If f(x)=∫0sin2x sin−1tdt and g(x)=
∫0cos2x cos−1tdt then the value of f(x)+g(x)
π
π/4
π/2
sin2x+sinx+x
f′(x)+g′(x)=sin−1(sinx)2sinxcosx−cos−1(cosx)2sinxcosx
=xsin2x−xsin2x=0
for all x∈R. Hence f(x)+g(x)= constant =C (say)
Putting x=π/4,C=∫01/2 sin−1tdt+∫01/2 cos−1tdt
=∫01/2 π2dt=π4
Hence f(x)+g(x)=π/4