Trigonometric Integration ${\int_{0}^{2\pi}} \sin(2x)\cos(3x)\, dx$ $${\int_{0}^{2\pi}} \sin(2x)\cos(3x)\, dx$$
I think you need to use a double angle formula but I am not sure how I am supposed to split the $3x$.
 A: If you would like to avoid any of those trig formulas and know the complex definition of $\sin$ and $\cos$ this is the easy to compute integral:
$$
\frac{1}{4i}\int_0^{2\pi}(e^{2i\theta}-e^{-2i\theta})(e^{3i\theta}+e^{-3i\theta})d\theta
$$
which you will find is zero by multiplying stuff and evaluating easy integrals by substitution.
A: Addition formula
$$
 \sin a \cos b = \frac{1}{2} \left(\sin (a+b) + \sin ( a - b)\right)
\tag{1}
$$

Recast integrand
For the integrand, use the addition formula $(1)$ with $a=2x$ and $b=3x$ to find
$$
\sin (2 x) \cos (3 x) = \frac{1}{2} \left(\sin (5 x)-\sin (x)\right)
$$

Transform integral
$$
\int \sin (2 x) \cos (3 x) dx = \frac{1}{2} \int \sin (5 x)-\sin (x)\, dx = \frac{1}{2} \cos x - \frac{1}{10} \cos (5x)
$$


A: Using substitution $t=x-\pi$ we get
$$\int_{-\pi}^\pi \sin 2t \cos(3t+\pi) \mathrm{d}t = - \int_{-\pi}^\pi \sin 2t \cos 3t \mathrm{d}t.$$
So we are integrating and odd function and the interval is symmetric w.r.t. origin, therefore the integral is zero.
This is basically the same thing as saying that half of the Fourier coefficients of an even function are zero. (The even function here being $\cos 3t$.)
A: Use
$$\sin(a)\cos(b) = \frac{1}{2}\left(\sin(a+b) + \sin(a-b)\right)$$
