How can trigonometric substitution be used to simplify a complex integral?

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Hello, I have been wokring on this problem for some hours now, and I get the wrong answer, but I can't understand why, could you guys please look at it?
http://img219.imageshack.us/my.php?image=utennavnwv7.jpg

 

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I should probably mention that the answer is supposed to be:

2*arctan(2x)+4x/(4x^2+1) +C

 

[HallsofIvy]

I didn't go through that in detail but it looks like a very strange way to attack the problem! You have a square of a square and you write it as a fourth power of a square root of a square so you can apply a trig substitution!
You don't need the square root to apply a trig substitution. Let 2x= tan t and 4x²+ 1= tan² t+ 1= sec². (4x²+ 1)²= sec⁴ t and 2dx= sec² t dt. Your integral becomes
[tex]\int\frac{8dx}{(4x^2+ 1)^2}= \int \frac{4dt}{sec^2 t}= 4\int cos^2 t dt[/itex]<br /> That should be easy.[/tex]