Evaluate: `int_0^pi(xsin2xsin(pi/2cosx))/(2x-pi)dx`

Answer

`(8)/(pi^(2))`

Explanation

Let `I=int_(0)^(pi)(x sin(2x)*sin((pi)/(2)cosx))/((2x-pi))dx` . . . (i )
Then `I=int_(0)^(pi)((pi-x)*sin2(pi-x)* sin[(pi)/(2)cos(pi-x)])/(2(pi-x)-pi)dx` . . . (ii)
`rArr I= int_(0)^(pi)((pi-x)*sin2x*sin((pi)/(2)cosx))/(pi-2x)dx`
`rArr I= int_(0)^(pi)((pi-x)sin2xsin((pi)/(2)cosx))/((2x - pi))dx` . . . (iii)
On adding Eqs . (i) and (ii) , we get
`2I=int_(0)^(pi)sin 2x*sin ((pi)/(2)cosx)dx`
`rArr2I=int_(0)^(pi)sin 2xsin ((pi)/(2)cosx)dx`
`rArrI=int_(0)^(pi)sin xcosx*sin ((pi)/(2)cosx)dx`
`["put" (pi)/(2)cos x = t rArr - (pi)/(2)sin x dx = dt rArr sin x dx =- (2)/(pi)dt]`
`:. I=- (2)/(pi) int_(pi//2)^(-pi//2)(2t)/(pi)*sin t dt`
` = (4)/(pi^(2))int_(-pi//2)^(pi//2)t sin t dt`
`rArr I=(4)/(pi^(2))[-t cos t +sin t ] _(-pi//2)^(pi//2)=(4)/(pi^(2))xx2 = (8)/(pi^(2))`