Integration by Trigonometric Substitution

  • Thread starter James98765
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  • #1
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Homework Statement


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


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:

[tex]Let \tan{(x)} = 2\sin{(\theta)}[/tex]

[tex]x = \tan^{-1}{(2\sin{(\theta)})}[/tex]

[tex]dx = \frac{4\sin{(\theta)}\cos{(\theta)}}{1 + 4\sin^2{(\theta)}}d\theta[/tex]

[tex]\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[/tex]

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

[tex]=\frac{1}{2}\int\frac{\sin{(\theta)}\cos{(\theta)}}{(1 + 4\sin^2{(\theta)})\cos^3{(\theta)}}d\theta[/tex]

[tex]=\frac{1}{2}\int\frac{\sin{(\theta)}\sec^2{(\theta)}}{1 + 4\sin^2{(\theta)}}d\theta[/tex]

[tex]=\frac{1}{2}\int\frac{\sec{(\theta)}\tan{(\theta)}}{1 + 4\sin^2{(\theta)}}d\theta[/tex]

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!
 

Answers and Replies

  • #2
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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
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Did the substitution you used work out? It doesn't look like it produced anything that would be useful.
 
  • #4
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Your substitution tanx = 2sinΘ doesn't look valid at all. I would have tried x = tanΘ. What did you get for your answer?
 
  • #5
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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.
 

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