If f(x)=∫0sin2xsin-1tdt+∫0cos2xcos-1tdt, then which of the following is not true?

f1(x)=sin-1sinx2sinxcosx+cos-1cosx-2sinxcosx=0⇒f(x)= constant ⇒f(x)=fπ4=∫012sin-1tdt+∫012cos-1tdt                         =∫012sin-1t+cos-1tdt                        =π2∫012dt                        =π4∴f(x)=π4∀x∈RNow fπ4+fπ2=π2