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f ( g ( x ) ) g ( x ) d x = F ( g ( x ) ) + C .

Then

a b f [ g ( x ) ] g ( x ) d x = F ( g ( x ) ) | x = a x = b = F ( g ( b ) ) F ( g ( a ) ) = F ( u ) | u = g ( a ) u = g ( b ) = g ( a ) g ( b ) f ( u ) d u ,

and we have the desired result.

Using substitution to evaluate a definite integral

Use substitution to evaluate 0 1 x 2 ( 1 + 2 x 3 ) 5 d x .

Let u = 1 + 2 x 3 , so d u = 6 x 2 d x . Since the original function includes one factor of x 2 and d u = 6 x 2 d x , multiply both sides of the du equation by 1 / 6 . Then,

d u = 6 x 2 d x 1 6 d u = x 2 d x .

To adjust the limits of integration, note that when x = 0 , u = 1 + 2 ( 0 ) = 1 , and when x = 1 , u = 1 + 2 ( 1 ) = 3 . Then

0 1 x 2 ( 1 + 2 x 3 ) 5 d x = 1 6 1 3 u 5 d u .

Evaluating this expression, we get

1 6 1 3 u 5 d u = ( 1 6 ) ( u 6 6 ) | 1 3 = 1 36 [ ( 3 ) 6 ( 1 ) 6 ] = 182 9 .
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Use substitution to evaluate the definite integral −1 0 y ( 2 y 2 3 ) 5 d y .

91 3

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Using substitution with an exponential function

Use substitution to evaluate 0 1 x e 4 x 2 + 3 d x .

Let u = 4 x 3 + 3 . Then, d u = 8 x d x . To adjust the limits of integration, we note that when x = 0 , u = 3 , and when x = 1 , u = 7 . So our substitution gives

0 1 x e 4 x 2 + 3 d x = 1 8 3 7 e u d u = 1 8 e u | 3 7 = e 7 e 3 8 134.568.
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Use substitution to evaluate 0 1 x 2 cos ( π 2 x 3 ) d x .

2 3 π 0.2122

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Substitution may be only one of the techniques needed to evaluate a definite integral. All of the properties and rules of integration apply independently, and trigonometric functions may need to be rewritten using a trigonometric identity before we can apply substitution. Also, we have the option of replacing the original expression for u after we find the antiderivative, which means that we do not have to change the limits of integration. These two approaches are shown in [link] .

Using substitution to evaluate a trigonometric integral

Use substitution to evaluate 0 π / 2 cos 2 θ d θ .

Let us first use a trigonometric identity to rewrite the integral. The trig identity cos 2 θ = 1 + cos 2 θ 2 allows us to rewrite the integral as

0 π / 2 cos 2 θ d θ = 0 π / 2 1 + cos 2 θ 2 d θ .

Then,

0 π / 2 ( 1 + cos 2 θ 2 ) d θ = 0 π / 2 ( 1 2 + 1 2 cos 2 θ ) d θ = 1 2 0 π / 2 d θ + 0 π / 2 cos 2 θ d θ .

We can evaluate the first integral as it is, but we need to make a substitution to evaluate the second integral. Let u = 2 θ . Then, d u = 2 d θ , or 1 2 d u = d θ . Also, when θ = 0 , u = 0 , and when θ = π / 2 , u = π . Expressing the second integral in terms of u , we have

1 2 0 π / 2 d θ + 1 2 0 π / 2 cos 2 θ d θ = 1 2 0 π / 2 d θ + 1 2 ( 1 2 ) 0 π cos u d u = θ 2 | θ = 0 θ = π / 2 + 1 4 sin u | u = 0 u = θ = ( π 4 0 ) + ( 0 0 ) = π 4 .
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Key concepts

  • Substitution is a technique that simplifies the integration of functions that are the result of a chain-rule derivative. The term ‘substitution’ refers to changing variables or substituting the variable u and du for appropriate expressions in the integrand.
  • When using substitution for a definite integral, we also have to change the limits of integration.

Key equations

  • Substitution with Indefinite Integrals
    f [ g ( x ) ] g ( x ) d x = f ( u ) d u = F ( u ) + C = F ( g ( x ) ) + C
  • Substitution with Definite Integrals
    a b f ( g ( x ) ) g ' ( x ) d x = g ( a ) g ( b ) f ( u ) d u

Why is u -substitution referred to as change of variable ?

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2. If f = g h , when reversing the chain rule, d d x ( g h ) ( x ) = g ( h ( x ) ) h ( x ) , should you take u = g ( x ) or u = h ( x ) ?

u = h ( x )

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In the following exercises, verify each identity using differentiation. Then, using the indicated u -substitution, identify f such that the integral takes the form f ( u ) d u .

x x + 1 d x = 2 15 ( x + 1 ) 3 / 2 ( 3 x 2 ) + C ; u = x + 1

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x 2 x 1 d x ( x > 1 ) = 2 15 x 1 ( 3 x 2 + 4 x + 8 ) + C ; u = x 1

f ( u ) = ( u + 1 ) 2 u

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Source:  OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2
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