Integration by Trigonometric Substitution

Homework Statement

Evaluate:
$$\int\frac{1}{(4 - \tan^2{x})^{3/2}}dx$$

Homework Equations

I must integrate the above equation using only trigonometric subtitutions of algebraic equations.

The Attempt at a Solution

Here is what I have so far:

$$Let \tan{(x)} = 2\sin{(\theta)}$$

$$x = \tan^{-1}{(2\sin{(\theta)})}$$

$$dx = \frac{4\sin{(\theta)}\cos{(\theta)}}{1 + 4\sin^2{(\theta)}}d\theta$$

$$\int\frac{1}{(4 - \tan^2{x})^{3/2}}dx = \int\frac{1}{(4 - (2\sin{(\theta)})^2)^{3/2}}\frac{4\sin{(\theta)}\cos{(\theta)}}{1 + 4\sin^2{(\theta)}}d\theta$$

$$= \frac{4}{8}\int\frac{\sin{(\theta)}\cos{(\theta)}}{(1 + 4\sin^2{(\theta)})(1 - \sin^2{\theta)})^{3/2}}d\theta$$

$$=\frac{1}{2}\int\frac{\sin{(\theta)}\cos{(\theta)}}{(1 + 4\sin^2{(\theta)})\cos^3{(\theta)}}d\theta$$

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

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

I can't seem to integrate the final integral above. Can anybody help me get past this step or can anybody tell me if I've made a mistake. Thanks!
-James

P.S. Sorry if it's messy!