In the pool of difficult (at least to me) integrals I've been trying to solve this one: $$\int\frac{x}{(1-x^3)\sqrt{1-x^2}}dx$$

Since Wolfram Alpha has been helpful with all the other integrals ( at least I can verify that my solution is correct), I turned to it this time as well. But the result seemed quite ...odd to me. My attemp in solving this was using the substitution $x=\sin{u}$ which leaves me with the following integral: $$\int\frac{\sin{u}}{1-\sin^3{u}}du$$

But this is as far as I could get . And again, even for this integral, WA returnes something which isn't likely to be a solution to an integral in introductory calculus course.

  • $\begingroup$ use $\arcsin$ ? $\endgroup$
    – Maman
    Apr 22 '16 at 19:11
  • 1
    $\begingroup$ the result looks terrible $\endgroup$ Apr 22 '16 at 19:16
  • $\begingroup$ @Maman That came to mind, but arcsin and partial integration didn't really help me here - it made matters only worse :( $\endgroup$ Apr 22 '16 at 19:19

The quadratic under the radical term $\sqrt{1-x^{2}}=\sqrt{\left(1-x\right)\left(1+x\right)}$ possesses two real roots, which suggests that an Euler substitution of the third kind may be the way to go. Consider the substitution relation,

$$\sqrt{\frac{1+x}{1-x}}=t\implies x=\frac{t^{2}-1}{t^{2}+1};~~~\small{-1<x<1}.$$

Taking differentials of both sides, we obtain


or equivalently,


Now, your problem could be recast as the definite integral,


Transforming the integral using the substitution rule above yields an integral with rational integrand: for $-1<z<1$,

$$\begin{align} \int_{0}^{z}\frac{x}{\left(1-x^{3}\right)\sqrt{1-x^{2}}}\,\mathrm{d}x &=\int_{0}^{z}\frac{x}{\left(1+x+x^{2}\right)\left(1-x\right)\sqrt{1-x^{2}}}\,\mathrm{d}x\\ &=\int_{1}^{\sqrt{\frac{1+z}{1-z}}}\frac{t^{4}-1}{3t^{4}+1}\,\mathrm{d}t.\\ \end{align}$$

From there, you would of course proceed by partial fractions and integrate term by term in the usual manner. Can you take it from here?

  • $\begingroup$ Of course! But can you explain what happend to result on Wolfram Alpha? Why is it their result is different ( I assume it has something to do with ranges in which we're integrating ) $\endgroup$ Apr 22 '16 at 19:58
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    $\begingroup$ @Transcendental Although it would be a pain to verify, WRA's result is ultimately the same. The steps WRA uses weren't available, but based previous experience my guess is that it 1) starts by subbing $x=\sin{\theta}$ like you did, then 2) uses the tangent-half-angle substitution. The main difference now is that WRA's default assumption is that parameters can be any complex number, and it will use the full set of complex roots of the denominator to fully decompose the integrand by partial fractions. The end result is that WRA gets a lot of complex log terms instead of arctan ones. $\endgroup$
    – David H
    Apr 22 '16 at 20:24

The substitution $u = \sqrt{\dfrac{1-x}{1+x}}$ gives you $$ \int \dfrac{u^4-1}{u^6+3u^2}\; du$$ which you can do with partial fractions: the factorization

$$ u^4 + 3 = \left(u^2 + \sqrt{2} \sqrt[4]{3}\; u + \sqrt{3}\right) \left(u^2 - \sqrt{2} \sqrt[4]{3}\; u + \sqrt{3}\right)$$ may be useful.


