# 3.3 Trigonometric substitution  (Page 3/5)

 Page 3 / 5

## Problem-solving strategy: integrating expressions involving $\sqrt{{a}^{2}+{x}^{2}}$

1. Check to see whether the integral can be evaluated easily by using another method. In some cases, it is more convenient to use an alternative method.
2. Substitute $x=a\phantom{\rule{0.1em}{0ex}}\text{tan}\phantom{\rule{0.1em}{0ex}}\theta$ and $dx=a\phantom{\rule{0.1em}{0ex}}{\text{sec}}^{2}\theta d\theta .$ This substitution yields
$\sqrt{{a}^{2}+{x}^{2}}=\sqrt{{a}^{2}+{\left(a\phantom{\rule{0.1em}{0ex}}\text{tan}\phantom{\rule{0.1em}{0ex}}\theta \right)}^{2}}=\sqrt{{a}^{2}\left(1+{\text{tan}}^{2}\theta \right)}=\sqrt{{a}^{2}{\text{sec}}^{2}\theta }=|a\phantom{\rule{0.1em}{0ex}}\text{sec}\phantom{\rule{0.1em}{0ex}}\theta |=a\phantom{\rule{0.1em}{0ex}}\text{sec}\phantom{\rule{0.1em}{0ex}}\theta .$ (Since $-\frac{\pi }{2}<\theta <\frac{\pi }{2}$ and $\text{sec}\phantom{\rule{0.1em}{0ex}}\theta >0$ over this interval, $|a\phantom{\rule{0.1em}{0ex}}\text{sec}\phantom{\rule{0.1em}{0ex}}\theta |=a\phantom{\rule{0.1em}{0ex}}\text{sec}\phantom{\rule{0.1em}{0ex}}\theta .\right)$
3. Simplify the expression.
4. Evaluate the integral using techniques from the section on trigonometric integrals.
5. Use the reference triangle from [link] to rewrite the result in terms of $x.$ You may also need to use some trigonometric identities and the relationship $\theta ={\text{tan}}^{-1}\left(\frac{x}{a}\right).$ ( Note : The reference triangle is based on the assumption that $x>0;$ however, the trigonometric ratios produced from the reference triangle are the same as the ratios for which $x\le 0.\right)$ A reference triangle can be constructed to express the trigonometric functions evaluated at θ in terms of x .

## Integrating an expression involving $\sqrt{{a}^{2}+{x}^{2}}$

Evaluate $\int \frac{dx}{\sqrt{1+{x}^{2}}}$ and check the solution by differentiating.

Begin with the substitution $x=\text{tan}\phantom{\rule{0.1em}{0ex}}\theta$ and $dx={\text{sec}}^{2}\theta d\theta .$ Since $\text{tan}\phantom{\rule{0.1em}{0ex}}\theta =x,$ draw the reference triangle in the following figure. The reference triangle for [link] .

Thus,

$\begin{array}{ccccc}\hfill \int \frac{dx}{\sqrt{1+{x}^{2}}}& =\int \frac{{\text{sec}}^{2}\theta }{\text{sec}\phantom{\rule{0.1em}{0ex}}\theta }d\theta \hfill & & & \begin{array}{c}\text{Substitute}\phantom{\rule{0.2em}{0ex}}x=\text{tan}\phantom{\rule{0.1em}{0ex}}\theta \phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}dx={\text{sec}}^{2}\theta d\theta .\phantom{\rule{0.2em}{0ex}}\text{This}\hfill \\ \text{substitution makes}\phantom{\rule{0.2em}{0ex}}\sqrt{1+{x}^{2}}=\text{sec}\phantom{\rule{0.1em}{0ex}}\theta .\phantom{\rule{0.2em}{0ex}}\text{Simplify.}\hfill \end{array}\hfill \\ & ={\int }^{\text{​}}\text{sec}\phantom{\rule{0.1em}{0ex}}\theta d\theta \hfill & & & \text{Evaluate the integral.}\hfill \\ & =\text{ln}|\text{sec}\phantom{\rule{0.1em}{0ex}}\theta +\text{tan}\phantom{\rule{0.1em}{0ex}}\theta |+C\hfill & & & \begin{array}{c}\text{Use the reference triangle to express the result}\hfill \\ \text{in terms of}\phantom{\rule{0.2em}{0ex}}x.\hfill \end{array}\hfill \\ & =\text{ln}|\sqrt{1+{x}^{2}}+x|+C.\hfill & & & \end{array}$

To check the solution, differentiate:

