How does one evaluate the following integral?

$$\int_0^1 \frac{x^3}{2(2-x^2)(1+x^2) + 3\sqrt{(2-x^2)(1+x^2)}}\,\mathrm dx$$

This is a homework problem and I have been evaluating this integral for hours yet no success so far. I have tried to rationalize the integrand by multiplying it with $$\frac{2(2-x^2)(1+x^2) - 3\sqrt{(2-x^2)(1+x^2)}}{2(2-x^2)(1+x^2) - 3\sqrt{(2-x^2)(1+x^2)}}$$ but the integrand is getting worse. I have tried to use trigonometric substitutions like $x=\tan\theta$ and $x=\sqrt{2}\sin\theta$, but I cannot rid off the square root form. I have also tried to use hyperbolic trigonometric substitutions but the thing does not get any easier neither also substitution $y=x^2$ nor $y=\sqrt{(2-x^2)(1+x^2)}$. Using integration by parts is almost impossible for this one. I have also tried to use the tricks from this thread, but still did not get anything. No clue is given. My professor said, we must use clever substitutions but I cannot find them. Any idea or hint? Any help would be appreciated. Thanks in advance.

Edit :

The answer I got from my Prof is $\dfrac{3-2\sqrt{2}}{6}$.

  • $\begingroup$ Hello, Venus. IIRC in the original version, posted on November 17, the integrand had a factor $x^2$ (instead of $x^3$) in the numerator. Apparently that was a mistake. If so, that must have been the very reason that no-one here (me included) for a long time could help you solve the problem. I would very much like to know when you edited your question, thereby allowing your good friend Integrator to "solve" the problem. $\endgroup$ – M. Wind Dec 22 '14 at 4:22
  • $\begingroup$ I'm really sorry. I've made a mistake in the previous post. Please don't take this personally & sorry I didn't tell you because I've edited my problem. I'll take this as a lesson for the improvement of myself in the future. $\endgroup$ – Venus Dec 22 '14 at 5:06

By clever substitutions your professor probably meant Euler Substitutions.

$$\begin{align} I &=\int_0^1 \frac{x^3}{2(2-x^2)(1+x^2) + 3\sqrt{(2-x^2)(1+x^2)}}\,\mathrm dx\tag{1}\\ &=\frac12\int_0^1 \frac{t}{2(2-t)(1+t) + 3\sqrt{(2-t)(1+t)}}\,\mathrm dt\tag{2}\\ &=\frac13\int_{\sqrt2}^{1/\sqrt2} \frac{u^2-2}{(1+u)^2(u^2+1)}\,\mathrm du\tag{3}\\ &=\frac13\left[\int_{\sqrt2}^{1/\sqrt2} \frac{3 u}{2 \left(u^2+1\right)}\,\mathrm du -\int_{\sqrt2}^{1/\sqrt2} \frac{3}{2 (u+1)}\,\mathrm du -\int_{\sqrt2}^{1/\sqrt2} \frac{1}{2 (u+1)^2}\,\mathrm du\right]\tag{4}\\ &=\frac13\left[\frac{3}{4} \log \left(u^2+1\right)-\frac{3}{2} \log (u+1)+\frac{1}{2 (u+1)}\right]_{\sqrt2}^{1/\sqrt2}\tag{5}\\ &=\frac{1}{3}\Bigg[\frac{3-2 \sqrt{2}}{2}\Bigg]\tag{6}\\ \end{align}$$

$$\large\int_0^1 \frac{x^3}{2(2-x^2)(1+x^2) + 3\sqrt{(2-x^2)(1+x^2)}}\,\mathrm dx=\frac{1}{6}\left(3-2 \sqrt{2}\right)$$


$(2)$ Substitute $x^2=t\iff2x\,\mathrm dx=\,\mathrm dt$

$(3)$ Using Type $\rm III$ Euler Substitution

$$\small \sqrt{(2-t)(t+1)}=(t+1)u \iff t+1=\frac{3}{u^2+1} \iff t=\frac{2-u^2}{u^2+1} \iff \,\mathrm dt=-\frac{6 u}{\left(u^2+1\right)^2}\,\mathrm du$$

$(4)$ Using Partial Fraction Decomposition

$$\small\frac{2-u^2}{(1+u)^2(u^2+1)}=-\frac{3 u}{2 \left(u^2+1\right)}+\frac{3}{2 (u+1)}+\frac{1}{2 (u+1)^2}$$

  • 2
    $\begingroup$ Finally, you post your answer. Thanks. (+1) $\endgroup$ – Venus Dec 16 '14 at 9:02
  • $\begingroup$ @Venus Glad you liked it! $\endgroup$ – Aditya Hase Dec 16 '14 at 9:08

Perhaps the most promising approach is to rewrite the term $(2-x^2)(1+x^2)$ as follows:

$$(2 - x^2)(1 + x^2) = 2 + x^2 - x^4 = \frac{9}{4} - \left(x^2 - \frac{1}{2}\right)^2 = \frac{9}{4}\left(1 - \frac{4}{9}\left(x^2 - \frac{1}{2}\right)^2 \right)$$

This observation leads one to consider the substitution $y = \frac{2}{3}\left(x^2-\frac{1}{2}\right)$.

Result of the substitution is that the denominator cleans up considerably, becoming $(1 - y^2) + \sqrt{1 - y^2}$. Later on in the evaluation setting $y = \sin t$ may well be a promising idea.

  • $\begingroup$ Okay, I'll take your answer into consideration $\endgroup$ – Venus Nov 17 '14 at 20:01
  • 1
    $\begingroup$ I have the strong impression that the problem, which now appears to be satisfactorily solved, is in fact not the same as the one you posted a month ago!!! I am pretty sure that in the original version posted by you the numerator was $x^2$. Now it is $x^3$, which is obviously much nicer, e.g. in view of the $x^2 -> t$ substitution. $\endgroup$ – M. Wind Dec 21 '14 at 15:02

Let $x=\sin y$, and note that $(2-x^2)(1+x^2)$ simplifies to $2+\sin^2y\cos^2y$. Then integral becomes $$I=\int_0^{\pi/2}\frac{\sin^3y\cos y}{2(2+\sin^2y\cos^2y)+3\sqrt{2+\sin^2y\cos^2y}}dy.$$ Now set $y=z-\frac{\pi}{2}$, to observe that $$I=I_0=\int_0^{\pi/2}\frac{\cos^3y\sin y}{2(2+\sin^2y\cos^2y)+3\sqrt{2+\sin^2y\cos^2y}}dy.$$ With $\cos^3y\sin y+\cos y\sin^3 y=\cos y\sin y$ you obtain (with $2I=I+I_0$) $$\begin{align}I&=\frac{1}{2}\int_0^{\pi/2}\frac{\sin y\cos y}{2(2+\sin^2y\cos^2y)+3\sqrt{2+\sin^2y\cos^2y}}dy\\ &=\frac{1}{2}\int_0^{\pi/2}\frac{\sin y}{8+\sin^2y+3\sqrt{8+\sin^2y}}dy\end{align}$$ The rest should be manageable.

Addendum: $$\begin{align}I&=\int_0^{\pi/2}\frac{2\sin y}{17-\cos 2y+3\sqrt{34-2\cos2y}}dy\\ &=\int_0^{\pi/2}\Big(\frac{3\sec y \tan y}{\sqrt{34-2\cos 2y}}-\frac{1}{2}\sec y \tan y \Big) dy\\ &=0-\lim_{y\rightarrow 0}(-\frac{1}{2}\sec y+\frac{\sqrt{17-\cos 2y}\sec y}{6\sqrt{2}}) \end{align}$$ which results in what we need. The first part of the second integral is clear. For the the second part note that $$\begin{align}\sec y\frac{3 \tan y}{\sqrt{34-2\cos 2y}}&=\sec y \Big( \frac{\sin 2y}{6\sqrt{34-2\cos 2y}}+\frac{\sqrt{34-2\cos 2y}\tan y}{12}\Big)\\ &=\Big(\frac{d \frac{\sqrt{17-\cos 2y}}{6\sqrt{2}}}{dy}\Big)\sec y + \Big(\frac{d \sec y}{dy}\Big) \frac{\sqrt{17-\cos 2y}}{6\sqrt{2}} \end{align}$$

  • $\begingroup$ Could you please elaborate a bit more because the rest doesn't seem manageable to me? $\endgroup$ – Venus Dec 15 '14 at 17:09
  • $\begingroup$ @Venus: it is managable: substitute $\cos y = t$, then get rid of the root in the denominator and you are left with two terms, each of which has a simple antiderivative. $\endgroup$ – user111187 Dec 15 '14 at 18:09
  • $\begingroup$ @Venus: See Addendum. $\endgroup$ – Math-fun Dec 16 '14 at 8:36
  • $\begingroup$ @Mehdi Thank you very much for adding the details. I appreciate it. (+1) $\endgroup$ – Venus Dec 16 '14 at 9:01
  • $\begingroup$ @ Venus: you are welcome! $\endgroup$ – Math-fun Dec 16 '14 at 9:35

Let us first substitute $y = x^{2} + 1$. Then

\begin{align*} I &:= \int_{0}^{1} \frac{x^{3}}{2(2-x^{2})(1+x^{2}) + 3\sqrt{(2-x^{2})(1+x^{2})}} \, dx \\ &= \frac{1}{2} \int_{1}^{2} \frac{y - 1}{2y(3-y) + 3\sqrt{y(3-y)}} \, dy \tag{1} \end{align*}

