1708\hspace{-20}$Given $\bf{\int f(x)\cdot \sin (2x)dx = \frac{2}{b^2-a^2}\cdot \ln (f(x))+\mathbb{C}}$\\\\\\ Now Diff. both side w. r.to $\bf{x\;,}$ we get\\\\\\ $\bf{\frac{d}{dx}\left\{\int f(x)\cdot \sin (2x)dx\right\} =\frac{2}{b^2-a^2}\cdot \frac{d}{dx}\left\{\ln(f(x))+\mathbb{C}\right\}}$\\\\\\ $\bf{\Rightarrow f(x)\sin (2x) = \frac{2}{b^2-a^2}\cdot \frac{1}{f(x)}\cdot f^{'}(x)}$\\\\\\ $\bf{\Rightarrow \frac{f^{'}(x)}{(f(x))^2}dx = \frac{b^2-a^2}{2}\sin 2x}$\\\\\\ $\bf{\Rightarrow \int \frac{f^{'}(x)}{(f(x))^2}dx = \frac{b^2-a^2}{2}\int \sin 2xdx}$\\\\\\ Now Let $\bf{f(x)=t\;,}$ Then $\bf{f^{'}(x)dx = dt}$\\\\\\ $\bf{\Rightarrow \int\frac{1}{t^2}dt = -\frac{b^2-a^2}{4}\cos (2x)}$\\\\\\ $\bf{\Rightarrow \frac{1}{f(x)} = \frac{b^2-a^2}{4}\cos (2x)}$\\\\\\ So $\bf{\boxed{\bf{f(x)=\frac{4}{\left(b^2-a^2\right)}\cdot \sec (2x)}}}$