21-122 Homework # 1— Solutions
1. Problem: Evaluate the indefinite integral
Z
(2 + 4x)3/2 dx.
Solution: Use the substitution u = 2 + 4x. We have
Z
(2 + 4x)3/2 dx =
Z
u3/2
du
1
=
4
4
Z
u3/2 du =
du
1
= 4 so dx = du. So
dx
4
1 2 5/2
1
· u + C = (2 + 4x)5/2 + C,
4 5
10
where C is an arbitrary constant.
2. Problem: Evaluate the definite integral
Z
π/4
0
sin x
dx.
cos2 x
du
= − sin x so sin(x)dx = −du. Now we
dx
apply the substitution rule for definite integrals (page 411 in Stewart). The lower limit x = 0
corresponds
to u = cos(0) = 1 and the upper limit x = π/4 corresponds to u = cos(π/4) =
√
1/ 2, so
Solution: Use the substitution u = cos x. Then
Z 1/√2
Z 1/√2
Z 1
sin x
1
1
1
dx =
(−du) = −
du =
du
√
2
2
2
2
cos x
u
u
0
1
1
1/ 2 u
√
1 u=1
1 √
= −
= − + 2 = 2 − 1.
√
u u=1/ 2
1
Z
π/4
Note that we do not have to change back to x when using the substitution rule. It is still
possible to evaluate this integral without using the substitution rule, but that is not always
be the case.
3. Problem: Evaluate the definite integral
Z 2
x6
−2
sin5 x
dx.
cos4 x
Remark on solution: As mentioned in class on Friday 5/20, this integral is divergent1 .
π
The function has asymptotes at x = ± . I didn’t realize this when I typed up the homework.
2
The “spirit” of the problem is to notice that the function is odd and quickly deduce that the
integral is zero. The grading does not require you to check for asymptotes, only to apply
symmetry.
1
The reason that it diverges is slightly too difficult to go through at this point in the course. I might come back
to it later.
1
Solution using symmetry, ignoring asymptotes: See page 412 in Stewart. Let f (x) =
sin5 x
x6 4 . Then
cos x
f (−x) = (−x)6
5
5
(sin(−x))5
6 (− sin x)
6 sin x
=
(−x)
=
−x
= −f (x),
(cos(−x))4
(− cos x)4
cos4 x
so f is odd and we have
Z
2
sin5 x
dx = 0.
cos4 x
x6
−2
4. Problem: Evaluate the definite integral
Z
2
xe2x dx.
0
Solution: Use integration by parts. The integral becomes easier if we differentiate x and
integrate e2x , so we let
u = x, dv = e2x dx,
1
du = dx, v = e2x .
2
Then
Z
Z
1 2x
1 2x
2x
(x)(e dx) = x e −
e dx
2
2
x
1
= e2x − e2x
2
4
and so
Z
0
2
xe2x dx =
x 2x 1 2x
e − e
2
4
2
x=0
1
1
3
1
= e4 − e4 − 0 + = e4 + .
4
4
4
4
5. Problem: Evaluate the indefinite integral
Z
e2x sin x dx.
Solution: Here we use a standard trick – see Example 4 on page 466 of Stewart. Let
u = e2x ,
dv = sin x dx,
du = 2e2x dx, v = − cos x.
Let I(x) denote the integral we are evaluating. Then
Z
Z
I(x) = e2x sin x dx = −e2x cos x + 2 e2x cos x dx.
Now set
u = e2x ,
dv = cos x dx,
2x
du = 2e dx,
v = sin x
2
so
2x
Z
e2x cos x dx + C
Z
2x
2x
2x
= −e cos x + 2 e sin x − 2 e sin x dx + C
I(x) = −e
cos x + 2
= −e2x cos x + 2e2x sin x − 4I(x) + C
and we can solve for I(x):
1
I(x) = e2x (2 sin x − cos x) + C.
5
(Note that C is always an arbitrary constant – technically we should write C/5 in the last
step but it is understood that C and C/5 are essentially the same thing when C is arbitrary.)
3