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The following are solutions to the Math 229 Integration Worksheet - Substitution Method. Here’s
the link to that worksheet http://www.math.niu.edu/courses/math229/misc/int_prac.pdf
(5x + 4)^5 dx
(a) Let u = 5x + 4
(b) Then du = 5 dx or
du = dx.
(c) Now substitute ∫ (5x + 4)^5 dx =
u^5 ·
du
=
u^5 du
=
u^6 + C
=
(5x + 4)^6 + C
3 t^2 (t^3 + 4)^5 dt
(a) Let u = t^3 + 4
(b) Then du = 3t^2 dt
(c) Now substitute ∫ 3 t^2 (t^3 + 4)^5 dt =
(t^3 + 4)^5 · 3 t^2 dt
=
u^5 · du
=
u^6 + C
=
(t^3 + 4)^6 + C
4 x − 5 dx
(a) Let u = 4x − 5
(b) Then du = 4 dx or
du = dx
(c) Now substitute ∫ (^) √ 4 x − 5 dx =
u ·
du
=
u^1 /^2 du
=
u^3 /^2 ·
+ C
(4x − 5)^3 /^2 + C
t^2 (t^3 + 4)−^1 /^2 dt
(a) Let u = t^3 + 4
(b) Then du = 3t^2 dt or
du = t^2 dt
(c) Now substitute ∫ t^2 (t^3 + 4)−^1 /^2 dt =
(t^3 + 4)−^1 /^2 · t^2 dt
=
u−^1 /^2 ·
du
=
u−^1 /^2 du
=
u^1 /^2 ·
+ C
u^1 /^2 + C
=
(t^3 + 4)^1 /^2 + C
cos(2x + 1) dx
(a) Let u = 2x + 1
(b) Then du = 2 dx or
du = dx
(c) Now substitute ∫ sin(x) (cos(x))^5 dx =
(cos(x)^5 · sin(x) dx
=
u^5 (−du)
=
−u−^5 + C
= − u−^4 − 4
+ C
= u−^4 + C
=
(cos(x))^4
+ C
x − 1)^2 √ x dx
(a) Let u =
x − 1
(b) Then du =
x dx or 2 du =
√^1
x dx
(c) Now substitute ∫ (
x − 1)^2 √ x dx =
x − 1)^2 ·
x dx
=
u^2 (2 du)
=
2 u^2 du
=
u^3 + C
=
x − 1)^3 + C
x^3 + x^2 (3x^2 + 2x) dx
(a) Let u = x^3 + x^2
(b) Then du = (3x^2 + 2x) dx
(c) Now substitute ∫ (^) √ x^3 + x^2 · (3x^2 + 2x) dx =
u du
=
u^1 /^2 du
=
u^3 /^2 + C
=
(x^3 + x^2 )^3 /^2 + C
− 1
x + 1 (x^2 + 2x + 2)^3 dx
(a) Let u = x^2 + 2x + 2
(b) Then du = (2x + 2) dx → du = 2(x + 1) dx or
du = (x + 1) dx
(c) If x = −1, then u = (−1)^2 + 2(−1) + 2 = 1
(d) If x = 1, then u = (1)^2 + 2(1) + 2 = 5
(e) Now substitute ∫ (^1)
− 1
(x + 1) (x^2 + 2x + 2)^3 dx =
− 1
(x^2 + 2x + 2)^3 · (x + 1) dx
1
u^3
du
1
u−^3 du
u−^2 − 2
5
1 = −
4 u^2
5
1
[
4(5)^2
]
[
4(1)^2
]
t
t^2
dt
(a) Let u = 1 +
t (b) Then du = −
t^2 dt or −du =
t^2 dt
(c) Now substitute ∫ ( 1 +
t
t^2 dt =
u^3 (−du)
= −
u^4 + C
= −
t
+ C
− 1
x^2
x^3 + 1 dx
(a) Let u = x^3 + 1
(b) Then du = 3x^2 dx or
du = x^2 dx
(c) If x = −1, then u = (−1)^3 + 1 = 0.
