Integration by Trigonometric Substitution

James98765
Messages
30
Reaction score
0

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!
 
Physics news on Phys.org
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
 
Did the substitution you used work out? It doesn't look like it produced anything that would be useful.
 
Your substitution tanx = 2sinΘ doesn't look valid at all. I would have tried x = tanΘ. What did you get for your answer?
 
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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top