$\displaystyle \int_{0}^{\pi}\frac{\texttt{sin}^2n\theta}{\texttt {sin}^2\theta}\texttt{d}\theta$
Let $\displaystyle I_n = \int_0^{\pi} \frac{ \sin^2{n\theta} }{ \sin^2{\theta}}~d\theta $
We have
$\displaystyle I_{n+1} - I_n = \int_0^{\pi} \frac{ \sin^2{(n+1) \theta} - \sin^2{n\theta} }{ \sin^2{\theta}}~d\theta $
Note that $\displaystyle \sin^2(A) - \sin^2(B) = \sin(A-B)\sin(A+B)$
$\displaystyle I_{n+1} - I_n = \int_0^{\pi} \frac{ \sin{\theta} \sin{ (2n+1)\theta}}{ \sin^2{\theta}}~d\theta $
Consider $\displaystyle J_n = \int_0^{\pi} \frac{ \sin{ (2n+1)\theta}}{ \sin{\theta}}~d\theta $
Again $\displaystyle J_{n} - J_{n-1} = $
$\displaystyle \int_0^{\pi} \frac{ \sin{ (2n+1)\theta} - \sin{ (2n-1)\theta} }{ \sin{\theta}}~d\theta $
$\displaystyle = 2\int_0^{\pi} \cos{2n\theta}~d\theta = 0 $
Therefore , $\displaystyle J_n = J_{n-1} = ... = J_1 = J_0 $
and $\displaystyle J_0 = \int_0^{\pi} d\theta = \pi $
so $\displaystyle I_{n+1} - I_n = \pi $ with $\displaystyle I_0 = 0 $
Finally $\displaystyle I_n = n\pi $