If f(x) = int_0^{sin^2 x} sin^{-1} sqrt{t} , | vedclass

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(B) Using the Leibniz rule for differentiation under the integral sign:$f'(x) = \sin^{-1}(\sqrt{\sin^2 x}) \cdot \frac{d}{dx}(\sin^2 x) = x \cdot (2 \

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$\frac{\pi}{4}$

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(B) Using the Leibniz rule for differentiation under the integral sign:$f'(x) = \sin^{-1}(\sqrt{\sin^2 x}) \cdot \frac{d}{dx}(\sin^2 x) = x \cdot (2 \sin x \cos x) = x \sin(2x)$$g'(x) = \cos^{-1}(\sqrt{\cos^2 x}) \cdot \frac{d}{dx}(\cos^2 x) = x \cdot (-2 \cos x \sin x) = -x \sin(2x)$Thus, $f'(x) + g'(x) = x \sin(2x) - x \sin(2x) = 0$.Since the derivative is zero, $f(x) + g(x) = C$ (a constant).To find $C$, substitute $x = \frac{\pi}{4}$:$f(\frac{\pi}{4}) = \int_0^{1/2} \sin^{-1} \sqrt{t} \, dt$$g(\frac{\pi}{4}) = \int_0^{1/2} \cos^{-1} \sqrt{t} \, dt$$f(\frac{\pi}{4}) + g(\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 $t \in [0, 1]$:$C = \int_0^{1/2} \frac{\pi}{2} \, dt = \frac{\pi}{2} [t]_0^{1/2} = \frac{\pi}{2} \cdot \frac{1}{2} = \frac{\pi}{4}$.

If $f(x) = \int_0^{\sin^2 x} \sin^{-1} \sqrt{t} \, dt$ and $g(x) = \int_0^{\cos^2 x} \cos^{-1} \sqrt{t} \, dt$, then the value of $f(x) + g(x)$ is

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