If int f(x) sin x cos x , dx = frac{1}{2(b^2 | vedclass

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(C) Given $\int f(x) \sin x \cos x \, dx = \frac{1}{2(b^2 - a^2)} \log(f(x)) + c$. Differentiating both sides with respect to $x$,we get: $f(x) \sin x

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$\frac{1}{a^2 \cos^2 x + b^2 \sin^2 x}$

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(C) Given $\int f(x) \sin x \cos x \, dx = \frac{1}{2(b^2 - a^2)} \log(f(x)) + c$. Differentiating both sides with respect to $x$,we get: $f(x) \sin x \cos x = \frac{1}{2(b^2 - a^2)} \cdot \frac{f'(x)}{f(x)}$. Rearranging the terms: $2(b^2 - a^2) \sin x \cos x = \frac{f'(x)}{f(x)^2}$. Integrating both sides with respect to $x$: $\int (2b^2 \sin x \cos x - 2a^2 \sin x \cos x) \, dx = \int \frac{f'(x)}{f(x)^2} \, dx$. Using the substitution $u = \sin^2 x$ and $v = \cos^2 x$,we have $du = 2 \sin x \cos x \, dx$ and $dv = -2 \cos x \sin x \, dx$. Thus,$b^2 \sin^2 x - a^2 \cos^2 x = -\frac{1}{f(x)} + C_1$. Simplifying,we get $f(x) = \frac{1}{a^2 \cos^2 x + b^2 \sin^2 x}$ (adjusting constants). Comparing with the given options,the correct form is $f(x) = \frac{1}{a^2 \cos^2 x + b^2 \sin^2 x}$.

If $\int f(x) \sin x \cos x \, dx = \frac{1}{2(b^2 - a^2)} \log(f(x)) + c$,then $f(x) = $

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