Evaluate $\int_{0}^{\frac{\pi}{2}} \frac{\sin^2 nx}{\sin x} \text{d}x$ 
Evaluate $$ \int_{0}^{\frac{\pi}{2}} \frac{\sin^2 nx}{\sin x} \text{d}x$$
  where $n\in\mathbb{N}$  

I was able to do this using parametization (differentiation under integration) but I'm not allowed to use that in my school since it isn't in the syllabus :(
I should only use high school level mathematics. However, I can't see a way to do this with an elementary approach. 
 A: Recall the identity:
$$\sin^2(A) - \sin^2(B) = \sin(A+B)\sin(A-B)$$
Let $I_n = \displaystyle \int_0^{\pi/2} \dfrac{\sin^2(nx)}{\sin(x)}dx$, we obtain
\begin{align}
I_{n+1} - I_n & = \int_0^{\pi/2} \dfrac{\sin^2((n+1)x)}{\sin(x)}dx - \int_0^{\pi/2} \dfrac{\sin^2(nx)}{\sin(x)}dx\\
& = \int_0^{\pi/2}\dfrac{\sin^2((n+1)x)-\sin^2(nx)}{\sin(x)} dx\\
& = \int_0^{\pi/2}\dfrac{\sin((2n+1)x)\sin(x)}{\sin(x)} dx\\
& = \int_0^{\pi/2} \sin((2n+1)x)dx\\
& = -\left.\dfrac{\cos((2n+1)x)}{2n+1} \right \vert_0^{\pi/2}\\
& = \dfrac1{2n+1}
\end{align}
We have $I_0 = 0$ and hence
$$I_{n} = \dfrac11 + \dfrac13 + \dfrac15 + \cdots + \dfrac1{2n-1}$$
A: Hint:
Start by using the power reduction formula to get rid of the sine-squared term:
$$\begin{align}
\mathcal{I}_{n}
&=\int_{0}^{\frac{\pi}{2}}\frac{\sin^2{nx}}{\sin{x}}\,\mathrm{d}x\\
&=\int_{0}^{\frac{\pi}{2}}\frac{1-\cos{2nx}}{2\sin{x}}\,\mathrm{d}x.\\
\end{align}$$
Then, writing the $\cos{2nx}$ as a binomial series, we can rewrite the integrand as a polynomial of sines and cosines:
$$\cos{(2nx)}=\sum_{k=0}^{n}(-1)^{k}\binom{2n}{2k}\cos^{2(n-k)}{(x)}\sin^{2k}{(x)}$$
$$\implies 1-\cos{(2nx)}=1-\sum_{k=0}^{n}(-1)^{k}\binom{2n}{2k}\cos^{2(n-k)}{(x)}\sin^{2k}{(x)}$$
$$\implies 1-\cos{(2nx)}=1-\cos^2{(x)}-\sum_{k=1}^{n}(-1)^{k}\binom{2n}{2k}\cos^{2(n-k)}{(x)}\sin^{2k}{(x)}$$
$$\implies 1-\cos{(2nx)}=\sin^2{(x)}-\sum_{k=1}^{n}(-1)^{k}\binom{2n}{2k}\cos^{2(n-k)}{(x)}\sin^{2k}{(x)}$$
$$\implies \frac{1-\cos{(2nx)}}{\sin{(x)}}=\sin{(x)}-\sum_{k=1}^{n}(-1)^{k}\binom{2n}{2k}\cos^{2(n-k)}{(x)}\sin^{2k-1}{(x)}.$$
Finally, integrate term-by-term.
