what is the integration of  xcotx

 to integrate 

xcotx u have to use integration by parts

let

u=x   => du=dx

cot(x)dx=dv => after integral 

v = ln(sinx)

now integral of 

 int xcot(x) = int u dv

              =  uv - int v du

              = x ln(sin(x)) - int ln(sin(x)) dx

now ln(sin(x)) cann't be integrate without limit if u have limie (0-90) u can procced like that 

I=∫ log(sin x).dx .............eq 1
I=∫logsin(π/2-x)dx
I=∫logcosx dx.......................eq2
add1&2
2I=∫logsinx+logcosx dx
=∫logsinx*cosx dx
=∫log(sin2x)/2 dx
=∫[logsin2x-log2] dx
=∫logsin2xdx - ∫log2dx .................eq 3
in the first integral put 2x=t
d.w.t.x
dx=dt/2
where x=0,t=0 where x=π/2 ,t=π
∫logsint dt with limit t= o & t=π
now
∫logsint dt =2 ∫log sint (dt /2 ) with limit t=0 t=π/2
we can write above eq in "x" also
=2 ∫log sinx (dx/2) 
put it in eq 3
2I=2 ∫log sinx (dx/2) - ∫log2dx 
but ∫log sinx dx =I
2I= I- ∫log2dx 
I= - ∫log2dx with limt x=0& x=π/2 
I= -π/2 log2