${\int }_0^{{\sin }^{2}x}{\sin }^{-1}\sqrt{t}dt+{\int }_0^{{\cos }^{2}x}{\cos }^{-1}\sqrt{t}dt=$

$\text{ Let }\mathrm{t}={\sin }^{2}u\text{ in the first integral and }t={\cos }^{2}u\text{ in the second integral. Then}$

$I_1+I_2={\int }_0^{x}u\sin 2udu-{\int }_{\pi /2}^{x}u\sin 2udu$

$\Rightarrow I_1+I_2={\int }_0^{x}u\sin 2udu+{\int }_x^{\pi /2}u\sin 2udu={\int }_0^{\pi /2}u\sin 2udu$

               $={\left(-\frac{u}{2}\cos 2u+\frac{1}{4}\sin 2u\right)}_0^{\frac{\mathrm{\pi }}{2}}$

               = $\frac{\mathrm{\pi }}{4}$