If f(x) = int_{pi^2/16}^{x^2} frac{sin x cdot | vedclass

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(A) Given function: $f(x) = \sin x \int_{\pi^2/16}^{x^2} \frac{\sin \sqrt{	heta}}{1 + \cos^2 \sqrt{	heta}} \, d	heta$. Using Leibniz's rule for dif

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$\pi$

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(A) Given function: $f(x) = \sin x \int_{\pi^2/16}^{x^2} \frac{\sin \sqrt{	heta}}{1 + \cos^2 \sqrt{	heta}} \, d	heta$. Using Leibniz's rule for differentiation: $f'(x) = \cos x \int_{\pi^2/16}^{x^2} \frac{\sin \sqrt{	heta}}{1 + \cos^2 \sqrt{	heta}} \, d	heta + \sin x \left[ \frac{\sin \sqrt{x^2}}{1 + \cos^2 \sqrt{x^2}} \cdot \frac{d}{dx}(x^2) ight]$. $f'(x) = \cos x \int_{\pi^2/16}^{x^2} \frac{\sin \sqrt{	heta}}{1 + \cos^2 \sqrt{	heta}} \, d	heta + \sin x \left[ \frac{\sin x}{1 + \cos^2 x} \cdot 2x ight]$. Now,substituting $x = \frac{\pi}{2}$: $f'(\frac{\pi}{2}) = \cos(\frac{\pi}{2}) \int_{\pi^2/16}^{\pi^2/4} \frac{\sin \sqrt{	heta}}{1 + \cos^2 \sqrt{	heta}} \, d	heta + \sin(\frac{\pi}{2}) \left[ \frac{\sin(\frac{\pi}{2})}{1 + \cos^2(\frac{\pi}{2})} \cdot 2(\frac{\pi}{2}) ight]$. Since $\cos(\frac{\pi}{2}) = 0$ and $\sin(\frac{\pi}{2}) = 1$: $f'(\frac{\pi}{2}) = 0 \cdot (	ext{integral value}) + 1 \cdot \left[ \frac{1}{1 + 0} \cdot \pi ight] = \pi$.

If $f(x) = \int_{\pi^2/16}^{x^2} \frac{\sin x \cdot \sin \sqrt{\theta}}{1 + \cos^2 \sqrt{\theta}} \, d\theta$,then the value of $f'(\frac{\pi}{2})$ is:

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