Integration by Substitution
November 22, 2013
Integration by Substitution
Introduction
Z
7x 2 cos(3x 3 )dx =?
Z
2xe x
2 +5
dx =?
Integration by Substitution
Chain rule
The chain rule:
d
(f (g (x))) = f 0 (g (x)) · g 0 (x).
dx
Use the chain rule to find f 0 (x) and then write the corresponding
anti-differentiation formula.
f (x) = e x
2 +4
Integration by Substitution
Chain rule
The chain rule:
d
(f (g (x))) = f 0 (g (x)) · g 0 (x).
dx
Use the chain rule to find f 0 (x) and then write the corresponding
anti-differentiation formula.
f (x) = e x
2 +4
f (x) = 71 (x 2 + 5)4
Integration by Substitution
Chain rule
The chain rule:
d
(f (g (x))) = f 0 (g (x)) · g 0 (x).
dx
Use the chain rule to find f 0 (x) and then write the corresponding
anti-differentiation formula.
f (x) = e x
2 +4
f (x) = 71 (x 2 + 5)4
f (x) = ln(x 2 + 1)
Integration by Substitution
Anti-differentiation formula
An anti-differentiation formula is of the form
Z
f (x)dx = F (x) + C
Integration by Substitution
Chain rule
The chain rule:
d
(f (g (x))) = f 0 (g (x)) · g 0 (x).
dx
Use the chain rule to find f 0 (x) and then write the corresponding
anti-differentiation formula.
f (x) = e x
2 +4
Integration by Substitution
Chain rule
The chain rule:
d
(f (g (x))) = f 0 (g (x)) · g 0 (x).
dx
Use the chain rule to find f 0 (x) and then write the corresponding
anti-differentiation formula.
f (x) = e x
2 +4
f 0 (x) = e x
2 +4
· 2x
Integration by Substitution
Chain rule
The chain rule:
d
(f (g (x))) = f 0 (g (x)) · g 0 (x).
dx
Use the chain rule to find f 0 (x) and then write the corresponding
anti-differentiation formula.
f (x) = e x
2 +4
2
f 0 (x) = e x +4 · 2x
R 0
f (x)dx = f (x) + C .
Integration by Substitution
Chain rule
The chain rule:
d
(f (g (x))) = f 0 (g (x)) · g 0 (x).
dx
Use the chain rule to find f 0 (x) and then write the corresponding
anti-differentiation formula.
f (x) = e x
2 +4
2
f 0 (x) = e x +4 · 2x
R 0
f (x)dx = f (x) + C .
R x 2 +4
2
e
· 2xdx = e x +4 + C .
Integration by Substitution
Chain rule
The chain rule:
d
(f (g (x))) = f 0 (g (x)) · g 0 (x).
dx
Use the chain rule to find f 0 (x) and then write the corresponding
anti-differentiation formula.
f (x) = 71 (x 2 + 5)4
Integration by Substitution
Chain rule
The chain rule:
d
(f (g (x))) = f 0 (g (x)) · g 0 (x).
dx
Use the chain rule to find f 0 (x) and then write the corresponding
anti-differentiation formula.
f (x) = 71 (x 2 + 5)4
R 4 2
1 2
3
4
7 (x + 5) · 2xdx = 7 (x + 5) + C .
Integration by Substitution
Chain rule
The chain rule:
d
(f (g (x))) = f 0 (g (x)) · g 0 (x).
dx
Use the chain rule to find f 0 (x) and then write the corresponding
anti-differentiation formula.
f (x) = ln(x 2 + 1)
Integration by Substitution
Chain rule
The chain rule:
d
(f (g (x))) = f 0 (g (x)) · g 0 (x).
dx
Use the chain rule to find f 0 (x) and then write the corresponding
anti-differentiation formula.
f (x) = ln(x 2 + 1)
R 1
· 2xdx = ln(x 2 + 1) + C .
x 2 +1
Integration by Substitution
Notice
The derivative of each inside function is 2x in the example.
Notice that the derivative of the inside function is a factor in
the integrand in each anti-differentiation formula.
Integration by Substitution
Notice
The derivative of each inside function is 2x in the example.
Notice that the derivative of the inside function is a factor in
the integrand in each anti-differentiation formula.
Finding an inside function whose derivative appears as a factor
in the integrand is key to the method of the substitution.
Integration by Substitution
Make a substitution in an integral
Let w (or v , t,...) be the inside function.
Integration by Substitution
Make a substitution in an integral
Let w (or v , t,...) be the inside function.
dw = w 0 (x)dx or dx =
1
w 0 (x) dw
Integration by Substitution
Make a substitution in an integral
Let w (or v , t,...) be the inside function.
dw = w 0 (x)dx or dx =
1
w 0 (x) dw
Express the integrand in terms of w .
Integration by Substitution
Examples
2
1
R
e x · 2xdx =?
2
R
(x 2 + 6)3 · 2xdx =?
3
R
1
x 2 +6
· 2xdx =?
Integration by Substitution
Example
In the previous examples the derivative of the inside function must
be present in the integrand for this method to work. Our method
works even when the derivative is missing a constant factor.
Z
2
4te t +3 dt =?
Integration by Substitution
Example
Z
x3
p
3
x 4 + 7dx =?
Integration by Substitution
Example
Z
x2
dx =?
1 + x3
Integration by Substitution
Examples
Z
sin(3 − x)dx =?
Z
e 2−4x dx =?
Integration by Substitution
Examples
Z
7x 2 cos(3x 3 )dx =?
Z
2xe x
2 +5
dx =?
Integration by Substitution
Example
Z
e cos θ sin θdθ =?
Integration by Substitution
Examples
√
cos x
√ dx =?
x
Z
√
Z
e x
√ dx =?
x
Integration by Substitution
Examples
Z
Z
ex
dx =?
ex + 1
ex + 1
√ x
dx =?
e +x
Integration by Substitution
Examples
Z
sinn (kx) cos(kx)dx =?
Z
cosn (kx) sin(kx)dx =?
Integration by Substitution
Harder Examples
Z
1
dx =?
1 + ex
Z
1
dx =?
2 + 3x
Z
x ln xdx =?
Integration by Substitution