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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 $\dfrac{3-2\sqrt{2}}{6}$.

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  • $\begingroup$ 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
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By clever substitutions your professor probably meant Euler Substitutions.

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

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


$\text{Explanations:}$

$(2)$ 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

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

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    $\begingroup$ 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
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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.

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  • $\begingroup$ Okay, I'll take your answer into consideration $\endgroup$ – Venus Nov 17 '14 at 20:01
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    $\begingroup$ 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
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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 $$\begin{align}\sec y\frac{3 \tan y}{\sqrt{34-2\cos 2y}}&=\sec y \Big( \frac{\sin 2y}{6\sqrt{34-2\cos 2y}}+\frac{\sqrt{34-2\cos 2y}\tan y}{12}\Big)\\ &=\Big(\frac{d \frac{\sqrt{17-\cos 2y}}{6\sqrt{2}}}{dy}\Big)\sec y + \Big(\frac{d \sec y}{dy}\Big) \frac{\sqrt{17-\cos 2y}}{6\sqrt{2}} \end{align}$$

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  • $\begingroup$ 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
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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*}

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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.

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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 $m=k^2$.

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  • $\begingroup$ 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
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    $\begingroup$ @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
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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.

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  • $\begingroup$ 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

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