<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] Yes and to complete: we have
$$\int f'(x)(f(x))^{-2}\;dx=C\int\sin(2x)\;dx$$
where $C=b^2-a^2$ so 
$$-\frac{1}{f(x)}=-\frac{C}{2}\cos(2x)+C'$$
and you can take $f(x)$ from it.
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