∫01 sin2tan−11+x1−xdx is equal to
π6
π4
π2
0
Let , I=∫01 sin2tan−11+x1−xdxput, x=cos2θThen, sin2tan−11+cos2θ1−cos2θ=sin2tan−1(cotθ)=sin2tan−1tanπ2−θ=sin2π2−θ=sin(π−2θ)=sin2θ=1−cos22θ=1−x2
Now,
∫01 sin(2tan−11+x1−xdx=∫01 1−x2dx=12x⋅1−x201+12sin−1x01=12[11−1−0]+12sin−1(1)−0=12[0]+12π2=π4