James98765
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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!