The value of int_{0}^{sin^2 x} sin^{-1} sqrt{ | vedclass

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(C) Let $I(x) = \int_{0}^{\sin^2 x} \sin^{-1} \sqrt{t} \, dt + \int_{0}^{\cos^2 x} \cos^{-1} \sqrt{t} \, dt$. Applying Leibniz's rule to differentiate

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$\frac{\pi}{4}$

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(C) Let $I(x) = \int_{0}^{\sin^2 x} \sin^{-1} \sqrt{t} \, dt + \int_{0}^{\cos^2 x} \cos^{-1} \sqrt{t} \, dt$. Applying Leibniz's rule to differentiate with respect to $x$: $I'(x) = \sin^{-1}(\sqrt{\sin^2 x}) \cdot \frac{d}{dx}(\sin^2 x) + \cos^{-1}(\sqrt{\cos^2 x}) \cdot \frac{d}{dx}(\cos^2 x)$. Since $x \in (0, \pi/2)$,$\sin x > 0$ and $\cos x > 0$,so $\sqrt{\sin^2 x} = \sin x$ and $\sqrt{\cos^2 x} = \cos x$. $I'(x) = (x) \cdot (2 \sin x \cos x) + (x) \cdot (-2 \cos x \sin x)$. $I'(x) = x \sin(2x) - x \sin(2x) = 0$. Since the derivative is $0$,$I(x)$ is a constant. To find the constant,evaluate at $x = \pi/4$: $I(\pi/4) = \int_{0}^{1/2} \sin^{-1} \sqrt{t} \, dt + \int_{0}^{1/2} \cos^{-1} \sqrt{t} \, dt$. Using the identity $\sin^{-1} \sqrt{t} + \cos^{-1} \sqrt{t} = \pi/2$: $I(\pi/4) = \int_{0}^{1/2} (\sin^{-1} \sqrt{t} + \cos^{-1} \sqrt{t}) \, dt = \int_{0}^{1/2} \frac{\pi}{2} \, dt = \frac{\pi}{2} [t]_{0}^{1/2} = \frac{\pi}{2} \cdot \frac{1}{2} = \frac{\pi}{4}$.

The value of $\int_{0}^{\sin^2 x} \sin^{-1} \sqrt{t} \, dt + \int_{0}^{\cos^2 x} \cos^{-1} \sqrt{t} \, dt$ for $x \in (0, \pi/2)$ is:

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