this q integration is too lenthy , here is the complete solution ...

I = xlog(sinx)dx     .........1                   lim 0 t0 pi

I =(pi-x)logsin(pi-x)dx                    lim 0 to pi                      (using property of integration)

I = -xlog(sinx)dx  + pilog(sinx)dx ................2              lim 0 to pi

adding 1 & 2

2I = pilogsinxdx or

2I/pi = logsinxdx  .....................3           lim 0 to pi    

 2Ipi = logsinxdx + logsinpi-x  dx      lim 0 to pi/2                (by property)

 2I/PI = 2logsinx dx ................4      lim 0 to pi/2

 2I/pi = 2logcosx dx ................5   lim 0 to pi/2           (by property)

 adding 4 & 5

 4I/pi =2log(sin2x)/2 dx       lim 0 to pi/2

  2I/pi = logsin2xdx - log2 dx        from 0 to pi/2    

  2I/pi =   I₁ - log2dx                      lim 0 to pi

from this eq 6 integrating logsin2x seperately

 I₁ = logsin2xdx             lim 0 to pi/2

put 2x =t

 I₁ = (logsintdt)/2  lim 0 to pi

now interchanging variable t with x

 I₁ = logsinxdx/2            lim 0 to pi ........................7

from eq 3 & 7

I₁ = I/pi .........8

putting eq 8 in 6

then

        2I/pi = I/Pi  - log2dx    lim 0 to pi

       I/pi = -(log2)x      lim 0 to pi

          I/pi   =-pilog2

hi. i'm not writting the integral sign and the limits manage with this

 xlog(sinx)=I

replace x by (0+∏)-x

I=[(0+∏)-x]log sin(∏-x)=(∏-x)log(sinx)

I=∏log(sinx)-xlog(sinx)

I=∏logsinx -I

2I=∏logsinx 

(2/∏)I=∫logsinx=F

then using another property of definte integration

i.e

_o ∫^∏ f(x)=_o∫^(∏/2) f(x) +_o∫^(∏/2) f(2a-x)

F= ₀∫?^(∏/2)logsinx+ ₀∫^(∏/2)logsin(∏-x)=2logsinx

F/2=-(∏/2)log2     (the above one is a standard integral)

F=-∏log2=(2/∏)I

I=-(∏²/2)log2

final answer