<s>[INST] 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:


*

*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.
 [/INST] try it with $$\sin(x)=\frac{2t}{1+t^2}$$
$$\cos(x)=\frac{1-t^2}{1+t^2}$$
$$dx=\frac{2dt}{1+t^2}$$
You will get this integral $$\int \frac{4 t
   \left(t^2-1\right)^2}{\left(t^2+1
   \right)^3 \left(t^2+2 t-1\right)}dt$$
and this is $$\int \left(1/2\,{\frac {-t+1}{{t}^{2}+1}}+{\frac {-4\,t+4}{ \left( {t}^{2}+1
 \right) ^{3}}}+1/2\,{\frac {t+1}{{t}^{2}+2\,t-1}}+{\frac {2\,t-4}{
 \left( {t}^{2}+1 \right) ^{2}}}\right)
 dt$$
</s>