I tried this one with trigonometric substitutions. The $4$ possibilities are $x=\sin\theta$, $x=\cos\theta$, $x=\tanh\theta$, and $x=\text{sech}\,\theta$. $x=\sin\theta$ and $x=\tanh\theta$ force you to complete the square at one point, and $x=\text{sech}\,\theta$ has a hole in the middle of its domain, so that leaves $x=\cos\theta$. Then $$\int\frac x{(1-x^3)\sqrt{1-x^2}}dx=\int\frac{\cos\theta}{\cos^3\theta-1}d\theta$$ Expanding by partial fractions, $$\frac x{x^3-1}=\frac x{\prod_{k=0}^2(x-\omega_k)}=\sum_{k=0}^2\frac{A_k}{x-\omega_k}$$ Where $\omega_k=e^{i\theta_k}$, $\theta_k=\frac{2\pi ik}3$. By L'Hopital's rule, $$\lim_{x\rightarrow\omega_k}\frac{(x-\omega_k)x}{x^3-1}=\frac{\omega_k}{3\omega_k^2}=\frac{\omega_k^2}3=A_k$$ So we can substitute $z=\tan\left(\frac{\theta}2\right)$ to get $$\begin{align}\int\frac x{(1-x^3)\sqrt{1-x^2}}dx&=\sum_{k=0}^2\frac{\omega_k^2}3\int\frac{d\theta}{\cos\theta-\omega_k}\\ &=\sum_{k=0}^2\frac{\omega_k^2}3\int\frac{2dz}{1-z^2-\omega_k(1+z^2)}\\ &=\sum_{k=0}^2\frac{2\omega_k^2}3\int\frac{dz}{1-\omega_k-(1+\omega_k)z^2}\end{align}$$ At this point it can be observed that $z=\sqrt{\frac{1-x}{1+x}}$, just like Robert Israel recommended. Here is where completing the square for $x=\sin\theta$ would come in. The $k=0$ term in the sum is $$-\frac13\int\frac{dz}{z^2}=\frac1{3z}+C_1$$ and the $k=2$ term = is complex conjugate of the $k=1$ term so we only have to integrate the latter term. Some arithmetic we will need is $1-\omega_1=\sqrt3e^{\frac{-i\pi}6}$, $1+\omega_1=e^{\frac{i\pi}3}$, and $\omega_1^2=e^{\frac{-2\pi i}3}$. Now we can sustitute $\sqrt{1+\omega_1}z=\sqrt{1-\omega_1}u$ transforming to $$\begin{align}\frac{2\omega_1^2}3\int\frac{dz}{1-\omega_1-(1+\omega_1)z^2}&=\frac{2\omega_1^2}3\int\frac{\frac{\sqrt{1-\omega_1}}{\sqrt{1+\omega_1}}du}{(1-\omega_1)(1-u^2)}\\ &=\frac23\frac{e^{\frac{-2\pi i}3}}{3^{1/4}e^{\frac{-\pi i}{12}}e^{\frac{\pi i}6}}\int\frac{du}{1-u^2}\\ &=\frac2{3^{5/4}}e^{\frac{-3\pi i}4}\frac12\ln\left(\frac{1+u}{1-u}\right)+C_2\\ &=\frac{(-1-i)}{2^{1/2}3^{5/4}}\ln\left(\frac{\frac{3^{1/4}}{2^{1/2}}(1-i)+z}{\frac{3^{1/4}}{2^{1/2}}(1-i)-z}\right)+C_2\\ &=\frac{(-1-i)}{2^{1/2}3^{5/4}}\left\{\frac12\ln\left(\left(z+\frac{3^{1/4}}{2^{1/2}}\right)^2+\frac{\sqrt3}2\right)-\\ \frac12\ln\left(\left(z-\frac{3^{1/4}}{2^{1/2}}\right)^2+\frac{\sqrt3}2\right)+\\ i\,\text{atan2}\left(-\frac{3^{1/4}}{2^{1/2}},\frac{3^{1/4}}{2^{1/2}}+z\right)\\ -i\,\text{atan2}\left(-\frac{3^{1/4}}{2^{1/2}},\frac{3^{1/4}}{2^{1/2}}-z\right)\right\}+C_2\end{align}$$ We rotate to $$\text{atan2}\left(-\frac{3^{1/4}}{2^{1/2}},\frac{3^{1/4}}{2^{1/2}}-z\right)=-\frac{\pi}2+\text{atan2}\left(\frac{3^{1/4}}{2^{1/2}}-z,\frac{3^{1/4}}{2^{1/2}}\right)=-\frac{\pi}2+\tan^{-1}\left(\frac{\frac{3^{1/4}}{2^{1/2}}-z}{\frac{3^{1/4}}{2^{1/2}}}\right)$$ $$\text{atan2}\left(-\frac{3^{1/4}}{2^{1/2}},\frac{3^{1/4}}{2^{1/2}}+z\right)=-\frac{\pi}2+\text{atan2}\left(\frac{3^{1/4}}{2^{1/2}}+z,\frac{3^{1/4}}{2^{1/2}}\right)=-\frac{\pi}2+\tan^{-1}\left(\frac{\frac{3^{1/4}}{2^{1/2}}+z}{\frac{3^{1/4}}{2^{1/2}}}\right)$$ Since we need only twice the real part of the $k=1$ term the overall integral is $$\begin{align}\int\frac x{(1-x^3)\sqrt{1-x^2}}dx&=\frac1{3z}-\\ &\frac1{2^{1/2}3^{5/4}}\left\{\ln\left(\left(z+\frac{3^{1/4}}{2^{1/2}}\right)^2+\frac{\sqrt3}2\right)-\ln\left(\left(z-\frac{3^{1/4}}{2^{1/2}}\right)^2+\frac{\sqrt3}2\right)\right\}+\\ &\frac{2^{1/2}}{3^{5/4}}\left\{\tan^{-1}\left(\frac{\frac{3^{1/4}}{2^{1/2}}+z}{\frac{3^{1/4}}{2^{1/2}}}\right)-\tan^{-1}\left(\frac{\frac{3^{1/4}}{2^{1/2}}-z}{\frac{3^{1/4}}{2^{1/2}}}\right)\right\}+C\end{align}$$ Where again $z=\sqrt{\frac{1-x}{1+x}}$. Verified by numerical quadrature. There seemed to be a bit of an advantage in getting to Robert Israel's substitution in two steps like this because in was dead easy to expand in partial fractions. Of course it was quite bloody to walk those complex constants back out through the logarithms. Wolfram|Alpha doesn't seem to clean up the messes it makes very efficiently.