$\begin{array}{cc}\hfill \frac{d}{dx}\left(\text{ln}|\sqrt{1+{x}^{2}}+x|\right)& =\frac{1}{\sqrt{1+{x}^{2}}+x}·\left(\frac{x}{\sqrt{1+{x}^{2}}}+1\right)\hfill \\ & =\frac{1}{\sqrt{1+{x}^{2}}+x}·\frac{x+\sqrt{1+{x}^{2}}}{\sqrt{1+{x}^{2}}}\hfill \\ & =\frac{1}{\sqrt{1+{x}^{2}}}.\hfill \end{array}$

Since $\sqrt{1+{x}^{2}}+x>0$ for all values of $x,$ we could rewrite $\text{ln}|\sqrt{1+{x}^{2}}+x|+C=\text{ln}\left(\sqrt{1+{x}^{2}}+x\right)+C,$ if desired.

## Evaluating $\int \frac{dx}{\sqrt{1+{x}^{2}}}$ Using a different substitution

Use the substitution $x=\text{sinh}\phantom{\rule{0.1em}{0ex}}\theta$ to evaluate $\int \frac{dx}{\sqrt{1+{x}^{2}}}.$

Because $\text{sinh}\phantom{\rule{0.1em}{0ex}}\theta$ has a range of all real numbers, and $1+{\text{sinh}}^{2}\theta ={\text{cosh}}^{2}\theta ,$ we may also use the substitution $x=\text{sinh}\phantom{\rule{0.1em}{0ex}}\theta$ to evaluate this integral. In this case, $dx=\text{cosh}\phantom{\rule{0.1em}{0ex}}\theta d\theta .$ Consequently,

$\begin{array}{ccccc}\hfill \int \frac{dx}{\sqrt{1+{x}^{2}}}& =\int \frac{\text{cosh}\phantom{\rule{0.1em}{0ex}}\theta }{\sqrt{1+{\text{sinh}}^{2}\theta }}d\theta \hfill & & & \begin{array}{c}\text{Substitute}\phantom{\rule{0.2em}{0ex}}x=\text{sinh}\phantom{\rule{0.1em}{0ex}}\theta \phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}dx=\text{cosh}\phantom{\rule{0.1em}{0ex}}\theta d\theta .\hfill \\ \text{Substitute}\phantom{\rule{0.2em}{0ex}}1+{\text{sinh}}^{2}\theta ={\text{cosh}}^{2}\theta .\hfill \end{array}\hfill \\ & =\int \frac{\text{cosh}\phantom{\rule{0.1em}{0ex}}\theta }{\sqrt{{\text{cosh}}^{2}\theta }}d\theta \hfill & & & \sqrt{{\text{cosh}}^{2}\theta }=|\text{cosh}\phantom{\rule{0.1em}{0ex}}\theta |\hfill \\ & =\int \frac{\text{cosh}\phantom{\rule{0.1em}{0ex}}\theta }{|\text{cosh}\phantom{\rule{0.1em}{0ex}}\theta |}d\theta \hfill & & & |\text{cosh}\phantom{\rule{0.1em}{0ex}}\theta |=\text{cosh}\phantom{\rule{0.1em}{0ex}}\theta \phantom{\rule{0.2em}{0ex}}\text{since}\phantom{\rule{0.2em}{0ex}}\text{cosh}\phantom{\rule{0.1em}{0ex}}\theta >0\phantom{\rule{0.2em}{0ex}}\text{for all}\phantom{\rule{0.2em}{0ex}}\theta .\hfill \\ & =\int \frac{\text{cosh}\phantom{\rule{0.1em}{0ex}}\theta }{\text{cosh}\phantom{\rule{0.1em}{0ex}}\theta }d\theta \hfill & & & \text{Simplify.}\hfill \\ & ={\int }^{\text{​}}1d\theta \hfill & & & \text{Evaluate the integral.}\hfill \\ & =\theta +C\hfill & & & \text{Since}\phantom{\rule{0.2em}{0ex}}x=\text{sinh}\phantom{\rule{0.1em}{0ex}}\theta ,\phantom{\rule{0.2em}{0ex}}\text{we know}\phantom{\rule{0.2em}{0ex}}\theta ={\text{sinh}}^{-1}x.\hfill \\ & ={\text{sinh}}^{-1}x+C.\hfill & & & \end{array}$

## Finding an arc length

Find the length of the curve $y={x}^{2}$ over the interval $\left[0,\frac{1}{2}\right].$

