∫01sin−1(2x1−x2)dx=

∫01sin−1(2x1−x2)dx,x=sinθ =∫0π2sin−1(sin2θ)cosθdθ =∫0π42θcosθdθ+∫π4π2(π−2θ)cosθdθ =2[θsinθ+cosθ]0π4+[(π−2θ)sinθ−2cosθ]π4π2 =π22+22−2−π22+22=2(2−1)