Diluting The Integrand

Relevant wiki: Integration Tricks

We have, $I = \displaystyle \int_0^\pi \frac{x\sin 2x \sin\left(\dfrac \pi 2 \cos x\right)}{2x-\pi}~dx$

We are here using the property below and adding, $\int_0^a f(x) ~dx = \int_0^a f(a-x)~dx$

So, let $f(x) =\dfrac{x\sin 2x \sin\left(\dfrac \pi 2 \cos x\right)}{2x-\pi}\\ f(\pi-x) =\dfrac{(x-\pi)\sin 2x \sin\left(\dfrac \pi 2 \cos x\right)}{2x-\pi}$

Adding both, we get $f(x) + f(\pi-x) = \sin 2x \sin\left(\dfrac \pi 2 \cos x\right)$

Thus, $\begin{aligned}2I &=\int_0^\pi f(x) + f(\pi-x) ~dx\\ &= \int_0^\pi\sin 2x \sin\left(\dfrac \pi 2 \cos x\right)~dx \end{aligned}$

which is easy to integrate by taking $\left(x' = \dfrac \pi2 \cos x\right)$ and further applying parts. Then you get $I = \dfrac{8}{\pi^2}$

So, $A = 8$ , $B = 1$ and $C = 2$ and $A+B+C = 11$ .

It still seems confusing!

$\begin{aligned} I & = \int_0^\pi \frac {x\sin 2x \sin \left(\frac \pi 2 \cos x\right)}{2x-\pi} dx \quad \quad \small \color{#3D99F6}{\text{By: }\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx} \\ & = \frac 12 \int_0^\pi \frac {x\sin 2x \sin \left(\frac \pi 2 \cos x\right)}{2x-\pi} + \frac {(\pi-x) \sin (2(\pi-x)) \sin \left(\frac \pi 2 \cos (\pi-x) \right)}{2(\pi-x)-\pi} dx \\ & = \frac 12 \int_0^\pi \frac {(2x-\pi)\sin 2x \sin \left(\frac \pi 2 \cos x\right)}{2x-\pi} dx \\ & = \int_0^\pi \cos x \color{#3D99F6}{\sin x \sin \left(\frac \pi 2 \cos x\right)} dx \quad \quad \small \color{#3D99F6}{\text{Note: }\frac d{dx} \cos \left(\frac \pi 2 \cos x\right) = \frac \pi 2 \sin x \sin \left(\frac \pi 2 \cos x\right)} \\ & = \cos x \cdot \color{#3D99F6}{\frac 2 \pi \cos \left(\frac \pi 2 \cos x\right)}\bigg|_0^\pi + \frac 2\pi \int_0^\pi \color{#D61F06}{\sin x \cos \left(\frac \pi 2 \cos x\right)} dx \\ & = 0 + \frac 2\pi \cdot \color{#D61F06}{\frac 2 \pi \sin \left(\frac \pi 2 \cos x\right)}\bigg|^0_\pi \quad \quad \small \color{#D61F06}{\text{Note: }\frac d{dx} \sin \left(\frac \pi 2 \cos x\right) = \frac \pi 2 \sin x \cos \left(\frac \pi 2 \cos x\right)} \\ & = \frac 8{\pi^2} \end{aligned}$

$\implies A+B+C = 8+1+2 = \boxed{11}$