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The question is to find

$$\displaystyle \int \frac {4\sin (x)}{5+4\cos^2x -8x}dx$$

Can anyone help me? I need all the steps, because I need to understand what to do. Many thanks in advance.

What I tried so far: substitute $t=\cos(x) \ldots$ no way to solve $\int \frac{-4}{4t^2-8\arccos t+ 5}dt$

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Why do you assume there is a simple answer for this integral? What is the exact wording of the homework problem... –  GEdgar Dec 7 '12 at 15:19
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Note that $\displaystyle \int \frac {4\sin (x)}{5+4\cos^2x -8\cos x}dx$ can be done easily. –  GEdgar Dec 7 '12 at 15:21
    
Just to solve it... –  Lorenzo Comoglio Dec 7 '12 at 15:36
    
This integral must be a very bad beast: WA can't handle it in regular time... –  DonAntonio Dec 7 '12 at 16:52
    
As GEdgar already hinted, very probably there's a typo in the question, as $$5+4\cos^2x-8\cos x=1+4(1-\cos x)^2$$ which together with the numerator given an almost immediate integral... –  DonAntonio Dec 7 '12 at 16:55
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