Evaluate: ∫π0xtanxsecx⋅cosecxdx

I=∫π0xtanx(secx.cosecx)dx=∫π0x×sinxcosx1cosx.1sinxdx⇒∫π0xsin2xdx=∫π0x×1−cos2x2dx⇒∫π0(x2−xcos2x2)dx=x24−(xsin2x4+18cos2x)⇒x24−xsin2x4󔼚cos2xSo substituting the limits, we get[π24󔼚]−[󔼚]=π24