<s>[INST] If $\int f(x) \sin{x} \cos{x}\,\mathrm dx = \frac {1}{2(b^2 - a^2)} \log f(x) +c $. Find $f(x)$ 
Problem: If $\int f(x) \sin{x}  \cos{x}\,\mathrm  dx = \frac {1}{2(b^2 - a^2)} \log f(x) +c $. Find $f(x)$
Solution: $\int f(x) \sin{x}  \cos{x}\,\mathrm  dx = \frac {1}{2(b^2 - a^2)} \log f(x) +c $
Differenting both sides,we get
$ f(x) \sin{x}  \cos{x} = \frac {f'(x)}{2(b^2 - a^2)f(x)}  $

Am I doing right ?
 [/INST] It is a good start. For simplicity write $y$ for $f(x)$. We can rewrite the result you got as 
$$\frac{y'}{y^2}=2(b^2-a^2)\sin x\cos x.$$
Integrate both sides. It may be handy to note that $2\sin x\cos x=\sin(2x)$. Or not, since it is clear that $2\sin x\cos x$ is the derivative of $\sin^2 x$. 
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