What’s the answer to $\int \frac{\cos^2x \sin x}{\sin x – \cos x} dx$

indefinite-integralsintegrationreal-analysissubstitutiontrigonometric-integrals

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.

Best Answer

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

Related Question