integral of cos(x)sin(2x)
the answer is (-1/6)cos(3x)-.5cos(x)
$\displaystyle sin(a + b) = sin(a)cos(b) + sin(b)cos(a)$
$\displaystyle sin(a - b) = sin(a)cos(b) - sin(b)cos(a)$
Thus
$\displaystyle sin(a + b) + sin(a - b) = 2sin(a)cos(b)$
So
$\displaystyle sin(a)cos(b) = \frac{sin(a + b) + sin(a - b)}{2}$
Your integral, then is
$\displaystyle
\int sin(2x)cos(x)~dx = \frac{1}{2} \int (sin(3x) + sin(x) )~dx$
-Dan
Yeah, but what you don't know is how many problems I ran into that I couldn't integrate before I finally started remembering this trick!
You should try kalagota's method also. It's always good to practice trig integrals (since they are, in my opinion, the nastiest kind.)
-Dan