<s>[INST] Evaluate $ \int_{0}^{\pi/2} \frac{\sin(nx)}{\sin(x)}\,dx $ For every $odd$ $n \geq  1$ the answer should be $\pi/2$
For every $even$ $n \geq  2$ the possible answers are :
$A )$ $ 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots + (-1)^{n/2+1}\frac{1}{n-1} $
$B )$ $ 3(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots + (-1)^{n/2+1}\frac{1}{n-1} )$
$C )$ $ 2(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots + (-1)^{n/2+1}\frac{1}{n-1} )$
$D )$ $ \frac{1}{2}(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots + (-1)^{n+1}\frac{1}{n-1} )$
$E )$ $ 1-\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\cdots - \frac{1}{n-1} $
Does the recurrence $ (n-1)(I_{n}-I_{n-2})=2sin(n-1)x $ help?
 [/INST] first we note that $I_0=0$ and $I_1 = \frac{\pi}2$.
now, using the trig identity
$$
\sin A - \sin B = 2\sin \frac{A-B}2 \cos \frac{A+B}2
$$
we obtain 
$$
I_{n+2} = I_n + \frac2{n+1} \sin \frac{(n+1)\pi}2
$$
if $N$ is odd, therefore, we have $I_N =I_{N-2}=\dots = I_1 = \frac{\pi}2$.
for even values of $N$, say $N=2M$, we have
$$
I_{2M}= I_0+2\bigg(\frac11\sin \frac{\pi}2 + \frac13\sin \frac{3\pi}2+\dots +\frac1{2M-1}\sin \frac{(2M-1)\pi}2 \bigg) = 2\sum_{k=1}^{M} \frac{(-1)^{k-1}}{2k-1}
$$
</s>