# 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:

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.

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

Hint:

Let $$\dfrac\pi4-x=y$$

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

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

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

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

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

• $2\cos^2 x=1+\cos2x$. – Thomas Shelby Jan 21 at 17:37

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

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

• Going by that method got me to a messy fraction, that lead me nowhere – Sashank Sriram Jan 21 at 17:48

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