# Trigonometric substitution

• Aug 9th 2010, 10:15 AM
SyNtHeSiS
Trigonometric substitution
Use the substitution $x = 2tan\theta, -\pi/2 < \theta < \pi/2$ to evaluate:

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

Attempt:

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

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

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

I dont know what to do from here.
• Aug 9th 2010, 10:33 AM
running-gag
Quote:

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

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

Attempt:

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

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

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

I dont know what to do from here.

= $\frac{1}{4}\int\frac{\cos\theta d\theta}{sin^2\theta}$

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