Answer
Using trig identities and integrals of odd functions it can be proved that $$\int_{-\pi}^\pi{\sin mx \cos nx}dx=0 $$
Work Step by Step
$$\int_{-\pi}^\pi{\sin mx \cos nx}dx $$
Using the trig identity
$$\int_{-\pi}^\pi{\sin mx \cos nx}dx =\frac{1}{2} \int_{-\pi}^\pi{ \sin (m+n)x +\sin (m-n)x}dx $$
since $ \sin (m+n)x $ and $\sin (m-n)x$ are odd functions
then their integral from $\pi$ to $-\pi$ equals zero
hence the whole integral equals zero
$$\int_{-\pi}^\pi{\sin mx \cos nx}dx=0 $$
