<s>[INST] A sine integral The integral
\begin{align}
\int_{0}^{\pi/2} \frac{ \sin(n\theta) }{ \sin(\theta) } \ d\theta
\end{align}
is claimed to not have a closed form expression. In this view find the series solution of the integral as a series involving of $n$. 
Editorial note: 
As described in the problem several series may be obtained, of which, all seem to hold validity. As a particular case, from notes that were made a long while ago, the formula
\begin{align}
\int_{0}^{\pi/2} \frac{ \sin(n\theta) }{ \sin(\theta) } \ d\theta = \sum_{r=1}^{\infty} (-1)^{r-1} \ \ln\left(\frac{2r+1}{2r-1}\right) \ \sin(r n \pi)
\end{align}
is stated, but left unproved. Can this formula be proven along with finding other series dependent upon $n$? 
 [/INST] Let $U_n(x)$ be the Chebyshev's polynomial of the second kind. We know
$$U_{n}(\cos\theta) = \frac{\sin(n+1)\theta}{\sin\theta}$$
and it has a generating function of the form
$$\sum_{n=0}^\infty U_n(x)t^n = \frac{1}{1-2xt+t^2}$$
Substitute $x$ by $\cos\theta$ in above expression, integrate $\theta$ over $[0,\frac{\pi}{2}]$ and uses a relatively easy to prove identity
$$\int_0^{\frac{\pi}{2}} \frac{d\theta}{K - \cos\theta} = \frac{2}{\sqrt{K^2-1}}\tan^{-1}\sqrt{\frac{K+1}{K-1}}\quad\text{ for }\quad K > 1$$
We have
$$\begin{align}
\sum_{n=0}^\infty t^n \int_0^{\frac{\pi}{2}} U_n(\cos\theta)d\theta
&= \frac{1}{2t}\int_0^{\frac{\pi}{2}}\frac{d\theta}{\frac{1+t^2}{2t} - \cos\theta}\\
&= \frac{1}{2t}\frac{2}{\sqrt{\left(\frac{1+t^2}{2t}\right)^2-1}}\tan^{-1}\sqrt{\frac{\frac{1+t^2}{2t}+1}{\frac{1+t^2}{2t}-1}}\\
&= \frac{2}{1-t^2}\tan^{-1}\left(\frac{1+t}{1-t}\right)\\
&= \frac{2}{1-t^2}\left(\frac{\pi}{4} + \tan^{-1} t\right)\\
&= \frac{1}{1-t^2}\left(\frac{\pi}{2} + 2\sum_{k=0}^\infty \frac{(-1)^k t^{2k+1}}{2k+1}\right)
\end{align}
$$
Compare the coefficients of both sides, we get
$$\int_0^{\frac{\pi}{2}}\frac{\sin((n+1)\theta)}{\sin\theta}d\theta
= \int_0^{\frac{\pi}{2}} U_n(\cos\theta)d\theta 
= \begin{cases} 
\frac{\pi}{2},& n = 2k\\
\displaystyle\;2\sum\limits_{j=0}^k \frac{(-1)^j}{2j+1},& n = 2k+1
\end{cases}$$
</s>