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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
$$ I=-\frac{x\sqrt{ax^2-2b}}{\sqrt b \sqrt{ax^4-bx^2}}\tan^{-1}\bigg(\frac{\sqrt{2b}}{\sqrt{ax^2-2b}}\bigg). $$

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Shouldn't you have $$I=-\dfrac{x\sqrt{cx^2-2d}}{\sqrt{d}\sqrt{cx^4-dx^2}}\tan^{-1}\left(\dfrac{\sqr‌​t{2d}}{\sqrt{cx^2-2d}}\right)?$$ –  SDevalapurkar May 9 at 3:22
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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 at 3:38

3 Answers 3

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

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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 $$I=\int \frac{dx}{\sqrt {ax^4-bx^2}}=-\frac{\tan ^{-1}\left(\frac{\sqrt{b}}{\sqrt{a x^2-b}}\right)}{\sqrt{b}}$$

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$$ \begin{aligned} \int\frac{\mathrm{d}x}{\sqrt{ax^4 - bx^2}}&=\int\frac{\mathrm{d}x}{\sqrt{bx^4\left(\frac{a}{b} - \frac{1}{x^2}\right)}}\\ &=\frac{1}{\sqrt{b}}\int\frac{1}{\sqrt{\frac{a}{b} - \left(\frac{1}{x}\right)^2}}\frac{1}{x^2}\,\mathrm{d}x \end{aligned} $$ 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, $$ \begin{aligned} I&=\frac{1}{\sqrt{b}}\arccos\left(u\sqrt{\frac{b}{a}}\right)+C\\ &=\frac{1}{\sqrt{b}}\arccos\left(\frac{1}{x}\sqrt{\frac{b}{a}}\right)+C \end{aligned} $$

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