(d) If x = 1, then u = (1)^3 + 1 = 2
(e) Now substitute ∫ (^1)
− 1
x^2
x^3 + 1 dx =
− 1
x^3 + 1 · x^2 dx
0
u ·
du
0
u^1 /^2 du
u^3 /^2 ·
2
0
u^3 /^2
2
0
[
(2)^3 /^2
]
[
(0)^3 /^2
]
3 x − 7
dx
(a) Let u = 3x − 7
(b) Then du = 3 dx or
du = dx
(c) Now substitute ∫ √^2 3 x − 7
dx =
√^2
u
du
=
u−^1 /^2 du
=
u^1 /^2 ·
+ C
u^1 /^2 + C
=
3 x − 7 + C
1
x(
x + 1)^2 dx
(a) Let u =
x + 1
(b) Then du =
x
dx or 2 du =
x
dx
(c) If x = 1, then u =
(d) If x = 4, then u =
(e) Now substitute ∫ (^1)
0
x √ x + 1
dx =
1
u − 1 √ u du
1
u u^1 /^2
u^1 /^2 du
1
u^1 /^2 − u−^1 /^2 du
u^3 /^2 − 2 u^1 /^2
2
1
[
(2)^3 /^2 − 2(2)^1 /^2
]
[
(1)^3 /^2 − 2(1)^1 /^2
]
x
2 x + 1 dx
(a) Let u = 2x + 1. Then x =
(u − 1) (need this for later)
(b) Then du = 2 dx or
du = dx
(c) Now substitute ∫ x
2 x + 1 dx =
(u − 1)
u ·
du
=
(u − 1) · u^1 /^2 du
=
u^3 /^2 −
u^1 /^2 du
=
u^5 /^2 ·
u^3 /^2 ·
+ C
u^5 /^2 −
u^3 /^2 + C
=
(2x + 1)^5 /^2 −
(2x + 1)^3 /^2 + C
x
x
x + 1 dx
You should rewrite the integral as ∫ x^1 /^2
x^3 /^2 + 1 dx
to help identify u.
(a) Let u = x^3 /^2 + 1
(b) Then du =
x^1 /^2 dx or
du = x^1 /^2 dx
(c) Now substitute ∫ x^1 /^2
x^3 /^2 + 1 dx =
x^3 /^2 + 1 · x^1 /^2 dx
=
u ·
du
=
u^1 /^2 du
=
u^3 /^2 ·
+ C
u^3 /^2 + C
=
(x^3 /^2 + 1)^3 /^2 + C
x^3
x^2 + 1 dx
(a) Let u = x^2 − 1. Then x^2 = u − 1 (need this for later)
(b) Then du = 2x dx or
du = x dx
x^2 + 2x x^2 + 2x + 1
x^2 + 2x + 1
(x + 1)^2
(b) So we are evaluating
(x + 1)^2 dx
(x + 1)^2 dx = x +
x + 1
+ C
x^2 + 6x + 9
(a) First, rewrite the integrand as
x^2 + 6x + 9
(x + 3)^2 (b) Rewrite the integral (^) ∫ 1 (x + 3)^2 dx
(c) Let u = x + 3
(d) Then du = dx ∫ 1 (x + 3)^2 dx =
u^2 du
=
u−^2 du
= −u−^1 + C
= −(x + 3)−^1 + C = −
x + 3
+ C
sec^2 (x) (1 + tan(x))^3 dx
(a) Let u = 1 + tan(x)
(b) Then du = sec^2 (x) dx
(c) Now substitute ∫ sec^2 (x) (1 + tan(x))^3 dx =
(1 + tan(x))^3 · sec^2 (x) dx
=
u^3 du
=
u−^3 du
= −
u−^2 + C
= −
2 u^2
+ C
2(1 + tan(x))^2
+ C
sin(x) (2 + 3 cos(x))^2
dx
(a) Let u = 2 + 3 cos(x)
(b) Then du = −3 sin(x) dx or −
du = sin(x) dx
(c) Now substitute ∫ sin(x) (2 + 3 cos(x))^2 dx =
(2 + 3 cos(x))^2 · sin(x) dx
=
u^2
du
=
u−^2 du
=
u−^1 + C
=
(2 + 3 cos(x))−^1 + C
x tan(x^2 ) sec(x^2 ) dx
(a) Let u = x^2
(b) Then du = 2x dx or
du = x dx
require some trig identities.
(tan(2x) + cot(2x))^2 = (tan(2x) + cot(2x)) · (tan(2x) + cot(2x)) = tan^2 (2x) + 2 tan(2x) cot(2x) + cot^2 (2x)
= tan^2 (2x) + 2 + cot^2 (2x)
= (sec^2 (2x) − 1) + 2 + (csc^2 (2x) − 1) = sec^2 (2x) + csc^2 (2x)
(b) The reason we used the trig identities at the end is because we know nice functions whose
derivatives are sec^2 () and csc^2 ().
(c) Let’s rewrite the integral (^) ∫ sec^2 (2x) + csc^2 (2x) dx
(d) Let u = 2x
(e) Then du = 2 dx or
du = dx
(f) Now substitute ∫ sec^2 (2x) + csc^2 (2x) dx =
(sec^2 (u) + csc^2 (u)) ·
du
=
sec^2 (u) + csc^2 (u) du
=
(tan(u) − cot(u)) + C
=
(tan(2x) − cot(2x)) + C
=
tan(2x) −
cot(2x) + C
