These problems demonstrate techniques for integration by substitution. Several integrals are solved by making appropriate substitutions to simplify the integrands, including substituting u = 3x - 5, u = x^2 + 9, u = √x, u = 5 + 2sin3x, and u = 4 - 3x. Trigonometric identities are also used to rewrite sin^2x and cos^2x terms before integrating. One integral is solved using long division and another using two different substitutions. The last integral is solved by multiplying the numerator and denominator by secx + tanx before making the substitution u = tanx + secx.

1. Solutions to Worksheet for Section 5.5
Integration by Substitution
V63.0121, Calculus I
Summer 2010
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
Solution. We have du = √ dx, which means that
2 x
√
e x
√ dx = 2 eu du = 2eu + C
x
1

2. 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
3
−1
7. x2 e3x dx
2

3. 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