Problem source

Given that ${\displaystyle I_{n}=\int_{0}^{\pi}\frac{x\sin^{2}(nx)}{\sin^{2}x}\,\mathrm{d}x,}$ where $n$ is a positive integer, show that $I_{n}-I_{n-1}=J_{n},$ where 
\[
J_{n}=\int_{0}^{\pi}\frac{x\sin(2n-1)x}{\sin x}\,\mathrm{d}x.
\]
Obtain also a reduction formula for $J_{n}.$
The curve $C$ is given by the cartesian equation 
\[
y=\dfrac{x\sin^{2}(nx)}{\sin^{2}x},
\]
 where $n$ is a positive integer and $0\leqslant x\leqslant\pi.$
Show that the area under the curve $C$ is $\frac{1}{2}n\pi^{2}.$

Solution source

\begin{align*}
I_n - I_{n-1} &= \int_0^{\pi} \frac{x \sin^2(nx)}{\sin ^2 x} \d x-\int_0^{\pi} \frac{x \sin^2((n-1)x)}{\sin ^2 x} \d x \\
&= \int_0^{\pi} \frac{x}{\sin^2 x} \left ( \sin^2 (nx) - \sin^2((n-1)x) \right) \d x \\
&= \int_0^{\pi} \frac{x}{\sin^2 x}\frac12 \left ( \cos (2(n-1)x) - \cos(2nx) \right) \d x \\
&= \int_0^{\pi} \frac{x}{\sin^2 x}\frac12 2 \sin ((2n-1)x )\sin x \d x \\
&= \int_0^{\pi} \frac{x\sin ((2n-1)x )}{\sin x}d x \\
&= J_n \\
\\
J_{n+1} - J_{n} &= \int_0^{\pi} \frac{x \left (\sin ((2n+1)x )-\sin ((2n-1)x )\right)}{\sin x} \d x \\
&=  \int_0^{\pi} \frac{x \left ( 2 \cos (\frac{4n x}{2}) \sin \frac{2x}{2} \right)}{\sin x} \d x \\
&= \int_0^{\pi}2x \cos (2n x) \d x \\
&= \left [ \frac{x}{2n} \sin (2n x) \right]_0^{\pi} - \int_0^{\pi} \frac{1}{2n} \sin (2n x) \d x \\
&= \left [ \frac{1}{4n^2} \cos (2n x)\right]_0^{\pi} \\
&= 0 \\
\\
J_1 &= \int_0^\pi x \d x \\
&= \frac{\pi^2}{2} \\
\Rightarrow J_n &= \frac{\pi^2}{2} \\
\end{align*}

And so $I_n = I_1 + (n-1) \frac{\pi^2}{2}$ and $I_1 = \frac{\pi^2}{2}$ so $I_n = \frac12 n \pi^2$.

But $I_n$ is exactly the area under the curve described.