Problem source

The integral $I_n$ is defined by
$$I_n=\int_0^\pi(\pi/2-x)\sin(nx+x/2)\,{\rm cosec}\,(x/2)\,\d x,$$
where $n$ is a positive integer.
Evaluate $I_n-I_{n-1}$,
and hence evaluate  $I_n$ leaving your
answer in the form of a sum.

Solution source

\begin{align*}
&& I_n - I_{n-1} &= \int_0^\pi \left ( \frac{\pi}{2} - x \right) \left ( \sin\left(nx + \frac{x}{2}\right) - \sin \left ((n-1)x + \frac{x}{2} \right)\right) \cosec \frac{x}{2} \d x \\
&&&= \int_0^\pi  \left ( \frac{\pi}{2} - x \right) \left (2 \sin \left ( \frac{nx + \frac{x}{2} - (n-1)x - \frac{x}{2} }{2}\right)\cos nx \right) \cosec \frac{x}{2} \d x \\
&&&= \int_0^\pi \left ( \frac{\pi}{2} - x \right) \left (2 \sin \left ( \frac{x}{2}\right)\cos nx \right) \cosec \frac{x}{2} \d x \\
&&&=2 \int_0^\pi \left ( \frac{\pi}{2} - x \right) \cos nx \d x \\
&&&=\pi \left [ \frac{\sin nx}{n}\right]_0^{\pi} - 2\int_0^\pi x \cos n x \d x \\
&&&= 0 - 2\left[ \frac{x \sin nx}{n} \right]_0^{\pi} + 2\int_0^\pi \frac{\sin nx}{n} \d x \\
&&&= 2\left[ -\frac{\cos nx}{n^2} \right]_0^{\pi} \\
&&&=2 \frac{1-(-1)^{n}}{n^2} \\
\\
&& I_0 &= \int_0^\pi (\pi/2 - x) \d x =0 \\
\Rightarrow && I_{2k+2} = I_{2k+1} &= 4 \left (\frac{1}{1^2} + \frac{1}{3^2} + \cdots + \frac{1}{(2k+1)^2} \right)
\end{align*}