# Evaluation of $\int_0^a \frac {\cos[(\alpha-\beta)x]}{\sin\beta x} dx$.

Can anyone explain me how to evaluate the following integral in detail.

$$\int_0^a \frac {\cos[(\alpha-\beta)x]}{\sin\beta x} dx$$

where $a\in[0,2\pi]$, $\alpha, \beta\in\mathbb R$ and $\alpha\neq \beta$.

Maybe it can't be done analytically, so what are alternative methods to do this integration. Or, at least, I did not succeed.

Thank you for your attention.

• As far as i remember such integrals can only be expressed in terms of hypergeometric functions. Writing the denominator as an geometric series, integrating term by term.... . – tired Apr 10 '15 at 16:59

We have $$\cos[(\alpha-\beta)x] = \cos(\alpha x)\cos(\beta x)+\sin(\beta x)\sin(\alpha x)$$ so \begin{align}\int_0^a \frac {\cos[(\alpha-\beta)x]}{\sin(\beta x)} \text{d}x = \int_0^a \frac{\cos(\alpha x)\cos(\beta x)+\sin(\beta x)\sin(\alpha x)}{\sin(\beta x)} \text{d}x \\ =\int_0^a \cot(\beta x)\cos(\alpha x) \text{d}x + \int_0^a \sin(\alpha x)\text{d}x\end{align} Can you integrate $\cot(\beta x)\cos(\alpha x)$?