What should I make u in this integration problem?

**Dana_Scully** · January 19th 2009, 11:53 AM

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

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

**skeeter** · January 19th 2009, 12:10 PM

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

**DeMath** · January 19th 2009, 12:10 PM

Originally Posted by Dana_Scully

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

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

$\int {\frac{{\sin \left( {32x} \right)}}{{1 + {{\cos }^2}\left( {32x} \right)}}dx} = \left\{ \begin{gathered}\cos (32x) = u, \hfill \\- 32\sin (32x)dx = du \hfill \\ \end{gathered} \right\} = - \frac{1}{{32}} \int {\frac{{du}}{{1 + {u^2}}}} =$ $- \frac{1}{{32}}\arctan \left( u \right) + C = - \frac{1}{{32}}\arctan \left( {\cos (32x)} \right) + C.$

**Dana_Scully** · January 19th 2009, 12:16 PM

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

**Bilbo Baggins** · January 19th 2009, 12:28 PM

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.

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