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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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    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...2012-12-07

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