
x+1
Solution
There are are at least two ways to solve this integral.
First way. We can use the rationalizing substitution approach of Section 7.4 and let u =
for x we obtain x = u3 − 1, whence dx = 3u2 du. Thus, using the substitution rule:
Z
Z 3
Z
x
u −1
2
√
dx =
· 3u du = 3 u4 − u du
3
u
x+1
!
5
(x + 1)5/3 (x + 1)2/3
u
u2
=3
−
+C =3
−
+C
5
2
5
2
√
3
x + 1. Solving
Second way. Or alternatively, we can substitute v = x + 1. We have dv = dx, so, by the substitution
rule:
Z
Z
Z
v−1
v
1
x
√
√
√
dx =
dv =
−√
dv =
3
3
3
3
v
v
v
x+1
2
1
Z
v 3 +1
v − 3 +1
2/3
−1/3
= v −v
dv = 2
− 1
+C =
−3 + 1
3 +1
!
3 5/3 3 2/3
(x + 1)5/3 (x + 1)2/3
= v − v +C =3
−
+C
5
2
5
2