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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