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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

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    See https://math.stackexchange.com/questions/2008044/how-do-i-prove-int-0-pi-frac-sin-nx2-sin-x2dx-n-pi – lab bhattacharjee Mar 01 '18 at 12:51
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    See also: https://math.stackexchange.com/questions/263705/compute-int-0-pi-2-frac-sin-2013x-sin-x-dx-space, https://math.stackexchange.com/questions/714706/solving-the-integral-int-0-pi-2-frac-sin2n1t-sin-t-mathrmdt – lab bhattacharjee Mar 01 '18 at 13:22

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