Integral trigonometry $\int\sin3x\sin^2x\,dx$
$=\int \sin3x \frac{(1-cos2x)}{2}dx$
$=\frac{1}{2}(\int \sin 3x dx - \int \sin 3x \cos 2x dx)$
$I=\frac{-1}{2}\frac{1}{3}cos 3x-1/4(\int \sin 5x dx+ \int \sin x dx)$
So my question is how do i get from:
$\int \sin (3x)\cos (2x) dx$   to $\frac{1}{2}(\int \sin (5x) dx+ \int \sin (x) dx)$
Thanks for fast answer i solved it.

by the formula $\sin { \left( x \right) \cos { \left( y \right) =\frac { 1 }{ 2 } \left[ \sin { \left( x+y \right) +\sin { \left( x-y \right)  }  }  \right]  }  } $
in your case we have $$\\ \sin { \left( 3x \right) \cos { \left( 2x \right) =\frac { 1 }{ 2 } \left[ \sin { \left( 3x+2x \right) +\sin { \left( 3x-2x \right)  }  }  \right] =\frac { 1 }{ 2 } \left[ \sin { \left( 5x \right) +\sin { \left( x \right)  }  }  \right]  }  } $$
