sorry it'd be
-x^2 cos x + \(\int\limits_{0}^{\pi/2} 2x \cos xdx\)
-x^2 cos x+2x sin x-\(\int\ limits_{0}^{pi/2} 2 sin x dx
so we get
-x^2 cos x+ 2x sin x+2 cos x
2x sin x- cos x(x^2 -2)
now insert the limits
[ 2* pi/2 * 1-0]-[0-1*(0-2)]
pi -2
sorry i thought it's x^2 sin x
let me put the correct one
-x^2 cos 2x /2+ ∫ 0 π/2 xcos2xdx
-x^2 cos 2x /2 + x sin 2x/2 -∫ 0 π/2 sin2x/2dx
-x^2 cos 2x/2 + x sin 2x /2 +cos 2x /4
now inserting the limits
[pi^2/8+0-1/4]-[1/4]
pi^2/8-1/2
π^2 /8-1/2