If f(x)=∫0sin2⁡x sin−1⁡tdt  and g(x)=∫0cos2⁡x cos−1⁡tdt then the value of f(x)+g(x)

f′(x)+g′(x)=sin−1⁡(sin⁡x)2sin⁡xcos⁡x−cos−1⁡(cos⁡x)2sin⁡xcos⁡x=xsin⁡2x−xsin⁡2x=0for all x∈R. Hence f(x)+g(x)= constant =C (say) Putting x=π/4,C=∫01/2 sin−1⁡tdt+∫01/2 cos−1⁡tdt=∫01/2 π2dt=π4Hence f(x)+g(x)=π/4