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Integration by Substitution
In this section we reverse the Chain rule of differentiation and derive a method for solving integrals
called the method of substitution. Recall the chain rule of differentiation says that
d
dx
f (g(x)) = f ′ (g(x))g ′ (x).
Reversing this rule tells us that
∫
f ′ (g(x))g ′ (x) dx = f (g(x)) + C
Example Use the chain rule to find the derivative of the composite function f (g(x)) = (x^2 + 1)^2 and
identify f and g in the expression.
Write the integral below as
f ′(g(x))g′(x) dx and evaluate it :
∫
4 x(x 2
The Substitution Rule says that if g(x) is a differentiable function whose range is the interval I and f is continuous on I, then ∫
f (g(x))g ′ (x) dx =
f (u) du
where u = g(x) and du = g ′ (x) dx.
When applying the method, we substitute u = g(x), integrate with respect to the variable u and then
reverse the substitution in the resulting antiderivative.
Example Find
2 x
x^2 + 1 dx.
Here we let g(x) = x^2 + 1. We have
2 x
x^2 + 1 dx =
g(x)g′(x) dx.
Now we let u = g(x) = x 2
- 1, giving us that du dx = 2x, giving us that^ du^ = 2xdx. Therefore, we have ∫
2 x
x^2 + 1 dx =
u du =
2 u^3 /^2
3
+ C.
We convert our answer back to an answer in terms of the variable x, to get
∫
2 x
x^2 + 1 dx =
2 u^3 /^2
3
+ C =
2(x^2 + 1)^3 /^2
3
+ C.
You should check that this the general antiderivative for 2x
x^2 + 1, by differentiating it using the chain
rule.
Sometimes your substitution may result in an integral of the form
f (u)c du for some constant c, which
is not a problem.
Example Find the following:
∫
x^3
x^4 + 1 dx,
sin^3 x cos x dx,
x sin(x^2 + 3) dx
Sometimes the appropriate substitution is non-obvious and you may have to work a little harder to put
the resulting integral in the form
f (u)du:
Example Find the following : ∫ x^3 √ x^2 + 1
dx
Example Evaluate the following definite integrals:
∫ (^) π 2
0
sin
x √ x
dx,
2
x
x^2 + 1 dx.
Even and Odd Functions
Sometimes we can use symmetry to make evaluation of integrals easier:
If f is an even function (f (x) = f (−x)), then
∫ (^) a −a f^ (x)dx^ = 2^
∫ (^) a 0 f^ (x)dx.
If f is an odd function (f (x) = −f (−x)), then
∫ (^) a −a f^ (x)dx^ = 0
Example Evaluate the following definite integrals:
∫ π 4
−π 4
tan 5 xdx,
− 1
x 4
Extra Examples (Please attempt these before you check the solutions)
Example Find the following indefinite integrals:
∫ x √ x^2 + 1
dx,
sin(2x + 1) dx
Example (tricky - ish) Find the following :
∫
sin 2 x cos 3 x dx
Example Evaluate the following definite integrals: ∫ π 3
π 4
sin 3 θ cos θ dθ,
1
x √ x^2 + 1
dx (Use results from previous example)
Example Evaluate the following definite integrals:
∫ π 3
π 4
sin 3 θ cos θ dθ,
1
x √ x^2 + 1
dx (Use results from previous example)
Ex 1: (^) ∫ π 3
π 4
sin 3 θ cos θ dθ
Let u = sin θ,
then du = cos θ dθ.
Changing the limits, we get
u( π 4 ) = sin^
π 4 =^
√^1 2
u( π 3 ) = sin^
π 3 =^
√ 3 2
∫ π 3
π 4
(sin θ) 3 cos θ dθ =
∫ (^) u( π 3 )
u( π 4 )
u 3 du =
√ 3 2
√^1 2
u 3 du =
u^4
4
√ 3 2
√^1 2
[
3)^4
2)^4
]
[
]
[
]
Ex. 2 (using method 1): Above, we saw that
∫ x √ x^2 + 1
dx =
x^2 + 1 + C
So ∫ (^2)
1
x √ x^2 + 1
dx =
x^2 + 1
2
1
