Integrating $\int_0^{\frac{\pi}{2}}\cos(2x)\sin(3x)\,dx$ How would I integrate the following 
$\int_0^{\frac{\pi}{2}}\cos(2x)\sin(3x)\,dx$
I did the following
$\frac{1}{2}\int \left(\sin(5x)-\sin(x)\right)\,dx$
$\frac{1}{10}-\cos(5x)+cos(x)$
$\frac{1}{10}(0)+0-0=0$
But I am not sure If I did it right.

Not quite. Using the identity
$$2 \cos{a}\sin{b} = \sin{(a + b)} - \sin{(a - b)}$$
we find that
\begin{align}
\int_0^{\pi/2} \cos{(2x)} \sin{(3x)} &= \frac 1 2 \int_0^{\pi/2} \sin{5x} - \sin{(-x)} \\
&= \frac 1 2 \int_0^{\pi/2} \sin{5x} + \sin{x} \\
&= \frac 1 2 \left(\frac{-1}{5} \cos{5x} - \cos{x}\right)_0^{\pi/2} \\
&= \frac 1 2 \left(\frac{-1}{5} \cos{\frac{5\pi}{2}} - \cos{\frac{\pi}{2}} - (\frac{-1}{5} \cos{0} - \cos{0})\right) \\
&= \frac 1 2 \left(0 - (\frac{-1}{5} - 1)\right) \\
&= \frac 3 5
\end{align}
