$\displaystyle \int \frac{\sin(32x)}{1+\cos(32x)^2} dx$

I tried 32x and sin(32x) but even if those are right I don't know where to go from there.

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- Jan 19th 2009, 11:53 AMDana_ScullyWhat should I make u in this integration problem?
$\displaystyle \int \frac{\sin(32x)}{1+\cos(32x)^2} dx$

I tried 32x and sin(32x) but even if those are right I don't know where to go from there. - Jan 19th 2009, 12:10 PMskeeter
is the denominator $\displaystyle 1+\cos^2(32x)$ or $\displaystyle 1 + \cos(32^2x^2) $ ?

- Jan 19th 2009, 12:10 PMDeMath
$\displaystyle \int {\frac{{\sin \left( {32x} \right)}}{{1 + {{\cos }^2}\left( {32x} \right)}}dx} = \left\{ \begin{gathered}\cos (32x) = u, \hfill \\- 32\sin (32x)dx = du \hfill \\ \end{gathered} \right\} = - \frac{1}{{32}} \int {\frac{{du}}{{1 + {u^2}}}} =$$\displaystyle - \frac{1}{{32}}\arctan \left( u \right) + C = - \frac{1}{{32}}\arctan \left( {\cos (32x)} \right) + C.$

- Jan 19th 2009, 12:16 PMDana_Scully
Skeeter, it was the former one. DeMath, thank you again!

- Jan 19th 2009, 12:28 PMBilbo Baggins