### Learning Outcome

- Use substitution to evaluate indefinite integrals

In the Other Strategies for Integration section, we will essentially use a table of integration formulas to evaluate integrals. The most important step when using an integration table is to build the exact integration table formula using the integral we are asked to evaluate. This can sometimes be tricky. Here we will review u-substitution techniques that can be useful when building the desired integration table formula.

## Use Substitution to Evaluate Indefinite Integrals

*Substitution* is where we substitute part of the integrand with the variable [latex]u[/latex] and part of the integrand with *du*. It is also referred to as **change of variables** because we are changing variables to obtain an expression that is easier to work with for applying the integration rules.

### Substitution with Indefinite Integrals

Let [latex]u=g(x),,[/latex] where [latex]{g}^{\prime }(x)[/latex] is continuous over an interval, let [latex]f(x)[/latex] be continuous over the corresponding range of [latex]g[/latex], and let [latex]F(x)[/latex] be an antiderivative of [latex]f(x).[/latex] Then,

[latex]\begin{array}{cc} {\displaystyle\int f\left[g(x)\right]{g}^{\prime }(x)dx}\hfill & = {\displaystyle\int f(u)du}\hfill \\ & =F(u)+C\hfill \\ & =F(g(x))+C.\hfill \end{array}[/latex]

The following steps should be followed when integrating by substitution:

- Look carefully at the integrand and select an expression [latex]g(x)[/latex] within the integrand to set equal to [latex]u[/latex]. Let’s select [latex]g(x).[/latex] such that [latex]{g}^{\prime }(x)[/latex] is also part of the integrand.
- Substitute [latex]u=g(x)[/latex] and [latex]du={g}^{\prime }(x)dx.[/latex] into the integral.
- We should now be able to evaluate the integral with respect to [latex]u[/latex]. If the integral can’t be evaluated we need to go back and select a different expression to use as [latex]u[/latex].
- Evaluate the integral in terms of [latex]u[/latex].
- Write the result in terms of [latex]x[/latex] and the expression [latex]g(x).[/latex]

### Example: Evaluating an Indefinite Integral Using Substitution

Use substitution to find the antiderivative of [latex]\displaystyle\int 6x{(3{x}^{2}+4)}^{4}dx.[/latex]

Show Solution
The first step is to choose an expression for [latex]u[/latex]. We choose [latex]u=3{x}^{2}+4.[/latex] because then [latex]du=6xdx.,[/latex] and we already have *du* in the integrand. Write the integral in terms of [latex]u[/latex]:

[latex]\displaystyle\int 6x{(3{x}^{2}+4)}^{4}dx=\displaystyle\int {u}^{4}du.[/latex]

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

[latex]\begin{array}{ll} {\displaystyle\int {u}^{4}du}\hfill & =\frac{{u}^{5}}{5}+C\hfill \\ \\ \\ & =\frac{{(3{x}^{2}+4)}^{5}}{5}+C.\hfill \end{array}[/latex]

**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 [latex]y=\frac{1}{5}{(3{x}^{2}+4)}^{5}+1.[/latex] We have

[latex]y=\frac{1}{5}{(3{x}^{2}+4)}^{5}+1,[/latex]

so

[latex]\begin{array}{}\\ \hfill {y}^{\prime }& =(\frac{1}{5})5{(3{x}^{2}+4)}^{4}6x\hfill \\ & =6x{(3{x}^{2}+4)}^{4}.\hfill \end{array}[/latex]

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

### Example: Evaluating an Indefinite Integral Using Substitution

Use substitution to find the antiderivative of [latex]\displaystyle\int z\sqrt{{z}^{2}-5}dz.[/latex]

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

[latex]\begin{array}{}\\ \hfill u& ={z}^{2}-5\hfill \\ \hfill du& =2zdz\hfill \\ \hfill \frac{1}{2}du& =\frac{1}{2}(2z)dz=zdz.\hfill \end{array}[/latex]

Write the integral in terms of [latex]u[/latex], but pull the [latex]\frac{1}{2}[/latex] outside the integration symbol:

[latex]\displaystyle\int z{({z}^{2}-5)}^{1\text{/}2}dz=\frac{1}{2}\displaystyle\int {u}^{1\text{/}2}du.[/latex]

Integrate the expression in [latex]u[/latex]:

[latex]\begin{array}{}\\ \frac{1}{2} {\displaystyle\int {u}^{1\text{/}2}du}\hfill & =(\frac{1}{2})\frac{{u}^{3\text{/}2}}{\frac{3}{2}}+C\hfill \\ \\ & =(\frac{1}{2})(\frac{2}{3}){u}^{3\text{/}2}+C\hfill \\ & =\frac{1}{3}{u}^{3\text{/}2}+C\hfill \\ & =\frac{1}{3}{({z}^{2}-5)}^{3\text{/}2}+C.\hfill \end{array}[/latex]

### Try It

Use substitution to find the antiderivative of [latex]\displaystyle\int {x}^{2}{({x}^{3}+5)}^{9}dx.[/latex]

Hint
Multiply the du equation by [latex]\frac{1}{3}.[/latex]

Show Solution
[latex]\frac{{({x}^{3}+5)}^{10}}{30}+C[/latex]