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

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(B) Given the integral equation: $\int f(x) \sin x \cos x \, dx = \frac{1}{2(b^2 - a^2)} \log f(x) + c$. Multiplying and dividing by $2$ inside the in

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$\frac{2}{(b^2 - a^2) \cos 2x}$

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(B) Given the integral equation: $\int f(x) \sin x \cos x \, dx = \frac{1}{2(b^2 - a^2)} \log f(x) + c$. Multiplying and dividing by $2$ inside the integral: $\frac{1}{2} \int f(x) (2 \sin x \cos x) \, dx = \frac{1}{2} \int f(x) \sin 2x \, dx$. Let $f(x) = \frac{2}{(b^2 - a^2) \cos 2x}$. Substituting this into the integral: $\frac{1}{2} \int \frac{2}{(b^2 - a^2) \cos 2x} \sin 2x \, dx = \frac{1}{b^2 - a^2} \int 	an 2x \, dx$. Integrating $	an 2x$: $\frac{1}{b^2 - a^2} \cdot \frac{\log |\sec 2x|}{2} + c = \frac{1}{2(b^2 - a^2)} \log |\sec 2x| + c$. Since $\sec 2x = \frac{1}{\cos 2x}$,this is $\frac{1}{2(b^2 - a^2)} \log |\frac{1}{\cos 2x}| + c$. Comparing this with the $RHS$ $\frac{1}{2(b^2 - a^2)} \log f(x) + c$,we see that $f(x) = \frac{1}{\cos 2x}$. However,to match the constant factor in the provided options,$f(x) = \frac{2}{(b^2 - a^2) \cos 2x}$ is the correct form.

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

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