If $\int f(x)\sin x\cos xdx = \frac{1}{2(a^2-b^2)}\ln \left|f(x)\right|+\mathcal{C}.$ Then $f(x)$ is 
If $\displaystyle \int f(x)\sin x\cos xdx = \frac{1}{2(a^2-b^2)}\ln \left|f(x)\right|+\mathcal{C}.$ Then $f(x)$ is
$\bf{Options:}$ $(a)\;\displaystyle \frac{1}{a^2\sin^2 x+b^2 \cos^2 x} \;\;\;\;\;\; (b)\; \frac{1}{a^2\sin^2 x-b^2 \cos^2 x}\;\;\;\;\;\;(c)\; \frac{1}{a^2\cos^2 x-b^2 \sin^2 x}$
$\displaystyle (d)\; \frac{1}{a^2\cos^2 x+b^2 \sin^2 x}$

$\bf{My\; Solution:}$ Given $\displaystyle \int f(x)\sin x\cos xdx = \frac{1}{2(a^2-b^2)}\ln \left|f(x)\right|+\mathcal{C}$
Now Diff. both side w. t. to $x\;,$ We Get $\displaystyle f(x)\sin x\cos x = \frac{1}{2(a^2-b^2)}\cdot \frac{f^{'}(x)}{f(x)}.$
Now We get $\displaystyle \frac{f^{'}(x)}{(f(x))^2} = (a^2-b^2)\cdot \sin 2x,$ Now Integrate both side w. r. to $x$
$\displaystyle \int \frac{f^{'}(x)}{(f(x))^2}dx = (a^2-b^2)\int  \sin 2xdx\Rightarrow --\frac{1}{f(x)} = -\frac{(a^2-b^2)}{2}\cos 2x$
So We get $\displaystyle f(x) = \frac{2}{(a^2-b^2)\cos 2x}$
But Answer Given is $(a)\;,$ I did not understand where i have make mistake, Help me
Thanks

$\displaystyle\int\frac{f^{'}(x)}{(f(x))^2}dx=(a^2-b^2)\int\sin 2xdx\\
\Rightarrow \frac{1}{f(x)}=\frac{(a^2-b^2)}{2}\cos 2x+C\\
=\frac{(a^2-b^2)}{2}(\cos^{2}x-\sin^{2}x)+\frac{(a^2+b^2)}{2}(\cos^{2}x+\sin^{2}x)+C\\
=a^2\cos^{2}x+b^2\sin^{2}x+C$
So We get $\displaystyle f(x) = \frac{1}{a^2\cos^{2}x+b^2\sin^{2}x+C}$ and the only possible answer is (d).
Answer (a) can't be correct, as $(a^2\sin^{2}x+b^2\cos^{2}x)'=2(a^2-b^2)\sin x\cos x$, so
$$\int\frac{2(a^2-b^2)\sin x\cos x}{a^2\sin^{2}x+b^2\cos^{2}x}
=\ln(a^2\sin^{2}x+b^2\cos^{2}x)+C
=-\ln\frac{1}{a^2\sin^{2}x+b^2\cos^{2}x}+C$$
