# How to evaluate the integral $\int_0^{2\pi} \frac{\sin{nx}\cos{nx}}{\sin{x}}dx$?

Would someone give me a hint or a solution ?

How to evaluate the integral $\displaystyle \int_0^{2\pi} \frac{\sin{nx}\cos{nx}}{\sin{x}}\mathrm dx$?

Thanks a lot.

• – FreezingFire Apr 15 '16 at 5:16

Such integral is just zero, because, given that: $$f(x) = \frac{\sin(2n x)}{\sin(x)},$$ we have: $$f(x)+f(x+\pi) = 0,$$ so: $$\int_{0}^{2\pi} f(x)\,dx = 0.$$
\begin{align*} \sin nx \times \cos nx &= \frac{1}{2} \, \sin 2nx \\ I_n &= \int_{0}^{2\pi} \frac{\sin 2nx}{2\sin x} \, dx \end{align*}
$$I_{n}=\frac{-\cos 2nx}{2n \sin x} - \int_{0}^{2\pi} \frac{-\cos 2nx}{4n} \times \frac{-\cos x}{\sin^2 x} \, dx$$
(for the domain $0$ to $2\pi$)