integration question

Quote:

Originally Posted by cinder

Evaluate the integral (in book, this is the "Integration by parts" section):

$\int x\cos(5x)dx$

Something like...

$x(\frac{1}{5})\sin5x - \int\frac{1}{5}\sin(5x)dx$

and then...? Or am I way off?

Wait, I think I got it. Is the answer:

$\frac{1}{5}x\sin5x + \frac{1}{25}\cos5x + C$

?

You can check your answer by differentiating it:

$\frac{d}{dx}\left[ \frac{1}{5}x\sin5x + \frac{1}{25}\cos5x + C \right]=$ $\frac{1}{5}\sin5x +x\cos5x - \frac{1}{5}\sin5x =x\cos5x$ ,

which is the integrand, so yes you have the correct integral.

RonL