# 1.5 Substitution  (Page 2/6)

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

$\int 3{x}^{2}{\left({x}^{3}-3\right)}^{2}dx=\frac{1}{3}{\left({x}^{3}-3\right)}^{3}+C$

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

## Using substitution with alteration

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

Rewrite the integral as $\int z{\left({z}^{2}-5\right)}^{1\text{/}2}dz.$ Let $u={z}^{2}-5$ and $du=2z\phantom{\rule{0.2em}{0ex}}dz.$ Now we have a problem because $du=2z\phantom{\rule{0.2em}{0ex}}dz$ and the original expression has only $z\phantom{\rule{0.2em}{0ex}}dz.$ 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,

$\begin{array}{}\\ \hfill u& ={z}^{2}-5\hfill \\ \hfill du& =2z\phantom{\rule{0.2em}{0ex}}dz\hfill \\ \hfill \frac{1}{2}du& =\frac{1}{2}\left(2z\right)dz=z\phantom{\rule{0.2em}{0ex}}dz.\hfill \end{array}$

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

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

Integrate the expression in u :

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

Use substitution to find the antiderivative of $\int {x}^{2}{\left({x}^{3}+5\right)}^{9}dx.$

$\frac{{\left({x}^{3}+5\right)}^{10}}{30}+C$

## Using substitution with integrals of trigonometric functions

Use substitution to evaluate the integral $\int \frac{\text{sin}\phantom{\rule{0.1em}{0ex}}t}{{\text{cos}}^{3}t}dt.$

We know the derivative of $\text{cos}\phantom{\rule{0.1em}{0ex}}t$ is $\text{−}\text{sin}\phantom{\rule{0.1em}{0ex}}t,$ so we set $u=\text{cos}\phantom{\rule{0.1em}{0ex}}t.$ Then $du=\text{−}\text{sin}\phantom{\rule{0.1em}{0ex}}tdt.$ Substituting into the integral, we have

$\int \frac{\text{sin}\phantom{\rule{0.1em}{0ex}}t}{{\text{cos}}^{3}t}dt=\text{−}\int \frac{du}{{u}^{3}}.$

Evaluating the integral, we get

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

Putting the answer back in terms of t , we get

$\begin{array}{cc}\int \frac{\text{sin}\phantom{\rule{0.1em}{0ex}}t}{{\text{cos}}^{3}t}dt\hfill & =\frac{1}{2{u}^{2}}+C\hfill \\ \\ & =\frac{1}{2{\text{cos}}^{2}t}+C.\hfill \end{array}$

Use substitution to evaluate the integral $\int \frac{\text{cos}\phantom{\rule{0.1em}{0ex}}t}{{\text{sin}}^{2}t}dt.$

$-\frac{1}{\text{sin}\phantom{\rule{0.1em}{0ex}}t}+C$

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.

## Finding an antiderivative using u -substitution

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

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 :

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

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

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

Use substitution to evaluate the indefinite integral $\int {\text{cos}}^{3}t\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.2em}{0ex}}dt.$

$-\frac{{\text{cos}}^{4}t}{4}+C$

## Substitution for definite integrals

Substitution can be used with definite integrals, too. However, using substitution to evaluate a definite integral requires a change to the limits of integration. If we change variables in the integrand, the limits of integration change as well.

## Substitution with definite integrals

Let $u=g\left(x\right)$ and let ${g}^{\text{′}}$ be continuous over an interval $\left[a,b\right],$ and let f be continuous over the range of $u=g\left(x\right).$ Then,

${\int }_{a}^{b}f\left(g\left(x\right)\right){g}^{\prime }\text{(}x\right)dx={\int }_{g\left(a\right)}^{g\left(b\right)}f\left(u\right)du.$

Although we will not formally prove this theorem, we justify it with some calculations here. From the substitution rule for indefinite integrals, if $F\left(x\right)$ is an antiderivative of $f\left(x\right),$ we have

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