Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am trying to Integrate $$ I=\int \frac{dx}{\sqrt {ax^4-bx^2}}, \qquad a,b\in \mathbb{R}. $$ Thanks. I tried to do $x=\sin \phi$ $$ \int \frac{\cos \phi\, d\phi}{\sqrt{a\sin^4 \phi-b\sin^2 \phi}}=\int \frac{\cot \phi \, b\phi}{\sqrt{a\sin^2\phi-b}} $$ but get stuck here. Mathematica gives a closed form result
$$ I=-\frac{x\sqrt{ax^2-2b}}{\sqrt b \sqrt{ax^4-bx^2}}\tan^{-1}\bigg(\frac{\sqrt{2b}}{\sqrt{ax^2-2b}}\bigg). $$

share|cite|improve this question
Shouldn't you have $$I=-\dfrac{x\sqrt{cx^2-2d}}{\sqrt{d}\sqrt{cx^4-dx^2}}\tan^{-1}\left(\dfrac{\sqr‌​t{2d}}{\sqrt{cx^2-2d}}\right)?$$ – Sanath K. Devalapurkar May 9 '14 at 3:22
The denominator can be rewritten as $x\sqrt{ax^2-b}$. An obvious substitution would be $x=\sqrt{\dfrac ba}\cdot\cosh t$ – Lucian May 9 '14 at 3:38

Hint: try $x= \sqrt{\frac{b}{a}}\tan{\theta}$.

share|cite|improve this answer

You have been given two possible changes of variable. A key point was also mentioned by Lucian for a rewrite of the denominator. After all of that, you should arrive to something much simpler that what Mathematica gave you (you did not simplify its result) since $$I=\int \frac{dx}{\sqrt {ax^4-bx^2}}=-\frac{\tan ^{-1}\left(\frac{\sqrt{b}}{\sqrt{a x^2-b}}\right)}{\sqrt{b}}$$

share|cite|improve this answer

$$ \begin{aligned} \int\frac{\mathrm{d}x}{\sqrt{ax^4 - bx^2}}&=\int\frac{\mathrm{d}x}{\sqrt{bx^4\left(\frac{a}{b} - \frac{1}{x^2}\right)}}\\ &=\frac{1}{\sqrt{b}}\int\frac{1}{\sqrt{\frac{a}{b} - \left(\frac{1}{x}\right)^2}}\frac{1}{x^2}\,\mathrm{d}x \end{aligned} $$ Now, set $u=\dfrac{1}{x}$ and $\mathrm{d}u=-\dfrac{1}{x^2}\,\mathrm{d}x$: $$ I=\frac{1}{\sqrt{b}} \int-\frac{1}{\sqrt{\left(\sqrt{\frac{a}{b}}\right)^{\!2} - u^2}}\,\mathrm{d}u $$ We have the integral $$ \int-\frac{\mathrm{d}u}{\sqrt{\alpha^2 - u^2}}=\arccos\left(\frac{u}{\alpha}\right)+C $$ Then, $$ \begin{aligned} I&=\frac{1}{\sqrt{b}}\arccos\left(u\sqrt{\frac{b}{a}}\right)+C\\ &=\frac{1}{\sqrt{b}}\arccos\left(\frac{1}{x}\sqrt{\frac{b}{a}}\right)+C \end{aligned} $$

share|cite|improve this answer


$\therefore x^2-\dfrac{b}{a}=t^2 \implies x\ dx=t\ dt$

$\therefore \displaystyle\int\dfrac{dx}{x\sqrt{x^2-\dfrac{b}{a}}}=\displaystyle\int\dfrac{dt}{\left(t^2+\dfrac{b}{a}\right)}$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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