# How do you integrate int 1/sqrt(x^2-49)dx using trigonometric substitution?

Jan 4, 2016

For $x > 7$,
$\int \frac{1}{\sqrt{{x}^{2} - 49}} \mathrm{dx} = \ln \left(\sqrt{{x}^{2} - 49} + x\right) + C$.

For $x < - 7$,
$\int \frac{1}{\sqrt{{x}^{2} - 49}} \mathrm{dx} = - \ln \left(\sqrt{{x}^{2} - 49} - x\right) + C$.

Where $C$ is the constant of integration.

#### Explanation:

Use the identity ${\sec}^{2} u - 1 \equiv {\tan}^{2} u$.

Substitute $x = 7 \sec u$.

For $x > 7$, let $0 < \setminus u < \setminus \frac{\pi}{2}$.

$\frac{\mathrm{dx}}{\mathrm{du}} = 7 \sec u \tan u$

$\int \frac{1}{\sqrt{{x}^{2} - 49}} \mathrm{dx} = \int \frac{1}{\sqrt{{x}^{2} - 49}} \frac{\mathrm{dx}}{\mathrm{du}} \mathrm{du}$

$= \int \frac{1}{\sqrt{{\left(7 \sec u\right)}^{2} - 49}} \left(7 \sec u \tan u\right) \mathrm{du}$

$= \int \frac{7 \sec u \tan u}{7 \sqrt{{\sec}^{2} u - 1}} \mathrm{du}$

$= \int \sec u \frac{\tan u}{\sqrt{{\tan}^{2} u}} \mathrm{du}$

Since $0 < \setminus u < \setminus \frac{\pi}{2}$, $\sqrt{{\tan}^{2} u} \equiv \tan u$.

$\int \sec u \frac{\tan u}{\sqrt{{\tan}^{2} u}} \mathrm{du} = \int \sec u \mathrm{du}$

$= \ln | \sec u + \tan u | + {C}_{1}$, where ${C}_{1}$ is an integration constant.

$= \ln \left(\sec u + \tan u\right) + {C}_{1}$

$= \ln \left(7 \sec u + 7 \tan u\right) + {C}_{2}$, where ${C}_{2} = {C}_{1} - \ln 7$.

$= \ln \left(x + \sqrt{{x}^{2} - 49}\right) + {C}_{2}$

For $x < - 7$, let $\frac{\pi}{2} < \setminus u < \setminus \pi$, then $\sqrt{{\tan}^{2} u} \equiv - \tan u$.

$\int \sec u \frac{\tan u}{\sqrt{{\tan}^{2} u}} \mathrm{du} = - \int \sec u \mathrm{du}$

$= - \ln | \sec u + \tan u | + {C}_{3}$, where ${C}_{3}$ is an integration constant.

$= - \ln \left(- \sec u - \tan u\right) + {C}_{3}$

$= - \ln \left(- 7 \sec u - 7 \tan u\right) + {C}_{4}$, where ${C}_{4} = {C}_{3} + \ln 7$.

$= - \ln \left(- x + \sqrt{{x}^{2} - 49}\right) + {C}_{4}$