Math 166
Quiz 3 Solution
Name:
Directions: This quiz is worth a total of 10 points. To receive full credit, all work must be shown.
Z
1.
a. Evaluate
2 tan 2x dx. Hint: recall that tan 2x =
sin 2x
and use substitution.
cos 2x
Let u = cos 2x. Then du = −2 sin 2x dx and thus
Z
Z
du
2 tan 2x dx = −
= − ln | cos 2x| + C.
u
Z
b. Evaluate
4x sec2 2x dx using integration by parts.
Let
u = 4x,
dv = sec2 2x dx.
Then
du = 4 dx,
v=
1
tan 2x.
2
Therefore
Z
2
Z
4x sec 2x dx = 2 tan 2x +
2 tan 2x dx
= 2 tan 2x − ln | cos 2x| + C.
Z
2. Evaluate
3
x5 ex dx. Hint: make a substitution before using integration by parts. Use a letter like y
rather than u when making your substitution.
Let y = x3 . Then dy = 3x2 dx and
Z
Z
Z
3
3
1
yey dy.
x5 ex dx = x3 ex x2 dx =
3
Now let
u = y,
dv = ey dy.
Then
du = dy,
v = ey .
Therefore
1
3
We conclude that
Z
3
Z
x5 ex dx =
Z
1
y
y
ye dy =
ye − e dy
3
1
= [yey − ey ] + C.
3
y
i
3
1 h 3 x3
1
[yey − ey ] + C =
x e − ex + C.
3
3