# How do you integrate int 1/(x^2sqrt(x^2-36)) by trigonometric substitution?

Apr 6, 2018

The integral is $\frac{\sqrt{{x}^{2} - 36}}{36 x} + C$.

#### Explanation:

Let $x = 6 \sec \theta$. This means that:

$\mathrm{dx} = 6 \sec \theta \tan \theta$ $d \theta$

Performing the substitution:

$\textcolor{w h i t e}{=} \int \frac{1}{{x}^{2} \sqrt{{x}^{2} - 36}} \mathrm{dx}$

$= \int \frac{1}{{\left(6 \sec \theta\right)}^{2} \sqrt{{\left(6 \sec \theta\right)}^{2} - 36}} 6 \sec \theta \tan \theta$ $d \theta$

$= \int \frac{1}{{\left(6 \sec \theta\right)}^{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}} \sqrt{{\left(6 \sec \theta\right)}^{2} - 36}} \textcolor{red}{\cancel{\textcolor{b l a c k}{6 \sec \theta}}} \tan \theta$ $d \theta$

$= \int \frac{1}{6 \sec \theta \sqrt{{\left(6 \sec \theta\right)}^{2} - 36}} \tan \theta$ $d \theta$

$= \int \frac{1}{6 \sec \theta \sqrt{36 {\sec}^{2} \theta - 36}} \tan \theta$ $d \theta$

$= \int \frac{1}{6 \sec \theta \sqrt{36 \left({\sec}^{2} \theta - 1\right)}} \tan \theta$ $d \theta$

$= \int \frac{1}{6 \sec \theta \cdot 6 \sqrt{{\sec}^{2} \theta - 1}} \tan \theta$ $d \theta$

By the Pythagorean identity ${\tan}^{2} \theta + 1 = {\sec}^{2} \theta$:

$= \int \frac{1}{36 \sec \theta \cdot \sqrt{{\tan}^{2} \theta + 1 - 1}} \tan \theta$ $d \theta$

$= \int \frac{1}{36 \sec \theta \cdot \sqrt{{\tan}^{2} \theta}} \tan \theta$ $d \theta$

$= \int \frac{1}{36 \sec \theta \cdot \tan \theta} \tan \theta$ $d \theta$

$= \int \frac{1}{36 \sec \theta \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{\tan}}} \theta} \textcolor{red}{\cancel{\textcolor{b l a c k}{\tan}}} \theta$ $d \theta$

$= \int \frac{1}{36 \sec \theta}$ $d \theta$

$= \frac{1}{36} \int \frac{1}{\sec} \theta$ $d \theta$

$= \frac{1}{36} \int \frac{1}{\frac{1}{\cos} \theta}$ $d \theta$

$= \frac{1}{36} \int \cos \theta$ $d \theta$

$= \frac{1}{36} \sin \theta + C$

Now we want $\sin \theta$ in terms of $x$. By our substitution earlier, we know:

$6 \sec \theta = x$

$\sec \theta = \frac{x}{6}$

$\frac{1}{\cos} \theta = \frac{x}{6}$

$\cos \theta = \frac{6}{x}$

${\cos}^{2} \theta = \frac{36}{x} ^ 2$

$- {\cos}^{2} \theta = - \frac{36}{x} ^ 2$

$- {\cos}^{2} \theta + 1 = - \frac{36}{x} ^ 2 + 1$

$1 - {\cos}^{2} \theta = 1 - \frac{36}{x} ^ 2$

$1 - {\cos}^{2} \theta = \frac{{x}^{2} - 36}{x} ^ 2$

${\sin}^{2} \theta = \frac{{x}^{2} - 36}{x} ^ 2$

$\sin \theta = \sqrt{\frac{{x}^{2} - 36}{x} ^ 2}$

$\sin \theta = \frac{\sqrt{{x}^{2} - 36}}{x}$

Plugging this back into the integral:

$\textcolor{w h i t e}{=} \frac{1}{36} \sin \theta + C$

$= \frac{1}{36} \cdot \frac{\sqrt{{x}^{2} - 36}}{x} + C$

$= \frac{\sqrt{{x}^{2} - 36}}{36 x} + C$

That's the whole integral. Hope this helped!