Math 2300: Calculus II Identifying Integral Substitutions Goal: To

Math 2300: Calculus II Identifying Integral Substitutions Goal: To

Math 2300: Calculus II
Identifying Integral Substitutions
Goal: To identify what (if any) u-substitutions are necessary to compute an integral and to practice
making such substitutions.
For each problem, identify what (if any) u-substitution(s) need to be made to evaluate each integral.
Make the substitution and simplify, but do not evaluate the integral.
Z
1.
x sin(x2 ) dx
Solution:
u = x2 ,Zdu = 2x dx,
Z
1
x sin(x2 ) dx =
sin(u) du
2
Z
2.
√
x(x + 3) dx
Solution:
No substitution
needed,
Z
Z
√
3/2
x(x + 3) dx = x + 3x1/2 dx
Z
3.
√
x x + 3 dx
Solution: u = xZ + 3, du = dx,
Z
Z
√
√
x x + 3 dx = (u − 3) u du = u3/2 − 3u1/2 du
Z p
ln (x)
4.
dx
x
Solution: u = ln(x), du =
Z p
Z
√
ln(x)
dx =
u du
x
1
x
dx,
1
Math 2300: Calculus II
Z
5.
Identifying Integral Substitutions
x+4
dx
x
Solution:
NoZ substitution needed,
Z
x+4
4
dx = 1 + dx
x
x
Z
6.
x
dx
x+4
Solution:
u=
dx,
Z x + 4, du = Z
Z
u−4
4
x
dx =
du = 1 − du
x+4
u
u
Z
7.
ex
√
dx
1 − e2x
Solution:
u = exZ, du = ex dx,
Z
x
1
e
√
√
dx =
du
2x
1−e
1 − u2
Z
8.
arctan(x)
dx
1 + x2
1
Solution: u = arctan(x), du =
dx,
1 + x2
Z
Z
arctan(x)
dx = u du
1 + x2
2
Math 2300: Calculus II
Z
9.
Identifying Integral Substitutions
x3
dx
(1 + x2 )2
Solution:
u = 1 + Zx2 , du = 2x dx,
Z
Z
Z
3
x
x2
u−1
1
1
1
1
1
dx =
· 2x dx =
du =
−
du
(1 + x2 )2
2
(1 + x2 )2
2
u2
2
u u2
Z
10.
x2
√
dx
1 − x3
Solution:
u = 1 − xZ3 , du = −3x2 dx,
Z
2
x
1
1
√
√ du
dx = −
3
3
u
1−x
Z
11.
sin(x)(3 cos4 (x) + 4 cos3 (x) − 9) dx
Solution:
u = cos(x), du = − sin(x) dx, Z
Z
sin(x)(3 cos4 (x) + 4 cos3 (x) − 9) dx = − 3u4 + 4u3 − 9 du
Z
12.
x3 + e3−x dx
Solution: Split into two integrals. The first doesn’t need a substitution.
For
Z the second, useZ u = 3 − x,Z du = − dx,
x3 + e3−x dx =
Z
13.
x3 dx −
eu du
√
sin( x + 1)
1
√
+ 2
dx
x +1
x
Solution: Again split into two integrals.
√
For the first, use u = x + 1, du = 2√1 x dx,
The
integral doesn’t needZ a substitution.
√
Z
Z second
sin( x + 1)
1
1
√
+ 2
dx = 2 sin(u) du +
dx
x +1
x2 + 1
x
3