The document provides solutions to 11 integrals using substitution methods. The integrals involve trigonometric, exponential, logarithmic and algebraic functions. Substitutions are given for some integrals and need to be determined for others. The solutions show setting up the substitution, finding the differential, integrating and solving for the antiderivative. Some integrals have multiple valid substitution options. One integral is solved using a trigonometric identity in addition to substitution.

1. Solutions to Worksheet for Section 5.5
Integration by Substitution
V63.0121, Calculus I
December 7, 2009
Find the following integrals. In the case of an indefinite integral, your answer should be the
most general antiderivative. In the case of a definite integral, your answer should be a number.
In these problems, a substitution is given.
1. (3x − 5)17 dx, u = 3x − 5
1
Solution. As suggested, we let u = 3x − 5. Then du = 3 dx so dx = du. Thus
3
1 1 1 18 1
(3x − 5)17 dx = u17 du = · u +C = (3x − 5)18 + C
3 3 18 54
4
2. x x2 + 9 dx, u = x2 + 9
0
Solution. We have du = 2x dx, which takes care of the x and dx that appear in the integrand.
The limits are changed to u(0) = 9 and u(4) = 25. Thus
4 25
1 25 √ 1 2
x x2 + 9 dx = u du = · u3/2
0 2 9 2 3 9
1 98
= · (125 − 27) = .
3 3
√
e x √
3. √ dx, u = x.
x
1

2. 1
Solution. We have du = √ dx, which means that
2 x
√
e x
√ dx = 2 eu du = 2eu + C
x
cos 3x dx
4. , u = 5 + 2 sin 3x
5 + 2 sin 3x
Solution. We have du = 6 cos 3x dx, so
cos 3x dx 1 du 1 1
= = ln |u| + C = ln |5 + 2 sin 3x| + C
5 + 2 sin 3x 6 u 6 6
In these problems, you need to determine the substitution yourself.
5. (4 − 3x)7 dx.
1
Solution. Let u = 4 − 3x, so du = −3x dx and dx = − du. Thus
3
1 u8 1
(4 − 3x)7 dx = − u7 du = − + C = − (4 − 3x)8 + C
3 24 24
π/3
6. csc2 (5x) dx
π/4
1
Solution. Let u = 5x, so du = 5 dx and dx = du. So
5
π/3 5π/3
1
csc2 (5x) dx = csc2 u du
π/4 5 5π/4
5π/3
1
=− cot u
5 5π/4
1 5π 5π
= cot − cot
5 4 3
√
1 3
= 1+
5 3
2

3. 3
−1
7. x2 e3x dx
1
Solution. Let u = 3x3 − 1, so du = 9x2 dx and dx = du. So
9
3
−1 1 1
x2 e3x dx = eu du = eu + C
9 9
1 3x3 −1
= e .
9
Sometimes there is more than one way to skin a cat:
x
8. Find dx, both by long division and by substituting u = 1 + x.
1+x
Solution. Long division yields
x 1
=1−
1+x 1+x
So
x dx
dx = x −
1+x 1+x
To find the leftover integral, let u = 1 + x. Then du = dx and so
dx du
= = ln |u| + C
1+x u
Therefore
x
dx = x − ln |x + 1| + C
1+x
Making the substitution immediately gives du = dx and x = u − 1. So
x u−1 1
dx = du = 1− du
1+x u u
= u − ln |u| + C
= x + 1 − ln |x + 1| + C
It may appear that the two solutions are “different.” But the difference is a constant, and we
know that antiderivatives are only unique up to addition of a constant.
2z dz
, both by substituting u = z 2 + 1 and u =
3
9. Find √
3
z 2 + 1.
z2 + 1
3

4. Solution. In the first substitution, du = 2z dz and the integral becomes
du 3 2/3 3
√ = u−1/3 du = u + C = (z 2 + 1)2/3
3
u 2 2
In the second, u3 = z 2 + 1 and 3u2 du = 2z dz. The integral becomes
3u2 3 3 3
du = 3u2 du = u + C = (z 2 + 1)2/3 + C.
u 2 2
The second one is a dirtier substitution, but the integration is cleaner.
Use the trigonometric identity
cos 2α = cos2 α − sin2 α = 2 cos2 α − 1 = 1 − 2 sin2 α
to find
10. sin2 x dx
Solution. Using cos 2α = 1 − 2 sin2 α, we get
1 − cos 2α
sin2 α =
2
So
1 1
sin2 x dx = − cos 2x dx
2 2
x 1
= − sin 2x + C.
2 4
11. cos2 x dx
Solution. Using cos 2α = 2 cos2 α − 1, we get
1 + cos 2α
sin2 α =
2
So
1 1
sin2 x dx = + cos 2x dx
2 2
x 1
= + sin 2x + C.
2 4
4

5. 12. Find
sec x dx
by multiplying the numerator and denominator by sec x + tan x.
Solution. We have
sec x(sec x + tan x)
sec x dx = dx
sec x + tan x
sec2 x + sec x tan x
= dx
tan x + sec x
Now notice the numerator is the derivative of the denominator. So the substitution u = tan x+sec x
gives
sec x dx = ln |sec x + tan x| + C.
5