Math 138 Midterm Solutions
[1] 1. (a) What substitution should you use to evaluate Z1
√b2x2+a2dx?
(You do not need to evaluate the integral).
Solution: x=a
btan θor θ= tan−1bx
a,−π
2< θ < π
2.
[2] (b) Find all constant solutions to the differential equation (x2+x+ 1)y0+y=y3.
Solution: Let y=k. We then have y0= 0, so we get (x2+x+ 1)(0) + k=k3. Thus,
0 = k3−k=k(k+ 1)(k−1). Hence, the constant solutions are k= 0, k= 1, and k=−1.
[2] (c) Use the direction field below to sketch the graph of the solution of the corresponding
differential equations that satisfies y(0) = 1.
[2] (d) Find d
dθ Zcos θ
sin θ
1
1−x2dx.
Solution: Let cbe a number between sin θand cos θ. Then
d
dθ Zcos θ
sin θ
1
1−x2dx =d
dθ Zc
sin θ
1
1−x2dx +d
dθ Zcos θ
c
1
1−x2dx
=−d
dθ Zsin θ
c
1
1−x2dx +d
dθ Zcos θ
c
1
1−x2dx
=−1
1−sin2θ(cos θ) + 1
1−cos2θ(−sin θ) by FTC and the Chain Rule
=−1
cos θ−1
sin θ
1
2. Find the following indefinite integrals.
[2] (a) Zxsin(x2)dx.
Solution: We have Zxsin(x2)dx =1
2sin(x2) + c
[3] (b) Ztan4θ dθ.
Solution: We have
Ztan4θ dθ =Ztan2θ(sec2θ−1) dθ =Zsec2θtan2θ−tan2θ dθ =Zsec2θtan2θ−Zsec2θ−1θ
Let u= tan θ, then du = sec2θ dθ. So, we have
Ztan4θ dθ =Zu2du −Zsec2θ−1θ=1
3u3−tan θ+θc =1
3tan3θ−tan θ+θ+c
[4] (c) Z4
(x2+ 1)(x−1)(x+ 1) dx.
