What's the answer to $\int \frac{\cos^2x \sin x}{\sin x - \cos x} dx$?

Question

I tried solving the integral $$\int \frac{\cos^2x \sin x}{\sin x - \cos x}\, dx$$ the following ways:

1. 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.

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

3. 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.

4. 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.

5. 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.

Answer 1

$$\int\frac{\cos^2x\sin{x}}{\sin{x}-\cos{x}}dx=$$ $$=\int\left(\frac{\cos^2x\sin{x}}{\sin{x}-\cos{x}}+\frac{1}{2}\sin{x}(\sin{x}+\cos{x})\right)dx-\frac{1}{2}\int\sin{x}(\sin{x}+\cos{x})dx=$$ $$=\frac{1}{2}\int\frac{\sin{x}}{\sin{x}-\cos{x}}-\frac{1}{2}\int\sin{x}(\sin{x}+\cos{x})dx=$$ $$=\frac{1}{2}\int\left(\frac{\sin{x}}{\sin{x}-\cos{x}}-\frac{1}{2}\right)dx-\frac{1}{2}\int\left(\sin{x}(\sin{x}+\cos{x})-\frac{1}{2}\right)dx=$$ $$=\frac{1}{4}\int\frac{\sin{x}+\cos{x}}{\sin{x}-\cos{x}}dx-\frac{1}{2}\int\left(\sin{x}(\sin{x}+\cos{x})-\frac{1}{2}\right)dx=$$ $$=\frac{1}{4}\ln|\sin{x}-\cos{x}|-\frac{1}{4}\int\left(2\sin^2{x}-1+\sin2x\right)dx.$$ Can you end it now?

Answer 2

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 $\cos (2x)=\cos^2 x-\sin^2 x.$

Answer 3

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 $$\int \left(1/2\,{\frac {-t+1}{{t}^{2}+1}}+{\frac {-4\,t+4}{ \left( {t}^{2}+1 \right) ^{3}}}+1/2\,{\frac {t+1}{{t}^{2}+2\,t-1}}+{\frac {2\,t-4}{ \left( {t}^{2}+1 \right) ^{2}}}\right) dt$$

Answer 4

A standard substitution for integrals of rational functions of trigonometric ones is $t=\tan \frac{x}{2}$.

Answer 5

Hint:

Let $\dfrac\pi4-x=y$

$\sin x-\cos x=\sqrt2\sin y$

$\sin x=\dfrac{\cos y-\sin y}{\sqrt2}$

$2\cos^2x=1+\cos2x=1+2\sin y\cos y$

$$\dfrac{(\cos y-\sin y)(1+2\sin y\cos y)}{\sin y}=2\cos^2y-2\sin y\cos y+\cot y-1=\cos2y-\sin2y+\cot y$$

Answer 6

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 $$\int \frac{\cos^2 x \sin x}{\sin x - \cos x} \, dx = \frac{1}{4} \ln |\sin x - \cos x| + \frac{1}{8} (\cos 2x + \sin 2x) + C.$$

Answer 7

$$\int\frac{\cos^2x\sin x}{\sin x-\cos x}dx$$

$$=\int\frac{-\sin x\cos^2x}{\cos x-\sin x}dx$$

$$=\int\frac{-\sin x\cos^2x}{\cos x-\sin x}\times\frac{\cos x+\sin x}{\cos x+\sin x}dx$$

$$=\int\frac{-\sin x\cos^2x\left(\cos x+\sin x\right)}{\cos^2x-\sin^2x}dx$$

$$=\int\frac{-\sin x\cos x\times\cos x\left(\cos x+\sin x\right)}{\cos^2x-\sin^2x}dx$$

$$=\int\frac{-\sin x\cos x\left(\cos^2x+\sin x\cos x\right)}{\cos^2x-\sin^2x}dx$$

$$=\int\frac{-\frac{\sin2x}{2}\left(\frac{1+\cos2x}{2}+\frac{\sin2x}{2}\right)}{\cos2x}dx$$

$$=\int\frac{-\frac{\sin2x}{2}\left(\frac{1+\cos2x+\sin2x}{2}\right)}{\cos2x}dx$$

