6
$\begingroup$

How can we evaluate the following integral?

$$ \mathcal{I} = \int \frac{dx}{(1-x^2)\sqrt[4]{(2x^2-1)}} $$


Some notes...

  1. Any standard integration methods (e.g., substitution method, integration by parts, etc.) does not work.

  2. I tried using integral-calculator.com, it does not work either.

  3. Trivially, I tried to set some possible limits (e.g., 0 $\rightarrow$ 1, $-\pi \rightarrow \pi$, etc.) — with no success.

  4. This is not a homework question. I'm open to the ideas given by experts here.

$\endgroup$
3
  • $\begingroup$ wolframalpha.com/… $\endgroup$ Feb 9, 2023 at 11:05
  • $\begingroup$ @SineoftheTime. This is awful ! Do you like it ? Cheers :-) $\endgroup$ Feb 9, 2023 at 11:16
  • $\begingroup$ @ClaudeLeibovici yor're indeed right, it's awful aesthetically. I've just added the link because the OP said that he did not find the antiderivative using a calculator $\endgroup$ Feb 9, 2023 at 11:18

2 Answers 2

9
$\begingroup$

Substitute $t=\sqrt[4]{2x^2-1}$ to simplify the integral first

$$ I= \int \frac{1}{(1-x^2)\sqrt[4]{2x^2-1}}dx = \frac1{\sqrt2}\int \frac{4 t^2}{(1-t^4)\sqrt{1+t^4}}dt $$ and break up the integrand $$ \frac{4 t^2}{(1-t^4)\sqrt{1+t^4}} = \frac{t^2-1}{(1+t^2)\sqrt{1+t^4}}+ \frac{ t^2+1}{(1-t^2)\sqrt{1+t^4}} $$ Then, the two resulting integrals are relatively easy to evaluate \begin{align} I= &\frac1{\sqrt2}\int \frac{t^2-1}{(1+t^2)\sqrt{1+t^4}}+ \frac{ t^2+1}{(1-t^2)\sqrt{1+t^4}}\ dt\\ =&\ \frac12\tan^{-1}\sqrt{\frac{1+t^4}{2t^2}}+ \frac12\coth^{-1}\sqrt{\frac{1+t^4}{2t^2}}+C \end{align} For example $$\int \frac{t^2-1}{(1+t^2)\sqrt{1+t^4}}dt = \int \frac{d(t+\frac1t)}{(t+\frac1t)\sqrt{(t+\frac1{t})^2-2}} $$

$\endgroup$
1
  • $\begingroup$ The key is the second line. Nicely done $\endgroup$ Feb 10, 2023 at 6:19
6
$\begingroup$

Mathematica gives $$I=\frac 12\left(\tanh ^{-1}\left(\frac{x}{\sqrt[4]{2 x^2-1}}\right)+\tan ^{-1}\left(\frac{x}{\sqrt[4]{2 x^2-1}}\right)\right)+C$$

$\endgroup$
6
  • $\begingroup$ Beautiful and symmetric! +1 $\endgroup$ Feb 9, 2023 at 12:00
  • $\begingroup$ @NinadMunshi. I found a mistake in my first answer. By the end, .... Mathematica. Cheers :-) $\endgroup$ Feb 9, 2023 at 12:02
  • $\begingroup$ @NinadMunshi. Even knowing the result, the substitution is not clear. Any idea ? $\endgroup$ Feb 9, 2023 at 12:19
  • $\begingroup$ Under the substitution $u = \sqrt{2x^2-1}$, it may be beneficial to consider the further substitution $$t = \sqrt{u +\frac{1}{u}}$$ $\endgroup$ Feb 9, 2023 at 13:01
  • $\begingroup$ looking at the answer, it is now obvious to apply the substitution $$u=\frac{x}{(2x^2-1)^{1/4}}$$ We can solve for x by taking the 4th power on each side and reducing it to a quadratic in $x^2$ $\endgroup$
    – phi-rate
    Feb 9, 2023 at 14:52

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .