Ulearngo

Proof

Let f, g, u, and F be as specified in the theorem. Then

\[\begin{aligned} \frac{d}{d x} F \left(\right. g \left(\right. x \left.\right) \left.\right) & = F^{'} ( g \left(\right. x \left.\right) \left.\right) g^{'} ( x \left.\right) \\ & = f \left[\right. g \left(\right. x \left.\right) \left]\right. g^{'} ( x \left.\right) . \end{aligned}\]

Integrating both sides with respect to x, we see that

\[\int f \left[\right. g \left(\right. x \left.\right) \left]\right. g^{'} ( x \left.\right) d x = F \left(\right. g \left(\right. x \left.\right) \left.\right) + C .\]

If we now substitute \(u = g \left(\right. x \left.\right) ,\) and \(d u = g ' \left(\right. x \left.\right) d x ,\) we get

\[\begin{aligned} \int f \left[\right. g \left(\right. x \left.\right) \left]\right. g^{'} ( x \left.\right) d x & = \int f \left(\right. u \left.\right) d u \\ & = F \left(\right. u \left.\right) + C \\ & = F \left(\right. g \left(\right. x \left.\right) \left.\right) + C . \end{aligned}\]

□

Returning to the problem we looked at originally, we let \(u = x^{2} - 3\) and then \(d u = 2 x d x .\) Rewrite the integral in terms of u:

\[\int \underset{u}{\underset{︸}{\left(\right. x^{2} - 3 \left.\right)}}^{3} \underset{d u}{\underset{︸}{\left(\right. 2 x d x \left.\right)}} = \int u^{3} d u .\]

Using the power rule for integrals, we have

\[\int u^{3} d u = \frac{u^{4}}{4} + C .\]

Substitute the original expression for x back into the solution:

\[\frac{u^{4}}{4} + C = \frac{\left(\right. x^{2} - 3 \left.\right)^{4}}{4} + C .\]

We can generalize the procedure in the following Problem-Solving Strategy.

Problem-Solving Strategy

Integration by Substitution

  1. Look carefully at the integrand and select an expression \(g \left(\right. x \left.\right)\) within the integrand to set equal to u. Let’s select \(g \left(\right. x \left.\right)\) such that \(g^{'} ( x \left.\right)\) is also part of the integrand.
  2. Substitute \(u = g \left(\right. x \left.\right)\) and \(d u = g^{'} ( x \left.\right) d x\) into the integral.
  3. We should now be able to evaluate the integral with respect to u. If the integral can’t be evaluated we need to go back and select a different expression to use as u.
  4. Evaluate the integral in terms of u.
  5. Write the result in terms of x and the expression \(g \left(\right. x \left.\right) .\)

Example 5.30

Using Substitution to Find an Antiderivative

Use substitution to find the antiderivative \(\int 6 x \left(\right. 3 x^{2} + 4 \left.\right)^{4} d x .\)

Solution

The first step is to choose an expression for u. We choose \(u = 3 x^{2} + 4\) because then \(d u = 6 x d x ,\) and we already have du in the integrand. Write the integral in terms of u:

\[\int 6 x \left(\right. 3 x^{2} + 4 \left.\right)^{4} d x = \int u^{4} d u .\]

Remember that du is the derivative of the expression chosen for u, regardless of what is inside the integrand. Now we can evaluate the integral with respect to u:

\[\begin{aligned} \int u^{4} d u & = \frac{u^{5}}{5} + C \\ \\ \\ & = \frac{\left(\right. 3 x^{2} + 4 \left.\right)^{5}}{5} + C . \end{aligned}\]

Analysis

We can check our answer by taking the derivative of the result of integration. We should obtain the integrand. Picking a value for C of 1, we let \(y = \frac{1}{5} \left(\right. 3 x^{2} + 4 \left.\right)^{5} + 1 .\) We have

\[y = \frac{1}{5} \left(\right. 3 x^{2} + 4 \left.\right)^{5} + 1 ,\]

so

\[\begin{aligned} \\ y^{'} & = \left(\right. \frac{1}{5} \left.\right) 5 \left(\left(\right. 3 x^{2} + 4 \left.\right)\right)^{4} 6 x \\ & = 6 x \left(\left(\right. 3 x^{2} + 4 \left.\right)\right)^{4} . \end{aligned}\]

This is exactly the expression we started with inside the integrand.

Checkpoint 5.25

Use substitution to find the antiderivative \(\int 3 x^{2} \left(\right. x^{3} - 3 \left.\right)^{2} d x .\)

Sometimes we need to adjust the constants in our integral if they don’t match up exactly with the expressions we are substituting.

