1. ## Evalute this DI

$\int_{0}^{\pi}\frac{\texttt{sin}^2n\theta}{\texttt {sin}^2\theta}\texttt{d}\theta$

2. Originally Posted by banku12
$\int_{0}^{\pi}\frac{\texttt{sin}^2n\theta}{\texttt {sin}^2\theta}\texttt{d}\theta$
What's $n$? Any real number or an integer?

3. Originally Posted by Danny
What's $n$? Any real number or an integer?
ya forgot to mention n is an integer

4. a hint will also help...if not full soln

5. Let $I_n = \int_0^{\pi} \frac{ \sin^2{n\theta} }{ \sin^2{\theta}}~d\theta$

We have

$I_{n+1} - I_n = \int_0^{\pi} \frac{ \sin^2{(n+1) \theta} - \sin^2{n\theta} }{ \sin^2{\theta}}~d\theta$

Note that $\sin^2(A) - \sin^2(B) = \sin(A-B)\sin(A+B)$

$I_{n+1} - I_n = \int_0^{\pi} \frac{ \sin{\theta} \sin{ (2n+1)\theta}}{ \sin^2{\theta}}~d\theta$

Consider $J_n = \int_0^{\pi} \frac{ \sin{ (2n+1)\theta}}{ \sin{\theta}}~d\theta$

Again $J_{n} - J_{n-1} =$

$\int_0^{\pi} \frac{ \sin{ (2n+1)\theta} - \sin{ (2n-1)\theta} }{ \sin{\theta}}~d\theta$

$= 2\int_0^{\pi} \cos{2n\theta}~d\theta = 0$

Therefore , $J_n = J_{n-1} = ... = J_1 = J_0$

and $J_0 = \int_0^{\pi} d\theta = \pi$

so $I_{n+1} - I_n = \pi$ with $I_0 = 0$

Finally $I_n = n\pi$