∫01 sin⁡2tan−1⁡1+x1−xdx is equal to

Let , I=∫01 sin⁡2tan−1⁡1+x1−xdxput, x=cos⁡2θThen, sin⁡2tan−1⁡1+cos⁡2θ1−cos⁡2θ=sin⁡2tan−1⁡(cot⁡θ)=sin⁡2tan−1⁡tan⁡π2−θ=sin⁡2π2−θ=sin⁡(π−2θ)=sin⁡2θ=1−cos2⁡2θ=1−x2Now,∫01 sin⁡(2tan−1⁡1+x1−xdx=∫01 1−x2dx=12x⋅1−x201+12sin−1⁡x01=12[11−1−0]+12sin−1⁡(1)−0=12[0]+12π2=π4