Evaluate:$\int_{0}^{\pi}\frac{x\sin2x\sin(\frac{\pi}{2}\cos x)}{2x-\pi}dx$ Here i have a problem
$$\int_{0}^{\pi}\frac{x\sin (2x)\sin(\frac{\pi}{2}\cos x)}{2x-\pi}dx$$
I have started applying the property of definite integral but stucked after going few steps. Please help. 

Let $A$ denote the value of the integral. First note that since $\sin(2x)/(2x - \pi) = -\sin(2x - \pi)/(2x - \pi) \to -1$ as $x \to \pi/2$, the integral is proper. By a change of variable $u = 2x - \pi$, we have
$$A = \int_{-\pi}^\pi \frac{u + \pi}{2u}\cdot \left[-\sin u \sin\left(\frac{\pi}{2}\cos\frac{\pi + u}{2}\right)\right]\, \frac{du}{2},$$
which is
$$-\frac{1}{4}\int_{-\pi}^\pi \sin u\sin\left(\frac{\pi}{2}\cos \frac{\pi + u}{2}\right)\, du - \frac{\pi}{4}\int_{-\pi}^\pi \sin u \sin\left(\frac{\pi}{2}\cos \frac{\pi + u}{2}\right)\, \frac{du}{u}.$$
Since $\cos (\pi + u)/2 = -\sin u/2$, this simplifies further to 
$$\frac{1}{4}\int_{-\pi}^\pi \sin u \sin\left(\frac{\pi}{2}\sin \frac{u}{2}\right)\, du + \frac{\pi}{4}\int_{-\pi}^\pi \sin u \sin\left(\frac{\pi}{2}\sin \frac{u}{2}\right)\, \frac{du}{u}.$$
By symmetry the second integral evaluates to $0$, so we are left with
$$\frac{1}{4}\int_{-\pi}^\pi \sin u \sin\left(\frac{\pi}{2}\sin \frac{u}{2}\right)\, du = \frac{1}{2}\int_0^\pi \sin u \sin\left(\frac{\pi}{2}\sin \frac{u}{2}\right)\, du.$$
Letting $t = \sin \frac{u}{2}$ we have
$$\int_0^\pi \sin u \sin\left(\frac{\pi}{2}\sin \frac{u}{2}\right)\, du = \int_0^1 2t\sqrt{1 - t^2} \sin\left(\frac{\pi}{2}t\right)\, \frac{2\, dt}{\sqrt{1 - t^2}} = 4 \int_0^1 t\sin\left(\frac{\pi}{2} t\right)\, dt.$$
By the change of variable $r = \pi t/2$, we obtain
$$4\int_0^1 t\sin\left(\frac{\pi}{2}t\right)\, dt = \frac{16}{\pi^2}\int_0^{\pi/2} r \sin r\, dr = \frac{16}{\pi^2}(-r\cos r + \sin r)\bigg|_{r = 0}^{\pi/2} = \frac{16}{\pi^2}.$$
Thus
$$A = \frac{1}{2}\cdot \frac{16}{\pi^2}  = \frac{8}{\pi^2}.$$