Using the substitution $y \mapsto 3-y$, it follows that

$$ I = \frac{1}{2} \int_{1}^{2} \frac{2-y}{2y(3-y) + 3\sqrt{y(3-y)}} \, dy. \tag{2} $$

Thus adding (1) and (2) dividing by 2, we obtain

\begin{align*} I &= \frac{1}{4} \int_{1}^{2} \frac{dy}{2y(3-y) + 3\sqrt{y(3-y)}} = \frac{1}{12} \left[ \frac{3 - 2\sqrt{(3-y)y}}{2y-3} \right]_{1}^{2} = \frac{3-2\sqrt{2}}{6}. \end{align*}

Of course this may not be a comprehensive answer. My original approach utilized a chain of substitutions:

\begin{align*} I &= \frac{1}{4} \int_{1}^{2} \frac{dy}{2y(3-y) + 3\sqrt{y(3-y)}} \\ &= \frac{1}{2} \int_{3/2}^{2} \frac{dy}{2y(3-y) + 3\sqrt{y(3-y)}}, \qquad (\because \text{ by symmetry}) \\ &= \frac{1}{2} \int_{0}^{1} \frac{dy}{(9-s^{2}) + 3\sqrt{9-s^{2}}}, \qquad (s = 2x-3) \\ &= \frac{1}{6} \int_{0}^{\arcsin(1/3)} \frac{d\theta}{1+\cos\theta}, \qquad (s = 3\sin\theta) \\ &= \frac{1}{6} \int_{0}^{3-2\sqrt{2}} dt, \qquad (t = \tan(\theta/2)) \\ &= \frac{3-2\sqrt{2}}{6}. \end{align*}


Here is what I have so far, so hopefully someone can take it further.. The above integral has a form of $$ -\frac{1}{8}\int \frac{df}{dx}\frac{dg}{dx}dx $$ Where $$ f(x) = \ln\left(1-2x^2\right)\\ g(x) = \ln\left(\frac{2}{3}\sqrt{\left(2-x^2\right)\left(1+x^2\right)} +1\right) $$

But as always double check the manipulation.


If you consider these imaginary parts of elliptic integrals as closed-forms, then

$$\frac{1}{2}-\frac{\sqrt 2}{3}-\frac{\sqrt{2} \log \left(\sqrt{2}-1\right)}{8}+\frac{\sqrt{2} \log \left(\sqrt{2}+1\right)}{8}-\frac{\sqrt 2}{8}\Im\left(F\left(i \log\left(\sqrt 2+1\right)|-\tfrac{1}{2}\right)\right)+\frac{\sqrt{2}}{6}\Im\left(E\left(i \log\left(\sqrt 2+1\right)|-\tfrac{1}{2}\right)\right)-\frac{3\sqrt{2}}{8 }\Im\left(\Pi \left(-2;i \log\left(\sqrt 2+1\right)|-\tfrac{1}{2}\right)\right),$$

where $F\left(x\,|\,m\right)$ is the elliptic integral of the first kind, $E\left(x\,|\,m\right)$ is the elliptic integral of the second kind, and $\Pi\left(n; x \, | \, m\right)$ is the elliptic integral of the third kind with parameters $m=k^2$.

  • $\begingroup$ There doesn't appear to be any imaginary portions of the given integral; how does this answer apply? $\endgroup$ – abiessu Nov 21 '14 at 4:50
  • $\begingroup$ @abiessu: The imaginary part of a complex number is a real quantity. $\endgroup$ – Lucian Nov 24 '14 at 3:38
  • $\begingroup$ @Lucian: True, but the answer still doesn't show any way to get from the given integral to the very long result... $\endgroup$ – abiessu Nov 24 '14 at 15:41
  • 2
    $\begingroup$ @abiessu Nobody said that this is the best answer. I've worked on it and I think the answer is correct, but I'm also waiting for a better answer. Feel free to post your solution. Until that time I think this is more than nothing. By the way there are many situation when we have to use imaginary parts in real solutions. $\endgroup$ – user153012 Nov 24 '14 at 17:37

It's easy. Differentiate

$$\frac{1}{4 (-1 + 2 x^2)} - \frac{\sqrt{2 + x^2 - x^4}}{6 (-1 + 2 x^2)} - {1\over 4} \log[3 + 2 \sqrt{2 + x^2 - x^4}]$$

To get your integral.

  • $\begingroup$ I don't think this answers how to evaluate (integrate) the given integral. It merely states a proposed answer, and suggests the OP verify by differentiation. $\endgroup$ – Namaste Dec 16 '14 at 14:07

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.