$$=\int\frac{-\sin2x\left(1+\cos2x+\sin2x\right)}{4\cos2x}dx$$

$$=\int\frac{-\tan2x\left(1+\cos2x+\sin2x\right)}{4}dx$$

$y=2x$

$dy=2\ dx$

$-\frac{dy}{8}=-\frac{dx}{4}$

$$=\int\frac{-\tan y\left(1+\cos y+\sin y\right)}{8}dy$$

$$=\frac{1}{8}\int\left(-\tan y-\sin y-\sin y\tan y\right)dy$$

$$=\frac{1}{8}\int\left(-\tan y-\sin y-\frac{\sin^2y}{\cos y}\right)dy$$

$$=\frac{1}{8}\int\left(-\tan y-\sin y-\frac{1-\cos^2y}{\cos y}\right)dy$$

$$=\frac{1}{8}\int\left(-\tan y-\sin y+\frac{\cos^2y-1}{\cos y}\right)dy$$

$$=\frac{1}{8}\int\left(-\tan y-\sin y+\cos y-\sec y\right)dy$$

$$=\frac{\ln\left|\cos y\right|+\cos y+\sin y-\ln\left|\sec y+\tan y\right|}{8}+c$$

$$=\frac{\ln\left|\cos2x\right|+\cos2x+\sin2x-\ln\left|\sec2x+\tan2x\right|}{8}+c$$

$$=\frac{\ln\frac{\left|\cos2x\right|}{\left|\sec2x+\tan2x\right|}+\cos2x+\sin2x}{8}+c$$

$$=\frac{\ln\left|\frac{\cos2x}{\sec2x+\tan2x}\right|+\cos2x+\sin2x}{8}+c$$

$$=\frac{\ln\left|\frac{\cos^2\left(2x\right)}{1+\sin2x}\right|+\cos2x+\sin2x}{8}+c$$

$$=\ln\sqrt[8]{\frac{\cos^2\left(2x\right)}{1+\sin2x}}+\frac{\cos2x+\sin2x}{8}+c$$

$$=\ln\sqrt[8]{\frac{1-\sin^2\left(2x\right)}{1+\sin2x}}+\frac{\cos2x+\sin2x}{8}+c$$

$$=\ln\sqrt[8]{1-\sin2x}+\frac{\cos2x+\sin2x}{8}+c$$

Answer 8

$$ \begin{aligned} I & =\int \frac{\cos ^2 x \sin x}{\sin x-\cos x} d x \\ & =-\frac{1}{\sqrt{2}} \int \frac{\cos ^2 x \sin x}{\sin \left(\frac{\pi}{4}-x\right)} d x \\ & =-\frac{1}{2 \sqrt{2}} \int \frac{\sin (2 x) \cos x}{\sin \left(\frac{\pi}{4}-x\right)} d x \end{aligned} $$ Now let’s transform the integral by the substitution $y=\frac{\pi}{4} -x$. $$ \begin{aligned} I & =\frac{1}{2 \sqrt{2}} \int \frac{\sin \left(\frac{\pi}{2}-2 y\right) \cos \left(\frac{\pi}{4}-y\right)}{\sin y} d y \\ & =\frac{1}{4} \int \frac{\cos (2 y)(\cos y+\sin y)}{\sin y} d y \\ & =\frac{1}{4}\left[\int \frac{\cos (2 y) \cos y}{\sin y} d y+\int \cos 2 y d y\right] \\ & =\frac{1}{4}\left[\int \frac{1-2 s^2}{s} d s+\frac{\sin (2 y)}{2}\right] \quad \textrm{ where } s=\sin y\\ & =\frac{1}{4}\left[\ln |s|+s^2+ \frac{\cos (2x)}{2}\right]+C\\&= \frac{1}{4}\left[\ln \left|\frac{1}{\sqrt{2}}(\cos x-\sin x)\right|-\frac{1}{2}(\cos x-\sin x)^2+\frac{\cos (2 x)}{2}\right]+C\\&= \frac{1}{8}[2 \ln |\cos x-\sin x|+\sin (2 x)+\cos (2 x)]+C \end{aligned} $$

Answer 9

Notice that, if we multiply both numerator and denominator of the integrand by $\cos x+\sin x$, the latter simplifies to $\cos2x$. Thus $$I:=\int\frac{\cos^2x\sin x}{\cos x-\sin x}\mathrm dx=\int\frac{(\cos x\sin x+\sin^2x)\cos^2x}{\cos2x}\mathrm dx.$$Moreover, the components in the denominator can be written in terms of $\cos2x$ and $\sin2x\,$: $$I=\int\frac{(\sin2x+1-\cos2x)(1+\cos2x)}{4\cos2x}\mathrm dx=\int\frac{\sin2x+1}{4\cos2x}\mathrm dx+\int\frac{\sin2x-\cos2x}{4}\mathrm dx.$$ Then standard integral formulae give (after a little tidying of the first integral) $$I=\frac18\left(\ln\frac{1+\sin2x}{1+\cos4x}-\cos2x-\sin2x\right)+C.$$