For 0 le x le frac{pi}{2},the value of int_{0 | vedclass

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(A) Let $f(x) = \int_{0}^{\sin^{2}x} \sin^{-1}(\sqrt{t}) \, dt + \int_{0}^{\cos^{2}x} \cos^{-1}(\sqrt{t}) \, dt$.Using the Leibniz integral rule,we di

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$\frac{\pi}{4}$

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(A) Let $f(x) = \int_{0}^{\sin^{2}x} \sin^{-1}(\sqrt{t}) \, dt + \int_{0}^{\cos^{2}x} \cos^{-1}(\sqrt{t}) \, dt$.Using the Leibniz integral rule,we differentiate $f(x)$ with respect to $x$:$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)$$f'(x) = x \cdot (2 \sin x \cos x) + (\frac{\pi}{2} - x) \cdot (-2 \cos x \sin x)$$f'(x) = x \sin(2x) - (\frac{\pi}{2} - x) \sin(2x)$$f'(x) = x \sin(2x) - \frac{\pi}{2} \sin(2x) + x \sin(2x) = (2x - \frac{\pi}{2}) \sin(2x)$.Since $f'(x)$ is not zero,let us evaluate $f(x)$ at $x = \frac{\pi}{4}$:$f(\frac{\pi}{4}) = \int_{0}^{1/2} \sin^{-1}(\sqrt{t}) \, dt + \int_{0}^{1/2} \cos^{-1}(\sqrt{t}) \, dt$$f(\frac{\pi}{4}) = \int_{0}^{1/2} (\sin^{-1}(\sqrt{t}) + \cos^{-1}(\sqrt{t})) \, dt$Since $\sin^{-1}(\sqrt{t}) + \cos^{-1}(\sqrt{t}) = \frac{\pi}{2}$ for $t \in [0, 1]$:$f(\frac{\pi}{4}) = \int_{0}^{1/2} \frac{\pi}{2} \, dt = \frac{\pi}{2} \cdot [t]_{0}^{1/2} = \frac{\pi}{2} \cdot \frac{1}{2} = \frac{\pi}{4}$.

For $0 \le x \le \frac{\pi}{2}$,the value of $\int_{0}^{\sin^{2}x} \sin^{-1}(\sqrt{t}) \, dt + \int_{0}^{\cos^{2}x} \cos^{-1}(\sqrt{t}) \, dt$ is equal to

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