Math 2300: Calculus II
Integrals: The Big Picture (solutions)
For each of these integrals, determine a strategy for evaluating. Don’t evaluate them,
just figure out which technique of integration will work, including what substitutions
you will use.
Z p
1.
x 9 − x2 dx
Set u = 9 − x2 so 21 du = xdx. Then
Z
Z p
1√
2
u du
x 9 − x dx =
2
1 2
= · u3/2 + C
2 3
1
= (9 − x2 )3/2 + C.
3
Z
2.
x2
p
9 − x2 dx
This is a trig substitution problem. Set x = 3 sin(ϑ), so dx = 3 cos(ϑ)dϑ. Then
Z
Z
p
2
2
x 9 − x dx = 9 sin2 (ϑ) · 3 cos(ϑ) · 3 cos(ϑ) dϑ
Z
= 81 sin2 (ϑ) cos2 (ϑ) dϑ
Z
81
=
(1 − cos2 (2ϑ)) dϑ
4
Z
81
=
sin2 (2ϑ) dϑ
4
Z
81
=
(1 − cos(4ϑ)) dϑ
8
81
81
= ϑ−
sin(4ϑ) + C
8
32
81
81
sin(2ϑ) cos(2ϑ) + C
= ϑ−
8
16
81
81
= ϑ−
sin(ϑ) cos(ϑ)(1 − 2 sin2 (ϑ)) + C
8
8
81
1 p
=
arcsin(x/3) − x 9 − x2 (9 − 2x2 ) + C.
8
8
Z
3.
sin6 (x) cos2 (x) dx
This is a trigonometric integral with multiple applications of the power reduction
formulae for sine and cosine. The solution is extremely lengthy, but otherwise
not any different from the above solution.
1
Math 2300: Calculus II
Z
4.
Integrals: The Big Picture (solutions)
sin5 (x) cos2 (x) dx
This is a trigonometric integral with a u-substitution. First, rewrite it as
Z
(1 − cos2 (x))2 cos2 (x) sin(x) dx.
Let u = cos(x), so −du = sin(x)dx. Then
Z
Z
2
2
2
(1 − cos (x)) cos (x) sin(x) dx = −(1 − u2 )2 u2 du
Z
= (−u2 + 2u4 − u6 ) du
1
2
1
= − u3 + u5 − u7 + C
3
5
7
2
1
1
= − cos3 (x) + cos5 (x) − cos7 (x) + C.
3
5
7
Z
5.
3
dx
x2 + 5x + 4
This is a partial fractions problem. Factoring the denominator, we have
3
A
B
=
+
,
(x + 4)(x + 1)
x+4 x+1
so by clearing the denominators in the above equation, we have
3 = A(x + 1) + B(x + 4).
Setting x = −4 yields 3 = −3A, so A = −1. Setting x = −1 yields 3 = 3B, so
B = 1. Thus,
Z
Z
Z
3
1
1
dx =
dx −
dx
2
x + 5x + 4
x+1
x+4
= ln|x + 1| − ln|x + 4| + C.
Z
6.
3
dx
+ 6x + 9
This looks like a partial fractions problem, but it is not. Factoring the denominator yields
Z
Z
3
3
dx =
dx
2
x + 6x + 9
(x + 3)2
x2
Set u = x + 3, so du = dx. Then
Z
=
3u−2 du
= −3u−1 + C
= −3(x + 3)−1 + C.
2
Math 2300: Calculus II
Integrals: The Big Picture (solutions)
Z
7.
arcsin(x) dx
This is an integration by parts problem. Set u = arcsin(x) and dv = dx. Then
1
du = √1−x
dx and v = x. Then
2
Z
Z
x
dx
arcsin(x) dx = x arcsin(x) − √
1 − x2
Set u = 1 − x2 , so − 21 du = xdx. Then
Z
1
1
√ du
= x arcsin(x) +
2
u
√
= x arcsin(x) + u + C
p
= x arcsin(x) + 1 − x2 + C.
Z
arctan(x)
dx
1 + x2
This is a u-substitution problem. Set u = arctan(x), so du =
Z
Z
arctan(x)
dx = u du
1 + x2
1
= u2 + C
2
1
= (arctan(x))2 + C.
2
Z
√
9.
x x + 2 dx
8.
1
1+x2
dx. Then
This is a clever u-substitution. Set u = x + 2 so du = dx and x = u − 2. Then
Z
Z
√
√
x x + 2 dx = (u − 2) u du
Z
= (u3/2 − 2u1/2 ) du
2
4
= u5/2 − u3/2 + C
5
3
2
4
= (x + 2)5/2 − (x + 2)3/2 + C.
5
3
Z
10.
√
(x + 2) x dx
We have
Z
√
Z
(x + 2) x dx =
(x3/2 + 2x1/2 ) dx
2
4
= x5/2 + x3/2 + C
5
3
3
Math 2300: Calculus II
Integrals: The Big Picture (solutions)
ex
dx
4 + e2x
This is a u-substitution and a formula from the table of integrals. Set u = ex , so
du = ex dx, and note that e2x = (ex )2 . Then
Z
Z
1
ex
dx =
du
2x
4+e
4 + u2
1
= arctan(u/2) + C
2
1
= arctan(ex /2) + C.
2
Z
1
dx
12.
x ln(x)
This is a u-substitution problem. Set u = ln(x), so du = x1 dx. Then
Z
Z
1
1
dx =
du
x ln(x)
u
= ln|u| + C
Z
11.
= ln|ln(x)| + C.
Z
13.
x2 cos(5x) dx
This is a double integration by parts. For the first application, set u = x2 and
dv = cos(5x) dx, so du = 2x dx and v = 51 sin(5x). Then
Z
Z
x2
2
x2 cos(5x) dx =
sin(5x) −
x sin(5x) dx.
5
5
For the integral on the right hand side, we must use integration by parts again,
this time with u = x and dv = sin(5x) dx. Then du = dx and v = − 15 cos(5x).
Hence,
Z
Z
x2
2
x
1
x2 cos(5x) dx =
sin(5x) −
− cos(5x) +
cos(5x) dx
5
5
5
5
Z
2
x
2x
2
=
sin(5x) +
cos(5x) −
cos(5x) dx
5
25
25
x2
2x
2
=
sin(5x) +
cos(5x) −
sin(5x) + C.
5
25
125
Z 2
x +1
14.
dx
x
Break the fraction up and integrate directly:
Z 2
Z x +1
1
dx =
x+
dx
x
x
1
= x2 + ln|x| + C.
2
4
Math 2300: Calculus II
Integrals: The Big Picture (solutions)
Z
15.
x+5
dx
x2 + 4
Break the fraction up, and we have two integrals. The first is u-substitution with
u = x2 + 4 (so du = 2x dx), while the second integral is a formula from the table
of integrals.
Z
Z
Z
x+1
x
5
dx
=
dx
+
dx
2
2
2
x +4
x +4
x +4
Z
1
1
5
=
du + arctan(x/2) + C
2
u
2
1
5
= ln|u| + arctan(x/2) + C
2
2
1
5
= ln(x2 + 4) + arctan(x/2) + C.
2
2
Z
16.
tan4 (x) sec2 (x) dx
This is a u-substitution trig integral. Set u = tan(x), so du = sec2 (x) dx. Then
Z
Z
tan4 (x) sec2 (x) dx = u4 du
1
= u5 + C
5
1
= tan5 (x) + C.
5
5