4
$\begingroup$

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.

$\endgroup$
6
$\begingroup$

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

$\endgroup$
3
$\begingroup$

Hint:

Let $\dfrac\pi4-x=y$

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

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

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

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

$\endgroup$
3
$\begingroup$

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

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

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

$\endgroup$
2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Going by that method got me to a messy fraction, that lead me nowhere $\endgroup$ – Sashank Sriram Jan 21 at 17:48
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.