The value of ∫0sin2⁡x sin−1⁡tdt+∫0cos2⁡x cos−1⁡tdt is

f(x)=∫0sin2⁡x sin−1⁡tdt+∫0cos2⁡x cos−1⁡tdtf′(x)=sin−1⁡sin2⁡x2sin⁡xcos⁡x+cos−1⁡cos2⁡x(−2cos⁡xsin⁡x)=sin⁡2xsin−1⁡(sin⁡x)−cos−1⁡(cos⁡x)=sin⁡2x(x−x)=0f (x) = constantFind f (x) for any value of x∫01/2 π2dx=π2⋅12=π4