Math 2260 Written HW #7 Solutions
1. Evaluate the indefinite integral
Z
(Hint: what is the derivative of
√
√
x3
dx.
4 + x2
4 + x2 ?)
Answer: Using the hint, notice that the derivative of
√
√
4 + x2 is
x
.
4 + x2
Then if we re-write the above integral as
Z
x
dx,
x2 √
4 + x2
it’s natural to do integration by parts with
x
dv = √
dx
4 + x2
p
v = 4 + x2 .
u = x2
du = 2x dx
Then the integral above is equal to
Z
p
p
2
2
x 4 + x − 2x 4 + x2 dx.
To evaluate the remaining integral, just let u = 4 + x2 , so that du = 2x dx. Then we have
Z
p
p
p
√
2
2
2
2
x 4+x −
u du = x2 4 + x2 − u3/2 + C = x2 4 + x2 − (4 + x2 )3/2 + C.
3
3
2. Evaluate the indefinite integral
Z
sin2 x cos3 x dx.
Answer: Notice that the power on the cosine is odd, so we will peel off one power of cosine
and convert the rest into sines:
Z
Z
2
3
sin x cos x dx = sin2 x cos2 x cos x dx
Z
= sin2 x(1 − sin2 x) cos x dx.
Now, let u = sin x so that du = cos x dx. Then the above integral becomes
Z
Z
u3 u5
u2 (1 − u2 ) du = (u2 − u4 ) du =
−
+ C.
3
5
Converting back to x yields the final answer:
sin3 x sin5 x
−
+ C.
3
5
1
3. Solve the initial-value problem
(x2 + 9)
dy
= 6,
dx
y(3) = 0.
Answer: As usual when solving differential equations, we first separate variables:
6
dx.
x2 + 9
dy =
Now we want to integrate both sides:
Z
Z
6
dx
+9
Z
6
y=
dx.
2
x +9
dy =
x2
To evaluate the x integral, make the trig substitution x = 3 tan θ. Then dx = 3 sec2 θ dθ and
the integral becomes
Z
Z
Z
sec2 θdθ
6
2
3
sec
θ
dθ
=
2
=
2
dθ = 2θ + C.
sec2 θ
9 tan2 θ + 9
Now, recall that x = 3 tan θ, so θ = arctan(x/3), and we conclude that
y = 2 arctan x/3 + C.
Now, we can solve for C using the initial condition y(3) = 0:
y(3) = 2 arctan(3/3) + C
0 = 2 arctan(1) + C
π
0 = 2 + C,
4
so we see that C = − π2 and so the solution to the initial-value problem is
y = 2 arctan(x/3) −
2
π
.
2