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The Method of Substitution for Integration: Reversing the Chain Rule, Study notes of Differential and Integral Calculus

The method of substitution for integration, which is a technique used to solve integrals by reversing the chain rule of differentiation. examples and explanations of how to apply the method, as well as the definite integral version of the rule.

What you will learn

  • How does the method of substitution for integration work?
  • How do you apply the substitution rule for definite integrals?
  • What is the chain rule of differentiation and how is it reversed for integration?

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2021/2022

Uploaded on 09/27/2022

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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
dxf(g(x)) = f0(g(x))g0(x).
Reversing this rule tells us that
Zf0(g(x))g0(x)dx =f(g(x)) + C
Example Use the chain rule to find the derivative of the composite function f(g(x)) = (x2+ 1)2and
identify fand gin the expression.
Write the integral below as Rf0(g(x))g0(x)dx and evaluate it :
Z4x(x2+ 1) dx
The Substitution Rule says that if g(x) is a differentiable function whose range is the interval I
and fis continuous on I, then
Zf(g(x))g0(x)dx =Zf(u)du
where u=g(x) and du =g0(x)dx.
When applying the method, we substitute u=g(x), integrate with respect to the variable uand then
reverse the substitution in the resulting antiderivative.
Example Find R2xx2+ 1 dx.
Here we let g(x) = x2+ 1. We have R2xx2+ 1 dx =Rpg(x)g0(x)dx.
Now we let u=g(x) = x2+ 1, giving us that du
dx = 2x, giving us that du = 2xdx. Therefore, we have
Z2xx2+ 1 dx =Zu du =2u3/2
3+C.
We convert our answer back to an answer in terms of the variable x, to get
Z2xx2+ 1 dx =2u3/2
3+C=2(x2+ 1)3/2
3+C.
You should check that this the general antiderivative for 2xx2+ 1, by differentiating it using the chain
rule.
1
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pf5

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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

    1. dx

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

  • x 2
  • 1dx.

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