Integration by Substitution:
I
Identify u (usually the inside of a composition);
use that to find du = u 0 dx
I
Replace the appropriate expressions of x in the integrand with u and
du.
Note: du can not be in the denominator, raised to any power, or
otherwise be inside any function.
I
Your substitution has been successful if (1) you now have only u and
du – no more x and dx – and (2) if your new integral is something
you can integrate.
I
Integrate the new simpler function.
I
Back substitute: Replace u with its original more complicated
formulation in terms of x.
Math 101-Calculus 1 (Sklensky)
In-Class Work
December 6, 2010
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In Class Work
Find the following indefinite integrals, and check your answers!!
Z
x
1.
dx
(u = 1 + x 2 )
1 + x2
Z √x+1
√
e
√
2.
dx
(u = x + 1)
x +1
Z
3.
sec2 x e tan x dx
Z
4.
12x cos(3x 2 + 5) dx
Z
5. (x 2 + 4x)(x 3 + 6x 2 + 4)100 dx
Z
sin(x)
6.
tan(x) dx
Hint: Write tan(x) as cos(x)
Z
1
7.
dx
x ln |x|
Math 101-Calculus 1 (Sklensky)
In-Class Work
December 6, 2010
2/9
Solutions:
Z
1.
I
I
x
dx
(u = 1 + x 2 )
1 + x2
Notice The composition is not obvious (you can see it if you write
x
= x(1 + x 2 )−1 ). When stuck, look for what portion of the
1+x 2
integrand is particularly troublesome.
Identify u and find du:
u = 1 + x2
I
I
I
⇒ u 0 = 2x
⇒ du = 2x dx
Replace appropriate expressions of x in the integrand with u and du:
Z
Z
Z
x
1
2x
1
1
dx
=
dx
=
du.
1 + x2
2
1 + x2
2
u
Integrate the new simpler function:
Z
1
1
1
= ln |u| + C
2
u
2
Z
x
1
dx = ln |1 + x 2 | + C
Substitute back in
2
1+x
2
Math 101-Calculus 1 (Sklensky)
In-Class Work
December 6, 2010
3/9
Solutions:
Z √
e
√
x+1
√
I
x + 1)
x +1
Notice Have a double-composition. Chose u to be biggest inside.
Identify u and find du:
I
1
1
1
x + 1 ⇒ u 0 = (x + 1)−1/2 ⇒ du = √
dx
2
2 x +1
Replace appropriate expressions of x in the integrand with u and du:
2.
I
dx
u=
(u =
√
√
Z
I
e
√
x+1
dx = 2
e
x+1 1
1
√
dx = 2
2 x +1
x +1
Integrate the new simpler function:
Z
2 e u = 2e u + C
√
Z
I
√
Z
Back substitute:
Math 101-Calculus 1 (Sklensky)
e
√
x+1
x +1
√
dx = 2e
In-Class Work
x+1
Z
e u du
+ C.
December 6, 2010
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Solutions
Z
3.
sec2 x e tan x dx
I
Choose u = inside of composition= tan(x).
Then u 0 = sec2 x, so du = sec2 (x) dx
I
Substitute:
Z
2
sec x e
I
Z
dx =
e u du
Integrate:
Z
2
sec x e
I
tan x
tan x
Z
dx =
e u du = e u + C
Back-substitute:
Z
Z
2
tan x
sec x e
dx = e u du = e u + C = e tan x + C
Math 101-Calculus 1 (Sklensky)
In-Class Work
December 6, 2010
5/9
Solutions
Z
12x cos(3x 2 + 5) dx
4.
I
Choose u = inside of composition= 3x 2 + 5.
Then u 0 = 6x, so du = 6x dx
I
Substitute:
Z
Z
Z
2
2
12x cos(3x + 5) dx = 2 6x cos(3x + 5) dx = 2 cos(u) du
I
Integrate:
Z
I
Z
12x cos(3x 2 + 5) dx = 2
Z
cos(u) du = 2 sin(u) + C
Back-substitute:
12x cos(3x 2 +5) dx = 2
Math 101-Calculus 1 (Sklensky)
Z
cos(u) du = 2 sin(u)+C = 2 sin(3x 2 +5)+C
In-Class Work
December 6, 2010
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Solutions
Z
(x 2 + 4x)(x 3 + 6x 2 + 4)100 dx
5.
I
I
Choose u = inside of composition= x 3 + 6x 2 + 4.
Then u 0 = 3x 2 + 12x, so du = (3x 2 + 12x) dx = 3(x 2 + 4x) dx
Substitute:
Z
Z
1
3(x 2 + 4x)(x 3 + 6x 2 + 4)100
(x 2 + 4x)(x 3 + 6x 2 + 4)100 dx =
3
Z
1
=
u 100 du
3
I
Integrate:
Z
Z
1
1
2
3
2
100
(x + 4x)(x + 6x + 4) dx =
u 100 du =
u 101 + C
3
3(101)
I
Back-substitute:
Z
1
(x 2 + 4x)(x 3 + 6x 2 + 4)100 dx =
(x 3 + 6x 2 + 4)101 + C
303
Math 101-Calculus 1 (Sklensky)
In-Class Work
December 6, 2010
7/9
Solutions
Z
6.
tan(x) dx
I
I
I
I
Hint: Write tan(x) as
Composition isn’t obvious, but seen this before, in #1.
Choose u = cos(x).
Then u 0 = − sin(x), so du = − sin(x) dx
Substitute:
Z
Z
Z
1
− sin(x)
dx = −
du
tan(x) dx = −
cos(x)
u
Integrate:
Z
Z
tan(x) dx = −
I
sin(x)
cos(x)
1
du = − ln |u| + C
u
Back-substitute:
Z
Z
1
tan(x) dx = −
du = − ln |u| + C = − ln | cos(x)| + C
u
Math 101-Calculus 1 (Sklensky)
In-Class Work
December 6, 2010
8/9
Solutions
Z
7.
I
I
I
I
I
1
dx
x ln |x|
Again have a fraction; this time, letting u be the entire denominator
isn’t going to work. Try letting u be just a part (the more
complicated part) of the denominator.
Choose u = ln |x|.
Then u 0 = x1 , so du = x1 dx
Substitute:
Z
Z
Z
1
1 1
1
dx =
dx =
du
x ln |x|
x ln |x|
u
Integrate:
Z
Z
1
tan(x) dx =
du = ln |u| + C
u
Back-substitute:
Z
Z
1
1
dx =
du = ln |u| + C = ln ln |x| + C
x ln |x|
u
Math 101-Calculus 1 (Sklensky)
In-Class Work
December 6, 2010
9/9
In Class Work
Z
1
2
xe x dx
1.
0
Z
(π/4)1/3
x 2 sec(x 3 ) tan(x 3 ) dx
2.
0
Z
4
1
dx
x(ln
|x|)2
2
Z 3
√
4.
x 1 + x dx
3.
1
Math 101-Calculus 1 (Sklensky)
In-Class Work
December 6, 2010
10 / 9