Finding value of $$\int^{\pi}_{0}\frac{\sin^2(10x)}{\sin^2 (x)}dx$$
Try: Using $\displaystyle \sin(10x)=\frac{e^{i(10x)}-e^{-i(10x)}}{2i}$ and $\displaystyle \sin (x)=\frac{e^{i(x)}-e^{-i(x)}}{2i}$
So $$\frac{\sin(10x)}{\sin x}=e^{-i(9x)}\cdot \frac{e^{i(20x)}-1}{e^{i(2x)}-1}=e^{-i(9x)}\sum^{9}_{n=1}e^{i(2nx)}$$
So $$\frac{\sin^2(10x)}{\sin^2(x)}=e^{-i(18x)}\bigg[\sum^{9}_{n=1}e^{i(2nx)}\bigg]^2$$
So $$\int^{\pi}_{0}\frac{\sin^2(10x)}{\sin^2(x)}=\frac{1}{2}\int^{\pi}_{-\pi}\frac{\sin^2(10x)}{\sin^2(x)}dx= \frac{1}{2}\int^{\pi}_{-\pi}e^{-i(18x)}\bigg[\sum^{9}_{n=1}e^{i(2nx)}\bigg]^2$$
Could some help me to proceed further, Thanks