# Thread: What should I make u in this integration problem?

1. ## What 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.

2. is the denominator $\displaystyle 1+\cos^2(32x)$ or $\displaystyle 1 + \cos(32^2x^2)$ ?

3. Originally Posted by Dana_Scully
$\displaystyle \int \frac{\sin(32x)}{1+\cos(32x)^2} dx$

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

4. Skeeter, it was the former one. DeMath, thank you again!

5. Originally Posted by Dana_Scully
Skeeter, it was the former one. DeMath, thank you again!
Well, that changes things considerably, doesn't it? I'm going to try and solve the original integral you posted without using a table or program.