# Trigonometric substitution

1. Oct 16, 2008

### hyper

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

2. Oct 16, 2008

### hyper

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

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

3. Oct 16, 2008

### HallsofIvy

Staff Emeritus
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]
That should be easy.