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.

Let $\displaystyle I =\frac{1}{2}\int\frac{2\cos^2 x\cdot \sin x}{\sin x-\cos x}dx=\frac{1}{2}\int\frac{(1+\cos 2x)\cdot \sin x}{\sin x-\cos x}dx$
So $\displaystyle I =\frac{1}{4}\int \frac{2\sin x}{\sin x-\cos x}dx+\frac{1}{2}\int\frac{\cos 2x\cdot \sin x}{\sin x-\cos x}dx$
Now writting 
$2\sin x=(\sin x+\cos x)+(\sin x-\cos x)$
and $\cos (2x)=\cos^2 x-\sin^2 x.$
