If $f\left(x\right)={\int }_0^{{\sin }^{2}x} {\sin }^{-1}\sqrt{t}dt$ and $g\left(x\right)=$

${\int }_0^{{\cos }^{2}x} {\cos }^{-1}\sqrt{t}$dt then the value of $f\left(x\right)+g\left(x\right)$

$f^{\mathrm{\prime }}\left(x\right)+g^{\mathrm{\prime }}\left(x\right)={\sin }^{-1}(\sin x)2\sin x\cos x-{\cos }^{-1}(\cos x)2\sin x\cos x$

$=x\sin 2x-x\sin 2x=0$

for all $x\in \mathbf{R}$. Hence $f\left(x\right)+g\left(x\right)=\text{ constant }=C\text{ (say) }$

Putting $x=\pi /4,C={\int }_0^{1/2} {\sin }^{-1}\sqrt{t}\mathrm{d}t+{\int }_0^{1/2} {\cos }^{-1}\sqrt{t}\mathrm{d}t$

$={\int }_0^{1/2} \frac{\pi }{2}\mathrm{d}t=\frac{\pi }{4}$

Hence $f\left(x\right)+g\left(x\right)=\pi /4$