Method 1: Using Leibniz Rule (Differentiation under Integral Sign)
Let \( f(x) = \int_{0}^{\sin ^{2} x} \sin ^{-1} \sqrt{t} \, d t + \int_{0}^{\cos ^{2} x} \cos ^{-1} \sqrt{t} \, d t \).
Differentiating w.r.t \( x \):
\( f'(x) = \sin^{-1}(\sqrt{\sin^2 x}) \cdot \frac{d}{dx}(\sin^2 x) + \cos^{-1}(\sqrt{\cos^2 x}) \cdot \frac{d}{dx}(\cos^2 x) \)
\( f'(x) = x \cdot (2\sin x \cos x) + x \cdot (-2\sin x \cos x) \)
\( f'(x) = x(\sin 2x) - x(\sin 2x) = 0 \)
Since \( f'(x) = 0 \), \( f(x) \) is a constant function.
Put \( x = \frac{\pi}{4} \):
\( f(\frac{\pi}{4}) = \int_{0}^{1/2} \sin^{-1} \sqrt{t} \, dt + \int_{0}^{1/2} \cos^{-1} \sqrt{t} \, dt \)
\( = \int_{0}^{1/2} (\sin^{-1} \sqrt{t} + \cos^{-1} \sqrt{t}) \, dt \)
Using \( \sin^{-1} \theta + \cos^{-1} \theta = \frac{\pi}{2} \):
\( = \int_{0}^{1/2} \frac{\pi}{2} \, dt = \frac{\pi}{2} [t]_{0}^{1/2} = \frac{\pi}{2} \cdot \frac{1}{2} = \frac{\pi}{4} \).

Method 2: Substitution
For \( I_1 \), put \( t = \sin^2 u \Rightarrow dt = 2\sin u \cos u \, du \). Limits: \( 0 \to x \).
For \( I_2 \), put \( t = \cos^2 v \Rightarrow dt = -2\cos v \sin v \, dv \). Limits: \( \frac{\pi}{2} \to x \).
Adding them yields \( \frac{\pi}{4} \).