# What should I make u in this integration problem?

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

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 PM
skeeter
is the denominator $1+\cos^2(32x)$ or $1 + \cos(32^2x^2)$ ?
• Jan 19th 2009, 12:10 PM
DeMath
Quote:

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.

$\int {\frac{{\sin \left( {32x} \right)}}{{1 + {{\cos }^2}\left( {32x} \right)}}dx} = \left\{ \begin{gathered}\cos (32x) = u, \hfill \\- 32\sin (32x)dx = du \hfill \\ \end{gathered} \right\} = - \frac{1}{{32}} \int {\frac{{du}}{{1 + {u^2}}}} =$ $- \frac{1}{{32}}\arctan \left( u \right) + C = - \frac{1}{{32}}\arctan \left( {\cos (32x)} \right) + C.$
• Jan 19th 2009, 12:16 PM
Dana_Scully
Skeeter, it was the former one. DeMath, thank you again!
• Jan 19th 2009, 12:28 PM
Bilbo Baggins
Quote:

Originally Posted by Dana_Scully
Skeeter, it was the former one. DeMath, thank you again!

Well, that changes things considerably, doesn't it? (Happy)I'm going to try and solve the original integral you posted without using a table or program.