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.
 A: You have to use the transformation formula
$\sin(a)\cos(b) = \frac{1}{2}(\sin(a+b) + \sin(a-b))$
A: Using the identities
$$\begin{cases}\cos(x\pm y)=\cos x\cos y\mp \sin x\sin y&(1)\\[1ex]\sin(x\pm y)=\sin x\cos y\pm \sin y\cos x&(2)\end{cases}$$
you can add/subtract the alternating forms of $(1)$ and $(2)$ to get equivalent expressions for $\sin x\cos y$ or $\cos x\cos y$.
A: From the compound angle formulae for sine, $$\sin A\cos B=\frac 12(\sin(A+B)+\sin(A-B))$$
A: 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 $$\\ \sin { \left( 3x \right) \cos { \left( 2x \right) =\frac { 1 }{ 2 } \left[ \sin { \left( 3x+2x \right) +\sin { \left( 3x-2x \right)  }  }  \right] =\frac { 1 }{ 2 } \left[ \sin { \left( 5x \right) +\sin { \left( x \right)  }  }  \right]  }  } $$
A: By Euler and de Moivre, from$$(z^3-z^{-3})(z-z^{-1})^2=z^5-z^{-5}-2(z^3-z^{-3})+z-z^{-1},$$
you deduce
$$2i\sin(3x)4i^2\sin^2(x)=2i\sin(5x)-4i\sin(3x)+2i\sin(x).$$
