What's the answer to $\int \frac{\cos^2x \sin x}{\sin x - \cos x} dx$? I tried solving the integral $$\int \frac{\cos^2x \sin x}{\sin x - \cos x}\, dx$$ the following ways:


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*Expressing each function in the form of $\tan \left(\frac{x}{2}\right)$, $\cos \left(\frac{x}{2}\right)\,$ and $\,\sin \left(\frac{x}{2}\right)\,$ independently, but that didn't go well for me.

*Multiplying and dividing by $\cos^2x$ or $\sin^2x$.

*Expressing $\cos^2x$ as $1-\sin^2x$ and splitting the integral, and I was stuck with $\int \left(\frac{\sin^3x}{\sin x - \cos x}\right)\, dx$ which I rewrote as  $\int \frac{\csc^2x}{\csc^4x (1-\cot x) } dx,\,$ and tried a whole range of substitutions only to fail.

*I tried to substitute   $\frac{1}{ \sin x - \cos x}$,  $\frac{\sin x}{ \sin x - \cos x}$, $\frac{\cos x \sin x}{ \sin x - \cos x}$ and  $\frac{\cos^2x \sin x}{\sin x - \cos x},$ independently, none of which seemed to work  out.

*I expressed the denominator as $\sin\left(\frac{\pi}{4}-x\right)$  and tried multiplying and dividing by $\sin\left(\frac{\pi}{4}+x\right)$, and carried out some substitutions. Then, I repeated the same with $\cos\left(\frac{\pi}{4}+x\right)$. Neither of them worked.

Hint:
Let $\dfrac\pi4-x=y$
$\sin x-\cos x=\sqrt2\sin y$
$\sin x=\dfrac{\cos y-\sin y}{\sqrt2}$
$2\cos^2x=1+\cos2x=1+2\sin y\cos y$
$$\dfrac{(\cos y-\sin y)(1+2\sin y\cos y)}{\sin y}=2\cos^2y-2\sin y\cos y+\cot y-1=\cos2y-\sin2y+\cot y$$
