# Trigonometric substitution

#### 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.

#### running-gag

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 $$\displaystyle u = \sin \theta$$

SyNtHeSiS