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Evaluate $$ \int_{0}^{\frac{\pi}{2}} \frac{\sin^2 nx}{\sin^2 x} \text{d}x$$

where $n\in\mathbb{N}$

This one is another intriguing question from my worksheet. I'm only allowed to use elementary methods and high school math. However, I can't see a way to do this without derivative under the integral (which is not allowed).
Please Help!
Thanks.

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Let, $$\text{I(n)}=\displaystyle \int_{0}^{\frac{\pi}{2}} \frac{\sin^2 nx}{\sin^2 x} \text{d}x$$

and

$\text{J}= \text{I(n) - I(n-1)}=\displaystyle \int_{0}^{\frac{\pi}{2}} \frac{\sin^2nx-\sin^2(n-1)x}{\sin^2 x} \text{d}x$

$=\displaystyle \int_{0}^{\frac{\pi}{2}} \frac{\sin(2n-1)x\times\sin x}{\sin^2 x} \text{d}x$ [Using $(\sin^2a - \sin^2b) = \sin(a+b)\times \sin(a-b)$]

$=\displaystyle \int_{0}^{\frac{\pi}{2}} \frac{\sin(2n-1)x}{\sin x} \text{d}x$

Now, consider

$$ \text{S}=\cos (2x) + \cos(4x) + \cos(6x) +......+ \cos(2n-2)x = \frac {\sin(n-1)x \times \cos nx}{\sin x}$$

$=\dfrac {2\sin(n-1)x \times \cos nx}{2\sin x}$

$=\dfrac {\sin(2n-1)x - \sin x}{2\sin x}$

$\implies \text{J}=\displaystyle \int_{0}^{\frac{\pi}{2}}(2\text{S}+1) \: \text{d}x$

$=\dfrac{\pi}{2}$

$\implies \text{I(1), I(2), I(3) ..... I(n)}$ form an Arithmetic Progression

Since $\text{I(1)}=\dfrac{\pi}{2}$ , $\text{I(n)}=\boxed{\dfrac{n\pi}{2}}$

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Recall the identity: $$\sin^2(A) - \sin^2(B) = \sin(A+B)\sin(A-B)$$ Let $I_n = \displaystyle \int_0^{\pi/2} \dfrac{\sin^2(nx)}{\sin^2(x)}dx$, we obtain \begin{align} I_{n+1} - I_n & = \int_0^{\pi/2} \dfrac{\sin^2((n+1)x)}{\sin^2(x)}dx - \int_0^{\pi/2} \dfrac{\sin^2(nx)}{\sin^2(x)}dx\\ & = \int_0^{\pi/2}\dfrac{\sin^2((n+1)x)-\sin^2(nx)}{\sin^2(x)} dx\\ & = \int_0^{\pi/2}\dfrac{\sin((2n+1)x)\sin(x)}{\sin^2(x)} dx\\ & = \int_0^{\pi/2}\dfrac{\sin((2n+1)x)}{\sin(x)}dx \end{align} Let $J_n = \displaystyle \int_0^{\pi/2} \dfrac{\sin((2n+1)x)}{\sin(x)}dx$, we obtain \begin{align} J_{n+1} - J_n & = \int_0^{\pi/2} \dfrac{\sin((2n+1)x)}{\sin(x)}dx - \int_0^{\pi/2} \dfrac{\sin((2n-1)x)}{\sin(x)}dx\\ & = \int_0^{\pi/2}\dfrac{\sin((2n+1)x)-\sin((2n-1)x)}{\sin(x)} dx\\ & = \int_0^{\pi/2}\dfrac{2\sin(x) \cos(2nx)}{\sin(x)} dx\\ & = 2\int_0^{\pi/2}\cos(2nx)dx = \left.2 \dfrac{\sin(2nx)}{2n} \right \vert_0^{\pi/2} = 0 \end{align} Hence, $J_n = J_0 = \dfrac{\pi}2$. Hence, we have $I_{n+1} - I_n = \dfrac{\pi}2$. Since $I_0 = 0$, we obtain $$I_n = \dfrac{n \pi}2$$

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Another way to evaluate this integral is to use residue. Note $$ \int_0^{\frac{\pi}{2}}\frac{\sin^2nx}{\sin^2x}dx=\frac{1}{4}\int_0^{2\pi}\frac{\sin^2nx}{\sin^2x}dx. $$ Letting $z=e^{i\theta}$, we have \begin{eqnarray} \int_0^{\frac{\pi}{2}}\frac{\sin^2nx}{\sin^2x}dx&=&\frac{1}{4}\int_{|z|=1}\frac{(z^n-z^{-n})^2}{(z-z^{-1})^2}\frac{1}{iz}dz\\ &=&\frac{1}{4i}\int_{|z|=1}\frac{(1-z^{2n})^2}{(1-z^2)^2}\frac{1}{z^{2n-1}}dz\\ \end{eqnarray} But \begin{eqnarray} \frac{(1-z^{2n})^2}{(1-z^2)^2}\frac{1}{z^{2n-1}}&=&\frac{1}{z^{2n-1}}\sum_{k=1}^\infty kz^{2(k-1)}(1-2z^{2n}+z^{4n})\\ &=&\sum_{k=1}^\infty kz^{2(k-n)-1}(1-2z^{2n}+z^{4n}) \end{eqnarray} whose coefficient of $\frac{1}{z}$ is $n$. Thus \begin{eqnarray} \int_0^{\frac{\pi}{2}}\frac{\sin^2nx}{\sin^2x}dx&=&\frac{1}{4}\int_{|z|=1}\frac{(z^n-z^{-n})^2}{(z-z^{-1})^2}\frac{1}{iz}dz\\ &=&\frac{1}{4i}\int_{|z|=1}\frac{(1-z^{2n})^2}{(1-z^2)^2}\frac{1}{z^{2n-1}}dz\\ &=&\frac{1}{4i}\cdot 2\pi i\cdot n\\ &=&\frac{n\pi}{2}. \end{eqnarray}

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