<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] In the case that $n=2k$ is an even positive integer we have the identity $$\frac{\sin 2kx}{\sin x} = 2 \cos(2k-1)x + 2 \cos(2k-3)x + \ldots + 2 \cos x.$$
Therefore,
$$ \begin{align} \int_{0}^{\pi /2} \frac{\sin 2kx}{\sin x} \ dx &= 2 \int_{0}^{\pi/2} \Big(\cos x + \cos 3x + \ldots + \cos(2k-1) \ x \Big) \ dx \\  &= 2 \left(\sin x + \frac{1}{3} \sin 3x + \ldots + \frac{1}{2k-1} \sin(2k-1) x \right)\Bigg|^{\pi/2}_{0} \\ &= 2 \left(1- \frac{1}{3} + \ldots + \frac{(-1)^{k+1}}{2k-1} \right).  \end{align}$$
EDIT:
To prove that $$ \frac{\sin 2kx}{\sin x} = 2 \cos(2k-1)x + 2 \cos(2k-3)x + \ldots + 2 \cos x,$$ notice that
$$\begin{align} \frac{\sin kx}{\sin x} &= \frac{e^{ikx}-e^{-ikx}}{e^{ix}-e^{-ix}} \frac{e^{-ix}}{e^{-ix}} \\ &= \frac{e^{i(k-1)x} - e^{i(-k-1)x}}{1-e^{-2ix}}  \\ &= e^{i(k-1)x} \frac{1-e^{-2ikx}}{1-e^{-2ix}} \\ &= e^{i(k-1)x} \left(1 + e^{-2ix} + e^{-4ix} + \ldots + e^{-2i(k-1)x} \right) \\ &= e^{i(k-1)x} + e^{i(k-3)x} + \ldots + e^{i(-k+1)x}. \end{align}$$
Therefore,
$$ \frac{\sin 2kx}{\sin x} = e^{i(2k-1)x} + e^{i(2k-3)x} + \ldots + e^{ix} + e^{-ix} + \ldots + e^{i(-2k+1)x}$$
$$ = 2 \cos(2k-1)x + 2 \cos(2k-3)x + \ldots + 2 \cos x.$$
</s>