Given $f'(x)=\csc{x}(\cot{x}–\sin{2x})$, find $f(x)$

I have a question on my integration assignment that I am not quite sure how to approach. I've looked at it and can't seem to think of a suitable place to use u-substitution.

All I can figure is that I can expand it out to be $f'(x)=\csc{x}\cot{x}-2\cos{x}$

I've considered maybe setting $u=2\sin{x}\cos{x}$ where $du=2(\cos^{2}{x}+\sin^{2}{x})\ dx$, which then simplifies to $2\ dx$, but then I can't get anywhere with that.

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Use the fact that the derivative of $\sin x$ is $\cos x$ and the derivative of $\csc x$ is $-\csc x\cot x$. Hence $f(x)=-\csc x-2\sin x+C$.