Evaluate ∫0π/2 xcot⁡xdx

Integrating by parts, taking cot⁡x as second function,  given integral becomes I=[xlog⁡sin⁡x]0π/2−∫0π/2 log⁡sin⁡xdx=0−limx→0 (xlog⁡sin⁡x)−∫0π/2 log⁡sin⁡xdx=12πlog⁡2 as limx→0 xlog⁡sin⁡x=limx→0 log⁡sin⁡x1/x=limx→0 cot⁡x−1/x2 (Using L'Hopital Rule) =limx→0 −x2tan⁡x=limx→0 −x×xtan⁡x=0×1=0