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.
 A: $$\int\frac{\cos^2x\sin{x}}{\sin{x}-\cos{x}}dx=$$
$$=\int\left(\frac{\cos^2x\sin{x}}{\sin{x}-\cos{x}}+\frac{1}{2}\sin{x}(\sin{x}+\cos{x})\right)dx-\frac{1}{2}\int\sin{x}(\sin{x}+\cos{x})dx=$$
$$=\frac{1}{2}\int\frac{\sin{x}}{\sin{x}-\cos{x}}-\frac{1}{2}\int\sin{x}(\sin{x}+\cos{x})dx=$$
$$=\frac{1}{2}\int\left(\frac{\sin{x}}{\sin{x}-\cos{x}}-\frac{1}{2}\right)dx-\frac{1}{2}\int\left(\sin{x}(\sin{x}+\cos{x})-\frac{1}{2}\right)dx=$$
$$=\frac{1}{4}\int\frac{\sin{x}+\cos{x}}{\sin{x}-\cos{x}}dx-\frac{1}{2}\int\left(\sin{x}(\sin{x}+\cos{x})-\frac{1}{2}\right)dx=$$
$$=\frac{1}{4}\ln|\sin{x}-\cos{x}|-\frac{1}{4}\int\left(2\sin^2{x}-1+\sin2x\right)dx.$$
Can you end it now?
A: 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 $\cos (2x)=\cos^2 x-\sin^2 x.$
A: Hint:
Let $\dfrac\pi4-x=y$
$\sin x-\cos x=\sqrt2\sin y$
$\sin x=\dfrac{\cos y-\sin y}{\sqrt2}$
$2\cos^2x=1+\cos2x=1+2\sin y\cos y$
$$\dfrac{(\cos y-\sin y)(1+2\sin y\cos y)}{\sin y}=2\cos^2y-2\sin y\cos y+\cot y-1=\cos2y-\sin2y+\cot y$$
A: 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 $$\int \left(1/2\,{\frac {-t+1}{{t}^{2}+1}}+{\frac {-4\,t+4}{ \left( {t}^{2}+1
 \right) ^{3}}}+1/2\,{\frac {t+1}{{t}^{2}+2\,t-1}}+{\frac {2\,t-4}{
 \left( {t}^{2}+1 \right) ^{2}}}\right)
 dt$$
A: A standard substitution for integrals of rational functions of trigonometric ones is $t=\tan \frac{x}{2}$.
A: For integrands that are a rational function of sine and cosine, as a guide to what trigonometric substitution one may try the Bioche Rules can be used. 
Here notice the differential form
$$w(x) = f(\sin x, \cos x) \, dx = \frac{\cos^2 x \sin x}{\sin x - \cos x} \, dx$$
is invariant under the substitution $x \mapsto \pi + x$, that is, $w(\pi + x) = w(x)$. This suggests a substitution of $t = \tan x$ can be used. As $dt = \sec^2 x \, dx$ we rewrite the integrand as the product between a rational function consisting of $\tan x$ terms and a $\sec^2 x$ terms. Doing so we have
\begin{align}
I &= \int \frac{\cos^2 x \sin x}{\sin x - \cos x} \, dx\\
&= \int \frac{\cos^2 x \sin x}{\sin x - \cos x} \cdot \frac{\sec^2 x}{\sec^2 x} \, dx\\
&= \int \frac{\tan x}{(\tan x - 1)(1 + \tan^2 x)^2} \cdot \sec^2 x \, dx.
\end{align}
Now let $t = \tan x$. Doing so yields
\begin{align}
I &= \int \frac{t}{(t - 1)(1 + t^2)^2} \, dt\\
&= \int \left [\frac{1}{4(t - 1)} - \frac{t + 1}{4(t^2 + 1)} + \frac{1 - t}{2(1 + t^2)^2} \right ] \, dt\\
&= \frac{1}{4} \ln |t - 1| - \frac{1}{8} \ln |1 + t^2| + \frac{t}{4(1 + t^2)} + C\\
&= \frac{1}{4}\ln |\tan x - 1| + \frac{1}{4} \ln |\cos x| + \frac{1}{4} (1 + \tan x) \cos^2 x + C.
\end{align}
or after playing around with a few trignometric identities
$$\int \frac{\cos^2 x \sin x}{\sin x - \cos x} \, dx = \frac{1}{4} \ln |\sin x - \cos x| + \frac{1}{8} (\cos 2x + \sin 2x) + C.$$ 
