# How do you integrate int x/sqrt(x^2-25) by trigonometric substitution?

Jan 14, 2017

$\int \frac{x}{\sqrt{{x}^{2} - 25}} \mathrm{dx} = \sqrt{{x}^{2} - 25} + C$

#### Explanation:

You do not need a trigonometric substitution to solve this integral.

Substituting:

$t = {x}^{2} - 25$
$\mathrm{dt} = 2 x \mathrm{dx}$

you have:

$\int \frac{x}{\sqrt{{x}^{2} - 25}} \mathrm{dx} = \frac{1}{2} \int \frac{\mathrm{dt}}{\sqrt{t}} = \sqrt{t} + C$

and substituting back $x$:

$\int \frac{x}{\sqrt{{x}^{2} - 25}} \mathrm{dx} = \sqrt{{x}^{2} - 25} + C$

Jan 14, 2017

As has been shown, trigonometric substitution is a waste of time and effort, but it can be used with the substitution $x = 5 \sec \theta \implies \mathrm{dx} = 5 \sec \theta \tan \theta d \theta$.

$\int \frac{x}{\sqrt{{x}^{2} - 25}} \mathrm{dx} = \int \frac{5 \sec \theta}{\sqrt{25 {\sec}^{2} \theta - 25}} \left(5 \sec \theta \tan \theta d \theta\right)$

Note that $\sqrt{25 {\sec}^{2} \theta - 25} = 5 \sqrt{{\sec}^{2} \theta - 1} = 5 \tan \theta$ via the form of the Pythagorean identity ${\tan}^{2} \theta + 1 = {\sec}^{2} \theta$.

$= \int \frac{5 \sec \theta}{5 \tan \theta} \left(5 \sec \theta \tan \theta d \theta\right) = 5 \int {\sec}^{2} \theta d \theta = 5 \tan \theta + C$

Note that $\tan \theta = \sqrt{{\sec}^{2} \theta - 1}$ and from our original substitution $\sec \theta = \frac{x}{5}$:

$= 5 \sqrt{{\sec}^{2} \theta - 1} + C = 5 \sqrt{{x}^{2} / 25 - 1} + C = 5 \sqrt{\frac{{x}^{2} - 25}{25}} + C$

$= \sqrt{{x}^{2} + 25} + C$