∫01 sin−1(2x1+x2)dx=
π2
π2+ln 2
π2−ln 2
ln 2
∫01Sin−1(2x1+x2)dx=∫012Tan−1x dx
=2[Tan−1x.x]01−2∫0111+x2x dx
=2(π4−0)−[log(1+x2)]01
=π2−log2