If $$I_n=\int_{-\pi}^{\pi}\frac{\sin^2\frac{nx}{2}}{\sin^2\frac{x}{2}}\,dx$$ for $n\in \Bbb Z$.
My attempt: I first plugged $\frac{x}{2}$ as t and then converted the whole integral according to the assumed substitution. Then as the function is even, so I changed the limits of the integral from (-$\frac{\pi}{2}$ and $\frac{\pi}{2}$) to (0 and $\frac{\pi}{2}$), so finally got a simpler form of this integral $$4\int_{0}^{\frac{\pi}{2}}\frac{\sin ^2({nt})}{\sin^2{t}}\,dt.$$
Then I changed the variable to x ($4\int_{0}^{\frac{\pi}{2}}\frac{\sin^2 ({nx})}{\sin^2{x}}\,dx$).
But after this stage, I am not able to proceed further. Neither I am able to solve this with integration by parts nor any further appropriate substitution.
The answer key states that $I_1, I_2,I_3….I_n $ forms an A.P.
I am even getting this answer by plugging the respective values of n directly into integral.
- But is it correct to directly plug in the values of n and solve?
- Is there any way by which we can simplify the given integral to a final form?
- Or can we use reduction formula to prove it is an A.P
Best Answer
$$I_n=2\int_{0}^{\pi}\frac{\sin^2 (nx/2)}{\sin^2 (x/2)} dx = \int_{0}^{\pi}\frac{1-\cos nx}{\sin^2(x/2)} dx$$ Take $$I_{n+1}-I_n=\int_{0}^{\pi} \frac{\cos(n+1)x-\cos nx}{\sin ^2(x/2)}=\int_{0}^{\pi}\frac{2\sin(n+1/2)x}{\sin(x/2)}dx=J_n$$ Next $$J_{n+1}-J_n=2\int_{0}^{\pi} \cos(n+1)x~dx=0.$$ Hence, $J_n$ is independent of of $n$, we have $J_n=J_0=2\pi$. This means that $I_n$ form an AP with common difference of $2\pi$, where $I_0=0$. Finally we get $I_n=2n\pi$