∫01x sin–1 x dx=

∫01xSin−1xdx=(Sin−1x.x22)01−∫0111−x2.x22dx                         =(π4−0)−12∫011−(1−x2)1−x2dx                          =π4−12∫01(11−x2−1−x2)dx                          =π4−12(Sin−1x)01+12×π×124                           =π8                 (∵∫0aa2−x2dx=πa24)