If int frac{f(x)}{log (sin x)} d x=log [log s | vedclass

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(A) Given the integral equation: $\int \frac{f(x)}{\log (\sin x)} d x = \log [\log \sin x] + c$. To find $f(x)$,we differentiate both sides with respe

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$\cot x$

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(A) Given the integral equation: $\int \frac{f(x)}{\log (\sin x)} d x = \log [\log \sin x] + c$. To find $f(x)$,we differentiate both sides with respect to $x$: $\frac{d}{d x} \left( \int \frac{f(x)}{\log (\sin x)} d x \right) = \frac{d}{d x} (\log [\log \sin x] + c)$. Using the Fundamental Theorem of Calculus on the left side,we get: $\frac{f(x)}{\log (\sin x)} = \frac{d}{d x} (\log [\log \sin x])$. Applying the chain rule on the right side: $\frac{d}{d x} (\log [\log \sin x]) = \frac{1}{\log \sin x} \cdot \frac{d}{d x} (\log \sin x)$. Since $\frac{d}{d x} (\log \sin x) = \frac{1}{\sin x} \cdot \cos x = \cot x$,we have: $\frac{f(x)}{\log \sin x} = \frac{1}{\log \sin x} \cdot \cot x$. Comparing both sides,we obtain $f(x) = \cot x$.

If $\int \frac{f(x)}{\log (\sin x)} d x=\log [\log \sin x]+c$,then $f(x)=$

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