If f(x) = int_0^{pi/2} frac{ln(1 + x sin^2 th | vedclass

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(D) Let $I(x) = \int_0^{\pi/2} \frac{\ln(1 + x \sin^2 	heta)}{\sin^2 	heta} d	heta$. Using Leibniz rule for differentiation under the integral sign

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Both $(A)$ and $(B)$

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(D) Let $I(x) = \int_0^{\pi/2} \frac{\ln(1 + x \sin^2 	heta)}{\sin^2 	heta} d	heta$. Using Leibniz rule for differentiation under the integral sign: $f'(x) = \int_0^{\pi/2} \frac{\partial}{\partial x} \left( \frac{\ln(1 + x \sin^2 	heta)}{\sin^2 	heta} ight) d	heta = \int_0^{\pi/2} \frac{1}{1 + x \sin^2 	heta} d	heta$. Multiply numerator and denominator by $\sec^2 	heta$: $f'(x) = \int_0^{\pi/2} \frac{\sec^2 	heta}{\sec^2 	heta + x 	an^2 	heta} d	heta = \int_0^{\pi/2} \frac{\sec^2 	heta}{1 + (1+x) 	an^2 	heta} d	heta$. Let $u = \sqrt{1+x} 	an 	heta$,then $du = \sqrt{1+x} \sec^2 	heta d	heta$. $f'(x) = \frac{1}{\sqrt{1+x}} \int_0^{\infty} \frac{du}{1+u^2} = \frac{1}{\sqrt{1+x}} [	an^{-1} u]_0^{\infty} = \frac{\pi}{2\sqrt{1+x}}$. Integrating $f'(x)$ with respect to $x$: $f(x) = \int \frac{\pi}{2\sqrt{1+x}} dx = \pi \sqrt{1+x} + C$. Since $f(0) = \int_0^{\pi/2} \frac{\ln(1)}{\sin^2 	heta} d	heta = 0$,we have $\pi \sqrt{1} + C = 0 \implies C = -\pi$. Thus,$f(x) = \pi(\sqrt{x+1} - 1)$. Both $(A)$ and $(B)$ are correct.

If $f(x) = \int_0^{\pi/2} \frac{\ln(1 + x \sin^2 \theta)}{\sin^2 \theta} d\theta$,$x \geq 0$,then:

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