# Integration by Trigonometric Substitution

1. Aug 31, 2009

### James98765

1. The problem statement, all variables and given/known data
Evaluate:
$$\int\frac{1}{(4 - \tan^2{x})^{3/2}}dx$$

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

3. 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!

2. Aug 31, 2009

### James98765

Sorry to reply to my own post but I've solved the problem and I do not know how to delete this. Thank you!
-James

3. Aug 31, 2009

### Staff: Mentor

Did the substitution you used work out? It doesn't look like it produced anything that would be useful.

4. Aug 31, 2009

### Bohrok

Your substitution tanx = 2sinΘ doesn't look valid at all. I would have tried x = tanΘ. What did you get for your answer?

5. Aug 31, 2009

### James98765

No actually I managed to waste a lot of time after I copied the problem down wrong from the book. I gues that doesn't change the fact that I still don't know how to integrate the problem I posted but it no longer matters.