Integration using Substitution: Learn It 1

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Integration using Substitution: Learn It 1

Identify when to use substitution to simplify and solve integrals
Apply substitution methods to find indefinite integrals
Apply substitution methods to find definite integrals

The Fundamental Theorem of Calculus gave us a method to evaluate integrals without using Riemann sums. The drawback of this method, though, is that we must be able to find an antiderivative, and this is not always easy. In this section we examine a technique, called integration by substitution, to help us find antiderivatives. Specifically, this method helps us find antiderivatives when the integrand is the result of a chain-rule derivative.

Substitution for Indefinite Integrals

At first, the approach to the substitution procedure may not appear very obvious. However, it is primarily a visual task—that is, the integrand shows you what to do; it is a matter of recognizing the form of the function.

So, what are we supposed to see? We are looking for an integrand of the form $f\left[g(x)\right]{g}^{\prime }(x)dx.$

For example, in the integral $\displaystyle\int {({x}^{2}-3)}^{3}2xdx,$ we have $f(x)={x}^{3},g(x)={x}^{2}-3,$ and $g\text{‘}(x)=2x.$ Then,

f [g (x)] g^{'} (x) = {(x^{2} - 3)}^{3} (2 x),

$f\left[g(x)\right]{g}^{\prime }(x)={({x}^{2}-3)}^{3}(2x),$

and we see that our integrand is in the correct form.

The method is called substitution because we substitute part of the integrand with the variable $u$ 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 $u=g(x),,$ where ${g}^{\prime }(x)$ is continuous over an interval, let $f(x)$ be continuous over the corresponding range of $g$ , and let $F(x)$ be an antiderivative of $f(x).$ Then,

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

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

Proof

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

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

$\begin{array}{cc}\frac{d}{dx}F(g(x))\hfill & ={F}^{\prime }(g(x)){g}^{\prime }(x)\hfill \\ & =f\left[g(x)\right]{g}^{\prime }(x).\hfill \end{array}$

Integrating both sides with respect to $x$ , we see that

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

$\displaystyle\int f\left[g(x)\right]{g}^{\prime }(x)dx=F(g(x))+C.$

If we now substitute $u=g(x),$ and $du=g\text{‘}(x)dx,$ we get

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

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

$_\blacksquare$

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

{\int (x^{2} - 3)      u}^{3} (2 x d x)      d u = \int u^{3} d u .

${\displaystyle\int \underset{u}{\underbrace{({x}^{2}-3)}}}^{3}\underset{du}{\underbrace{(2xdx)}}=\displaystyle\int {u}^{3}du.$

Using the power rule for integrals, we have

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

$\displaystyle\int {u}^{3}du=\frac{{u}^{4}}{4}+C$

Substitute the original expression for $x$ back into the solution:

\frac{u^{4}}{4} + C = \frac{{(x^{2} - 3)}^{4}}{4} + C

$\dfrac{{u}^{4}}{4}+C=\dfrac{{({x}^{2}-3)}^{4}}{4}+C$

How To: Integrate by Substitution

Look carefully at the integrand and select an expression $g(x)$ within the integrand to set equal to $u$ . Let’s select $g(x).$ such that ${g}^{\prime }(x)$ is also part of the integrand.
Substitute $u=g(x)$ and $du={g}^{\prime }(x)dx$ into the integral.
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 $u$ .
Evaluate the integral in terms of $u$ .
Write the result in terms of $x$ and the expression $g(x).$

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

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

$\displaystyle\int 6x{(3{x}^{2}+4)}^{4}dx=\displaystyle\int {u}^{4}du.$

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 $u$ :

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

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 $y=\frac{1}{5}{(3{x}^{2}+4)}^{5}+1.$ We have

$y=\frac{1}{5}{(3{x}^{2}+4)}^{5}+1,$

so

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

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

Watch the following video to see the worked solution to this example.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this segmented clip of “5.5 Substitution” here (opens in new window).

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

As long as you select a $g(x)$ for $u$ such that a multiple of $g'(x)$ exists in the integrand, it will work! In other words, make sure the exponents work – don’t worry about the constants.

For instance, in the example below, if we select ${u={z}^{2}-5}$ , $g'(x)={2z}$ . Although $g'(x)={2z}$ doesn’t appear in the integrand, $z$ does. Substitution can work here!

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

Rewrite the integral as $\displaystyle\int z{({z}^{2}-5)}^{1\text{/}2}dz.$ Let $u={z}^{2}-5$ and $du=2zdz.$

Now we have a problem because $du=2zdz$ and the original expression has only $zdz.$ 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,

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

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

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

Integrate the expression in $u$ :

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

Watch the following video to see the worked solution to this example.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this segmented clip of “5.5 Substitution” here (opens in new window).

Use substitution to evaluate the integral $\displaystyle\int \frac{ \sin t}{{ \cos }^{3}t}dt.$

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

$\displaystyle\int \frac{ \sin t}{{ \cos }^{3}t}dt=\text{−}\displaystyle\int \frac{du}{{u}^{3}}.$

Evaluating the integral, we get

$\begin{array}{}\\ \\ \text{−}{\displaystyle\int \frac{du}{{u}^{3}}}\hfill & =\text{−} {\displaystyle\int {u}^{-3}du}\hfill \\ & =\text{−}(-\frac{1}{2}){u}^{-2}+C.\hfill \end{array}$

Putting the answer back in terms of $t$ , we get

$\begin{array}{cc} {\displaystyle\int \frac{ \sin t}{{ \cos }^{3}t}dt}\hfill & =\frac{1}{2{u}^{2}}+C\hfill \\ \\ & =\dfrac{1}{2{ \cos }^{2}t}+C.\hfill \end{array}$

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.

Use substitution to find the antiderivative of $\displaystyle\int \frac{x}{\sqrt{x-1}}dx.$

If we let $u=x-1,$ then $du=dx.$ 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 $u$ :

$\begin{array}{ll} {\displaystyle\int \frac{x}{\sqrt{x-1}}dx}\hfill & = {\displaystyle\int \frac{u+1}{\sqrt{u}}du}\hfill \\ \\ & = {\displaystyle\int \sqrt{u}+\frac{1}{\sqrt{u}}du}\hfill \\ & = {\displaystyle\int ({u}^{1\text{/}2}+{u}^{-1\text{/}2})du}.\hfill \end{array}$

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

$\begin{array}{cc} {\displaystyle\int ({u}^{1\text{/}2}+{u}^{-1\text{/}2})du}\hfill & =\frac{2}{3}{u}^{3\text{/}2}+2{u}^{1\text{/}2}+C\hfill \\ \\ & =\frac{2}{3}{(x-1)}^{3\text{/}2}+2{(x-1)}^{1\text{/}2}+C\hfill \\ & ={(x-1)}^{1\text{/}2}\left[\frac{2}{3}(x-1)+2\right]+C\hfill \\ & ={(x-1)}^{1\text{/}2}(\frac{2}{3}x-\frac{2}{3}+\frac{6}{3})\hfill \\ & ={(x-1)}^{1\text{/}2}(\frac{2}{3}x+\frac{4}{3})\hfill \\ & =\frac{2}{3}{(x-1)}^{1\text{/}2}(x+2)+C.\hfill \end{array}$