Worksheet 2
Spring 2016
MATH 222, Week 2: I.6,I.8,I.10
Name: SOLUTIONS
Problem 1. Compute
R
x ln(x) dx.
Solution 1.
We integrate by parts. Let u = ln(x) and v 0 = x. Then u0 = 1/xdx and v = x2 /2. We then know:
Z
Z
Z
x2 ln(x)
x
0
2
x ln(x)dx = uv − vu = ln(x)(x /2) −
dx =
− x2 /4 + C
2
2
Problem 2.
(a) Compute
Rπ
0
cos(x)dx and
Rπ
0
x2 cos(x)dx
(b) Show that:
Z
n
n
x cos(x) dx = x sin(x) + nx
n−1
Z
cos(x) − n(n − 1)
(Hint: The steps are very similar to what you did in part (a) for
(c) Use the identity you just proved and part (a) to compute
Rπ
0
Rπ
0
xn−2 cos(x) dx
x2 cos(x) dx).
x4 cos(x) dx.
Solution 2.
(a)
Rπ
0
cos(x) = 0 and
Rπ
0
x2 cos(x)dx = −2π.
(b) Apply integration by parts twice with u = xn v 0 = cos(x) in the first step and u = xn−1 and v 0 = sin(x) in
the second step.
(c)
R4
0
x2 cos(x)dx = −4π(π 2 − 6)
Problem 3. Compute
R
x7 sin(2x4 ) dx.
Solution 3.
This is tricky and should serve as a reminder that the split for u and v 0 won’t always be obvious. Intuitively
you want to set v 0 = sin(2x4 ), but this is nearly impossible to antidifferentiate. The issue is that when you take
the derivative of − cos(2x4 ) you pick up an extra 8x3 thanks to the chain rule. You can fix this though! Choose
v 0 = x3 sin(2x4 ) and u = x4 . You’ll only need one round of integration by parts here (not seven)! u0 = 4x3 and
v = − cos(2x4 )/8, so we know:
Z
Z
Z
7
4
0
4
4
x sin(2x ) dx = uv − vu dx = −x cos(2x )/8 + x3 cos(2x4 )/2 dx
You can compute the integral by u-sub or just straight up guess and check to find:
Z
1
x7 sin(2x4 ) dx = (sin(2x4 ) − 2x4 cos(2x4 )) + C
16
Problem 4. Compute
R
1
x2 −4
dx
Solution 4.
Using partial fractions we can write
Z
Problem 5. Compute
R
x3
x2 +2
1
x2 −4
1/4
x−2
=
−
1/4
x+2
we can then integrate:
x3
1
dx = (ln(2 − x) − ln(2 + x)) + C
2
x +2
4
dx
Solution 5.
Use partial fractions again to write
x3
x2 +2
Z
Problem 6. Compute
R
1
2+e2t
=
−2x
x2 +2
+ x. We can then integrate:
x3
x2
dx
=
− ln(x2 + 2) + C
x2 + 2
2
dt
Solution 6.
We use partial fractions again, writing
Z
1
2+e2t
= 1/2 −
1/2e2t
2+e2t
we can then integrate:
t
1
1
dt = − ln(e2t + 2) + C
2 + e2t
2 4
2