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