2
$\begingroup$

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 $\int \frac{-4}{4t^2-8\arccos t+ 5}dt$

  • 0
    Why do you assume there is a simple answer for this integral? What is the exact wording of the homework problem...2012-12-07
  • 1
    Note that $\displaystyle \int \frac {4\sin (x)}{5+4\cos^2x -8\cos x}dx$ can be done easily.2012-12-07
  • 0
    Just to solve it...2012-12-07
  • 0
    This integral must be a very bad beast: WA can't handle it in regular time...2012-12-07
  • 0
    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

0 Answers 0