1. ## Trigonometric substitution

Use the substitution $\displaystyle x = 2tan\theta, -\pi/2 < \theta < \pi/2$ to evaluate:

$\displaystyle \int \frac{1}{x^2 \sqrt{x^2 + 4}} dx$

Attempt:

= $\displaystyle \int \frac{1}{4tan^2\theta\sqrt{4tan^2\theta+ 4} } dx$

$\displaystyle x = 2tan\theta$
$\displaystyle dx = 2sec^2\theta d\theta$

= $\displaystyle \int \frac {1}{(4tan^2\theta)|2sec\theta|}2sec^2\theta d\theta$
=$\displaystyle \int \frac {sec\theta}{4tan^2\theta} d\theta$
=$\displaystyle \frac{1}{4}\int\frac{sec\theta}{tan^2\theta}$

I dont know what to do from here.

2. Originally Posted by SyNtHeSiS
Use the substitution $\displaystyle x = 2tan\theta, -\pi/2 < \theta < \pi/2$ to evaluate:

$\displaystyle \int \frac{1}{x^2 \sqrt{x^2 + 4}} dx$

Attempt:

= $\displaystyle \int \frac{1}{4tan^2\theta\sqrt{4tan^2\theta+ 4} } dx$

$\displaystyle x = 2tan\theta$
$\displaystyle dx = 2sec^2\theta d\theta$

= $\displaystyle \int \frac {1}{(4tan^2\theta)|2sec\theta|}2sec^2\theta d\theta$
=$\displaystyle \int \frac {sec\theta}{4tan^2\theta} d\theta$
=$\displaystyle \frac{1}{4}\int\frac{sec\theta}{tan^2\theta}$

I dont know what to do from here.
=$\displaystyle \frac{1}{4}\int\frac{\cos\theta d\theta}{sin^2\theta}$

Then use the substitution $\displaystyle u = \sin \theta$