Because $\frac{dy}{dx}=2x,$ the arc length is given by

${\int }_{0}^{1\text{/}2}\sqrt{1+{\left(2x\right)}^{2}}\phantom{\rule{0.1em}{0ex}}dx={\int }_{0}^{1\text{/}2}\sqrt{1+4{x}^{2}}\phantom{\rule{0.1em}{0ex}}dx.$

To evaluate this integral, use the substitution $x=\frac{1}{2}\text{tan}\phantom{\rule{0.1em}{0ex}}\theta$ and $dx=\frac{1}{2}{\text{sec}}^{2}\theta d\theta .$ We also need to change the limits of integration. If $x=0,$ then $\theta =0$ and if $x=\frac{1}{2},$ then $\theta =\frac{\pi }{4}.$ Thus,

$\begin{array}{ccccc}\hfill {\int }_{0}^{1\text{/}2}\sqrt{1+4{x}^{2}}\phantom{\rule{0.1em}{0ex}}dx& ={\int }_{0}^{\pi \text{/}4}\sqrt{1+{\text{tan}}^{2}\theta }\frac{1}{2}{\text{sec}}^{2}\theta d\theta \hfill & & & \begin{array}{c}\text{After substitution,}\hfill \\ \sqrt{1+4{x}^{2}}=\text{tan}\phantom{\rule{0.1em}{0ex}}\theta .\phantom{\rule{0.2em}{0ex}}\text{Substitute}\hfill \\ 1+{\text{tan}}^{2}\theta ={\text{sec}}^{2}\theta \phantom{\rule{0.2em}{0ex}}\text{and simplify.}\hfill \end{array}\hfill \\ & =\frac{1}{2}{\int }_{0}^{\pi \text{/}4}{\text{sec}}^{3}\theta d\theta \hfill & & & \begin{array}{c}\text{We derived this integral in the}\hfill \\ \text{previous section.}\hfill \end{array}\hfill \\ & =\frac{1}{2}\left(\frac{1}{2}\text{sec}\phantom{\rule{0.1em}{0ex}}\theta \phantom{\rule{0.1em}{0ex}}\text{tan}\phantom{\rule{0.1em}{0ex}}\theta +\text{ln}|\text{sec}\phantom{\rule{0.1em}{0ex}}\theta +\text{tan}\phantom{\rule{0.1em}{0ex}}\theta |\right)|{}_{\begin{array}{c}\\ 0\end{array}}^{\begin{array}{c}\pi \text{/}4\\ \end{array}}\hfill & & & \text{Evaluate and simplify.}\hfill \\ & =\frac{1}{4}\left(\sqrt{2}+\text{ln}\left(\sqrt{2}+1\right)\right).\hfill & & & \end{array}$

Rewrite ${\int }^{\text{​}}{x}^{3}\sqrt{{x}^{2}+4}\phantom{\rule{0.1em}{0ex}}dx$ by using a substitution involving $\text{tan}\phantom{\rule{0.1em}{0ex}}\theta .$

${\int }^{\text{​}}32\phantom{\rule{0.1em}{0ex}}{\text{tan}}^{3}\theta \phantom{\rule{0.1em}{0ex}}{\text{sec}}^{3}\theta d\theta$

## Integrating expressions involving $\sqrt{{x}^{2}-{a}^{2}}$

The domain of the expression $\sqrt{{x}^{2}-{a}^{2}}$ is $\left(\text{−}\infty ,\text{−}a\right]\cup \left[a,\text{+}\infty \right).$ Thus, either $x<\text{−}a$ or $x>a.$ Hence, $\frac{x}{a}\le -1$ or $\frac{x}{a}\ge 1.$ Since these intervals correspond to the range of $\text{sec}\phantom{\rule{0.1em}{0ex}}\theta$ on the set $\left[0,\frac{\pi }{2}\right)\cup \left(\frac{\pi }{2},\pi \right],$ it makes sense to use the substitution $\text{sec}\phantom{\rule{0.1em}{0ex}}\theta =\frac{x}{a}$ or, equivalently, $x=a\phantom{\rule{0.1em}{0ex}}\text{sec}\phantom{\rule{0.1em}{0ex}}\theta ,$ where $0\le \theta <\frac{\pi }{2}$ or $\frac{\pi }{2}<\theta \le \pi .$ The corresponding substitution for $dx$ is $dx=a\phantom{\rule{0.1em}{0ex}}\text{sec}\phantom{\rule{0.1em}{0ex}}\theta \phantom{\rule{0.1em}{0ex}}\text{tan}\phantom{\rule{0.1em}{0ex}}\theta d\theta .$ The procedure for using this substitution is outlined in the following problem-solving strategy.

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