1. ## Evalute this DI

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

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

3. Originally Posted by Danny
What's $\displaystyle n$? 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 $\displaystyle I_n = \int_0^{\pi} \frac{ \sin^2{n\theta} }{ \sin^2{\theta}}~d\theta$

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 $\displaystyle I_n = n\pi$