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

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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: $f(x) \sin x \cos x = \frac{1}{2(b^2 - a^2)} \cdot \frac{f'(x)}{f(x)}$. Using $\sin x \cos x = \frac{\sin 2x}{2}$, we have: $f(x) \cdot \frac{\sin 2x}{2} = \frac{1}{2(b^2 - a^2)} \cdot \frac{f'(x)}{f(x)}$. Multiplying by $2$, we get: $f(x) \sin 2x = \frac{1}{b^2 - a^2} \cdot \frac{f'(x)}{f(x)}$. Rearranging the terms: $(b^2 - a^2) \sin 2x \, dx = \frac{f'(x)}{(f(x))^2} \, dx$. Integrating both sides: $(b^2 - a^2) \int \sin 2x \, dx = \int (f(x))^{-2} f'(x) \, dx$. $(b^2 - a^2) \left( -\frac{\cos 2x}{2} \right) = -\frac{1}{f(x)} + C_1$. Assuming the constant of integration $C_1 = 0$ for the functional form: $\frac{(b^2 - a^2) \cos 2x}{2} = \frac{1}{f(x)}$. Therefore, $f(x) = \frac{2}{(b^2 - a^2) \cos 2x}$.

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 equal to

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