# How do you integrate int x^2/sqrt(x^2+9) by trigonometric substitution?

Feb 6, 2017

$\frac{x \sqrt{{x}^{2} + 9}}{2} - \frac{9}{2} \ln | \frac{x + \sqrt{{x}^{2} + 9}}{3} | + C$

#### Explanation:

Use the substitution $x = 3 \tan \theta$. Then $\mathrm{dx} = 3 {\sec}^{2} \theta d \theta$.

$\implies \int {\left(3 \tan \theta\right)}^{2} / \sqrt{{\left(3 \tan \theta\right)}^{2} + 9} \cdot 3 {\sec}^{2} \theta d \theta$

$\implies \int \frac{9 {\tan}^{2} \theta}{\sqrt{9 {\tan}^{2} \theta + 9}} \cdot 3 {\sec}^{2} \theta d \theta$

$\implies \int \frac{9 {\tan}^{2} \theta}{\sqrt{9 {\sec}^{2} \theta}} \cdot 3 {\sec}^{2} \theta d \theta$

$\implies \int \frac{9 {\tan}^{2} \theta}{3 \sec \theta} \cdot 3 {\sec}^{2} \theta d \theta$

$\implies \int 9 {\tan}^{2} \theta \sec \theta d \theta$

$\implies 9 \int \tan \theta \left(\tan \theta \sec \theta\right) d \theta$

Use integration by parts for this one. Let $u = \tan \theta$ and $\mathrm{dv} = \tan \theta \sec \theta d \theta$. Then $\mathrm{du} = {\sec}^{2} \theta d \theta$ and $v = \sec \theta$.

$\implies 9 \tan \theta \sec \theta - 9 \int {\sec}^{3} \theta$

This is a (relatively) known integral. The proof can be found here

$\implies 9 \tan \theta \sec \theta - \frac{9}{2} \sec \theta \tan \theta - \frac{9}{2} \ln | \sec \theta + \tan \theta | + C$

$\implies \frac{9}{2} \tan \theta \sec \theta - \frac{9}{2} \ln | \sec \theta + \tan \theta | + C$

We know from our initial substitution that $\tan \theta = \frac{x}{3}$. This means that the hypotenuse of the right triangle would be $\sqrt{{x}^{2} + 9}$, and that $\sec \theta = \frac{\sqrt{{x}^{2} + 9}}{3}$.

$\implies \frac{9}{2} \left(\frac{x}{3}\right) \frac{\sqrt{{x}^{2} + 9}}{3} - \frac{9}{2} \ln | \frac{\sqrt{{x}^{2} + 9}}{3} + \frac{x}{3} | + C$

$\implies \frac{x \sqrt{{x}^{2} + 9}}{2} - \frac{9}{2} \ln | \frac{x + \sqrt{{x}^{2} + 9}}{3} | + C$

Hopefully this helps!