# How do you integrate int 1/(sqrt(x^2+16)) by trigonometric substitution?

Dec 2, 2016

The answer is $= \ln \left(\left\mid \frac{\sqrt{16 + {x}^{2}}}{4} + \frac{x}{4} \right\mid\right) + C$

#### Explanation:

We use ${\tan}^{2} x + 1 = {\sec}^{2} x$

Let $x = 4 \tan u$

$\mathrm{dx} = 4 {\sec}^{2} u \mathrm{du}$

$\int \frac{\mathrm{dx}}{\sqrt{{x}^{2} + 16}} = \int \frac{4 {\sec}^{2} u \mathrm{du}}{\sqrt{16 {\tan}^{2} u + 16}}$

$= \int \frac{4 {\sec}^{2} u \mathrm{du}}{4 \sec u}$

$= \int \sec u \mathrm{du}$

$= \int \frac{\sec u \left(\tan u + \sec u\right) \mathrm{du}}{\tan u + \sec u}$

$= \int \frac{\left(\sec u \tan u + {\sec}^{2} u\right) \mathrm{du}}{\tan u + \sec u}$

Let $\sec u + \tan u = w$

$\mathrm{dw} = \left(\sec u \tan u + {\sec}^{2} u\right) \mathrm{du}$

So,

$\int \frac{\left(\sec u \tan u + {\sec}^{2} u\right) \mathrm{du}}{\tan u + \sec u}$

$= \int \frac{\mathrm{dw}}{w}$

$= \ln \left(w\right)$

$= \ln \left(\sec u + \tan u\right)$

$\tan u = \frac{x}{4}$

${\sec}^{2} u = 1 + {\tan}^{2} u = 1 + {x}^{2} / 16$

$\ln \left(\sec u + \tan u\right) = \ln \left(\frac{\sqrt{16 + {x}^{2}}}{4} + \frac{x}{4}\right)$

$\int \frac{\mathrm{dx}}{\sqrt{{x}^{2} + 16}} = \ln \left(\left\mid \frac{\sqrt{16 + {x}^{2}}}{4} + \frac{x}{4} \right\mid\right) + C$