Evaluate: `int_0^(pi/2)xcotx dx`

Explanation

Integrating by parts, taking `cotx` as second function given integral becomes.
`I=[x log sin x]_(0)^(pi//2)-int_(0)^(pi//2)log sinx dx`
`=0-lim_(xto oo) (xlog sin x)-int_(0)^(pi//2)log sin x dx=1/2 pilog 2`
as `lim_(xto oo) x log sinx=lim_(xto oo) ((log sinx)/(1//x))`
`=lim_(xto oo)((cotx))/(-1//(x^(2)))`
`=lim_(xto oo) ((-x^(2))/(tanx))`
`=lim_(xto oo) (-x xx x/(tanx))`
`=0xx1=0`