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Integration By Substitution
Solutions To Selected Problems
Calculus 9th Edition Anton, Bivens, Davis
Matthew Staley
October 30, 2011
1. Evaluate the integrals using the indicated substitutions.
Z
(a)
2x(x2 + 1)23 dx u = x2 + 1
du = 2x dx
Z
23
u
Z
(b)
(x2 + 1)24
u24
du =
+C =
+C
24
24
cos3 (x) sin(x) dx u = cos(x) dx
du = − sin(x) dx
−du = sin(x) dx
Z
−
Z
(c)
2
(d)
cos4 (x)
u4
+C = −
+C
4
4
√
√
1
√ sin( x) dx u = x
x
1
du = √ dx
2 x
1
2 du = √ dx
x
Z
Z
u3 du = −
√
sin(u) du = −2 cos(u) + C = −2 cos( x) + C
cot(x) csc2 (x) dx u = cot(x)
du = − csc2 (x) dx
− du = csc2 (x) dx
Z
−
u du = −
u2
1
+ C = − cot2 (x) + C
2
2
1
(e)
Z
cos(2x) dx u = 2x
du = 2 dx
1
du = dx
2
(f)
1
2
Z
Z
√
x2 1 + x dx u = 1 + x → u − 1 = x
cos(u) du =
1
1
sin(u) + C = sin(2x) + C
2
2
du = dx
Z
2
(u − 1)
√
Z
u du =
Z
=
(u2 − 2u + 1)u1/2 du
u5/2 − 2u3/2 + u1/2 du
2
2
2
= u7/2 − 2u5/2 + u3/2 + C
7
5
3
4
2
2
= (1 + x)7/2 − (1 + x)5/2 + (1 + x)3/2 + C
7
5
3
2. Evaluate the integrals using appropriate substitutions.
(a)
Z
(4x − 3)9 dx Let u = 4x − 3
du = 4 dx
1
du = dx
4
1
4
Z
u9 du =
1 u1 0
1
+C =
(4x − 3)10 + C
4 10
40
2
(b)
Z
sec(4x) tan(4x) dx Let u = 4x
du = 4 dx
1
du = dx
4
1
4
(c)
Z
Z
sec(u) tan(u) du =
sin x5
dx
x2
1
1
sec(u) + C = sec(4x) + C
4
4
Let u =
5
x
du = −
5
dx
x2
1
1
− du = 2 dx
5
x
1
−
5
(d)
Z
Z
1
1
sin(u) du = − (− cos(u)) + C = cos(5/x) + C
5
5
x sec2 (x2 ) dx Let u = x2
du = 2x dx
1
dx = x du
2
1
2
Z
sec2 (u) du =
1
1
tan(u) + C = tan(x2 ) + C
2
2
3
(e)
Z
p
cos(4x) 2 − sin(4x) dx Let u = 2 − sin(4x)
du = −4 cos(4x) dx
1
− = cos(4x) dx
4
1
−
4
(f)
Z
Z
√
Z
√
1
u du = −
u1/2 du
4
1
12
= − u3/2 + C = − (2 − sin(4x))3/2 + C
43
6
y
dy
2y + 1
Let u = 2y + 1 →
1
(u − 1) = y
2
du = 2 dy
1
du = dy
2
1
2
Z
1
(u
2
Z
− 1)
u
1
1
√
√ − √ du
du =
4
u
u
u
Z
1
u1/2 − u−1/2 du
=
4
1 2 3/2
1/2
=
u − 2u
+C
4 3
=
1
1
(2y + 1)3/2 − (2y + 1)1/2 + C
6
2
4