How to evaluate this integral : $$\Omega =\int^{\frac{\pi}{2}}_{0} \cos^{a-1}(x)\frac{\sin(ax)}{\sin(x)} dx $$ where :$$a>0$$
If $a=1$: $$\Omega = \frac{\pi}{2}$$
If $a=2$ $$\Omega = 2\int^{\frac{\pi}{2}}_{0}\cos^2(x) dx = \frac{\pi}{2}$$
by using mathematica
If $a=1/3$ $$\Omega =\frac{\pi}{2}$$
If $a=3/2$ $$\Omega =\frac{\pi}{2}$$
I think that $$\Omega =\frac{\pi}{2}$$
But I can't prove it. Can I find some help on how to prove it?
\begin{align} &\int^{\frac{\pi}{2}}_{0} \cos^{a-1}x\ \frac{\sin ax}{\sin x} dx \\ =& \ \Im \int^{\frac{\pi}{2}}_{0}\frac{ \cos^{a-1}x}{\sin x}e^{i a x}dx \\ =& \ \Im \ \frac {1}{2^{a-1}}\int^{\frac{\pi}{2}}_{0}\frac{ (e^{i x} + e^{-ix})^{a-1} }{\sin x}e^{i a x}dx \\ = & \ \Im \ \frac {1}{2^{a-1}}\int^{\frac{\pi}{2}}_{0} \sum_{k=0}^\infty \frac{ \binom {a-1}k e^{i (2k+1)x}}{\sin x}dx \\ = & \ \frac {1}{2^{a-1}}\sum_{k=0}^\infty { \binom {a-1}k \int^{\frac{\pi}{2}}_{0}\frac{\sin[(2k+1)x]}{\sin x}dx }\\ = & \ \frac {1}{2^{a-1}}\sum_{k=0}^\infty \binom {a-1}k \frac\pi2 =\frac\pi2 \end{align}
