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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.

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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: $$ \int_{0}^{2\pi} f(x)\,dx = 0.$$

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$$\begin{align*} \sin nx \times \cos nx &= \frac{1}{2} \, \sin 2nx \\ I_n &= \int_{0}^{2\pi} \frac{\sin 2nx}{2\sin x} \, dx \end{align*}$$

Now try to solve the recursive formula, integration by parts:

$$I_{n}=\frac{-\cos 2nx}{2n \sin x} - \int_{0}^{2\pi} \frac{-\cos 2nx}{4n} \times \frac{-\cos x}{\sin^2 x} \, dx $$

(for the domain $0$ to $2\pi$)

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  • $\begingroup$ thanks for @darya khosrotash ,you can go on ? $\endgroup$ – Jacky Zhang Jan 25 '15 at 12:23

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