Integral of quotients of $\sin$ function

Consider, from where you left off, $$I_n=2\int\frac{\sin n\theta}{\sin\theta} d\theta, $$ where $n$ is odd. Noting that $$\frac {\sin n\theta-\sin(n-2)\theta}{\sin\theta}=2\cos(n-1)\theta, $$

We have $$I_n=I_{n-2}+4\int^{\frac{\pi}{2}}_0\cos(n-1)\theta d\theta$$ $$=I_{n-2}+0$$ since $n-1$ is even.

Hence $$I_n=I_{n-2}=...=I_1$$ and the value of is last integral is $\pi$