EDIT: After receiving a downvote I decided to look for avenues of improvement. I may have been too routine in my response in that normally the domain of $z$ would be $(-\infty,\infty)$ and so the two arctangents could not be combined into one because their difference could range from $-\pi$ to $\pi$. But here the domain of $z$ is $(0,\infty)$ so the two arctangents can be combined. Also the primitive should be defined at $x=-1$ so we need to express it in terms of $w=\frac1z=\sqrt{\frac{1+x}{1-x}}$, thus coming around full circle to David H's recommendation after $3$ steps! With $r=\frac{3^{1/4}}{2^{1/2}}$, $$\begin{align}\tan^{-1}\left(\frac{r+z}r\right)-\tan^{-1}\left(\frac{r-z}r\right)&=\tan^{-1}\left(\frac{2\frac zr}{1+\frac{r^2-z^2}{r^2}}\right)=\tan^{-1}\left(\frac{2rz}{2r^2-z^2}\right)\\ &=\cot^{-1}\frac{2r^2-z^2}{2rz}=\frac{\pi}2-\tan^{-1}\left(\frac{2r^2-z^2}{2rz}\right)\\ &=\frac{\pi}2+\tan^{-1}\left(\frac{z^2-2r^2}{2rz}\right)=\frac{\pi}2+\tan^{-1}\left(\frac{\frac1{2r^2}-\frac1{z^2}}{\frac1{rz}}\right)\\ &=\frac{\pi}2+\tan^{-1}\left(\frac{\frac1{\sqrt3}-w^2}{\frac{2^{1/2}}{3^{1/4}}w}\right)\end{align}$$ And we can clean up the logarithms a little, too: $$\begin{align}\ln\left((z\pm r)^2+r^2\right)&=\ln\left(z^2\pm2rz+2r^2\right)=\ln(2r^2z^2)+\ln\left(\frac1{2r^2}\pm\frac1{rz}+\frac1{z^2}\right)\\ &=\ln(2r^2z^2)+\ln\left(\left(\frac1z\pm\frac1{2r}\right)^2+\frac1{4r^2}\right)\\ &=\ln(2r^2z^2)+\ln\left((w\pm12^{-1/4})^2+12^{-1/4}\right)\end{align}$$ So our final expression for the integral has become $$\begin{align}\int\frac x{(1-x^3)\sqrt{1-x^2}}dx&=\frac13w+\frac1{2^{1/2}3^{5/4}}\ln\left(\frac{(w-12^{-1/4})^2+12^{-1/2}}{(w+12^{-1/4})^2+12^{-1/2}}\right)+\\ &\frac{2^{1/2}}{3^{5/4}}\tan^{-1}\left(\frac{\frac1{\sqrt3}-w^2}{\frac{2^{1/2}}{3^{1/4}}w}\right)+C_3 \end{align}$$ With $w=\sqrt{\frac{1+x}{1-x}}$, checked by comparison with the previous expression. Certainly this is an improvement. I would be curious to see the amount of work required if that substitution were made in the first place.

  • $\begingroup$ Why did I pick up a downvote? Doesn't everyone just love messy integrals like this? $\endgroup$ Apr 23 '16 at 6:08
  • $\begingroup$ OK, maybe what I had was in fact worthy of a downvote. Now the expression for the primitive has been cleaned up and $x=-1$ has been added to the domain. Thanks for the wake-up call. $\endgroup$ Apr 24 '16 at 0:51

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