Example 5.31

Using Substitution with Alteration

Use substitution to find \(\int z \sqrt{z^{2} - 5} d z .\)

Solution

Rewrite the integral as \(\int z \left(\right. z^{2} - 5 \left.\right)^{1 / 2} d z .\) Let \(u = z^{2} - 5\) and \(d u = 2 z d z .\) Now we have a problem because \(d u = 2 z d z\) and the original expression has only \(z d z .\) We have to alter our expression for du or the integral in u will be twice as large as it should be. If we multiply both sides of the du equation by \(\frac{1}{2} .\) we can solve this problem. Thus,

\[\begin{aligned} \\ u & = z^{2} - 5 \\ d u & = 2 z d z \\ \frac{1}{2} d u & = \frac{1}{2} \left(\right. 2 z \left.\right) d z = z d z . \end{aligned}\]

Write the integral in terms of u, but pull the \(\frac{1}{2}\) outside the integration symbol:

\[\int z \left(\right. z^{2} - 5 \left.\right)^{1 / 2} d z = \frac{1}{2} \int u^{1 / 2} d u .\]

Integrate the expression in u:

\[\begin{aligned} \\ \frac{1}{2} \int u^{1 / 2} d u & = \left(\right. \frac{1}{2} \left.\right) \frac{u^{3 / 2}}{\frac{3}{2}} + C \\ \\ & = \left(\right. \frac{1}{2} \left.\right) \left(\right. \frac{2}{3} \left.\right) u^{3 / 2} + C \\ & = \frac{1}{3} u^{3 / 2} + C \\ & = \frac{1}{3} \left(\left(\right. z^{2} - 5 \left.\right)\right)^{3 / 2} + C . \end{aligned}\]

Checkpoint 5.26

Use substitution to find \(\int x^{2} \left(\right. x^{3} + 5 \left.\right)^{9} d x .\)

Example 5.32

Using Substitution with Integrals of Trigonometric Functions

Use substitution to evaluate the integral \(\int \frac{\text{sin} t}{\text{cos}^{3} t} d t .\)

Solution

We know the derivative of \(\text{cos} t\) is \(− \text{sin} t ,\) so we set \(u = \text{cos} t .\) Then \(d u = − \text{sin} t d t .\) Substituting into the integral, we have

\[\int \frac{\text{sin} t}{\text{cos}^{3} t} d t = − \int \frac{d u}{u^{3}} .\]

Evaluating the integral, we get

\[\begin{aligned} \\ \\ − \int \frac{d u}{u^{3}} & = − \int u^{−3} d u \\ & = − \left(\right. - \frac{1}{2} \left.\right) u^{−2} + C . \end{aligned}\]

Putting the answer back in terms of t, we get

\[\begin{aligned} \int \frac{\text{sin} t}{\text{cos}^{3} t} d t & = \frac{1}{2 u^{2}} + C \\ \\ & = \frac{1}{2 \text{cos}^{2} t} + C . \end{aligned}\]

Checkpoint 5.27

Use substitution to evaluate the integral \(\int \frac{\text{cos} t}{\text{sin}^{2} t} d t .\)

Sometimes we need to manipulate an integral in ways that are more complicated than just multiplying or dividing by a constant. We need to eliminate all the expressions within the integrand that are in terms of the original variable. When we are done, u should be the only variable in the integrand. In some cases, this means solving for the original variable in terms of u. This technique should become clear in the next example.

Example 5.33

Finding an Antiderivative Using u-Substitution

Use substitution to find the antiderivative \(\int \frac{x}{\sqrt{x - 1}} d x .\)

Solution

If we let \(u = x - 1 ,\) then \(d u = d x .\) But this does not account for the x in the numerator of the integrand. We need to express x in terms of u. If \(u = x - 1 ,\) then \(x = u + 1 .\) Now we can rewrite the integral in terms of u:

\[\begin{aligned} \int \frac{x}{\sqrt{x - 1}} d x & = \int \frac{u + 1}{\sqrt{u}} d u \\ \\ & = \int \sqrt{u} + \frac{1}{\sqrt{u}} d u \\ & = \int \left(\right. u^{1 / 2} + u^{−1 / 2} \left.\right) d u . \end{aligned}\]

Then we integrate in the usual way, replace u with the original expression, and factor and simplify the result. Thus,

\[\begin{aligned} \int \left(\right. u^{1 / 2} + u^{−1 / 2} \left.\right) d u & = \frac{2}{3} u^{3 / 2} + 2 u^{1 / 2} + C \\ \\ & = \frac{2}{3} \left(\left(\right. x - 1 \left.\right)\right)^{3 / 2} + 2 \left(\left(\right. x - 1 \left.\right)\right)^{1 / 2} + C \\ & = \left(\left(\right. x - 1 \left.\right)\right)^{1 / 2} \left[\right. \frac{2}{3} \left(\right. x - 1 \left.\right) + 2 \left]\right. + C \\ & = \left(\left(\right. x - 1 \left.\right)\right)^{1 / 2} \left(\right. \frac{2}{3} x - \frac{2}{3} + \frac{6}{3} \left.\right) + C \\ & = \left(\left(\right. x - 1 \left.\right)\right)^{1 / 2} \left(\right. \frac{2}{3} x + \frac{4}{3} \left.\right) + C \\ & = \frac{2}{3} \left(\left(\right. x - 1 \left.\right)\right)^{1 / 2} \left(\right. x + 2 \left.\right) + C . \end{aligned}\]

Checkpoint 5.28

Use substitution to evaluate the indefinite integral \(\int \text{cos}^{3} t \text{sin} t d t .\)

This lesson is part of:

Integration

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