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Note that this question lacks a condition: $x \in [2\pi n , 2\pi n +\frac{\pi}{2} ] , n\in Z$ , Otherwise, the expression will be dependent on $x$ .

$\displaystyle f(x) = \int_0^{\sin ^2 x } \sin^{-1} (\sqrt{t}) dt + \int_0^{\cos ^2 x } \cos^{-1} (\sqrt{t}) dt$

Using the Fundamental Theorem of Calculus :

$\displaystyle f'(x) = (\sin ^2 x)' \sin^{-1} (\sqrt{\sin ^2 x}) + (\cos ^2 x)' \cos^{-1} (\sqrt{\cos ^2 x})$

$\displaystyle = (\sin ^2 x)' \sin^{-1} (|\sin x|) + (\cos ^2 x)' \cos^{-1} (|\cos x|)$

$\displaystyle = (2\sin x \cos x )(x) - (2\sin x \cos x )(x) =0$

Therefore: $f(x) = \text{constant}$ for all $x \in [2\pi n , 2\pi n +\frac{\pi}{2} ]$ .So, it's enough to find $f(\frac{\pi}{4})$ :

$\displaystyle f(\frac{\pi}{4}) = \int_0^{\frac{1}{2} } \sin^{-1} (\sqrt{t}) dt + \int_0^{\frac{1}{2} } \cos^{-1} (\sqrt{t}) dt$

$\displaystyle = = \int_0^{\frac{1}{2} } (\sin^{-1} (\sqrt{t}) +\cos^{-1} (\sqrt{t}) ) dt$

$\displaystyle = \int_0^{\frac{1}{2} } \frac{\pi}{2} dt = \boxed{\frac{\pi}{4} }$