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### Integration, Help needed!!

Posted: Tue Mar 18, 2014 12:34 am
Please solve this problem.....
$\int \frac {1}{\sqrt{x^4 -1} } dx$

When i used "http://integrals.wolfram.com/" they've given me the following answer:
$\frac {\sqrt{1-x^4} Elliptic F(sin^{-1}x|-1)}{\sqrt{x^4 -1} }$
Please explain...
what does this"Elliptic integral of the first kind" expression means?
what does this mean: $F(sin^{-1}x|-1)$
and how to get this answer?

### Re: Integration, Help needed!!

Posted: Fri Aug 22, 2014 12:06 am
Simple trig subs should do the trick. Put $$x=\sqrt{\sec z}$$ so that $$dx=\frac 1 2 \left(\sin z \sec^{3/2} z\right) dz$$. Put these to get: $\int \dfrac{dx}{\sqrt{x^4-1}}=\int\dfrac{\sin z\sec^{3/2} z}{2\tan z}dz=\int\dfrac{\cos z}{2\cos^{3/2} z}dz=\dfrac{1}{2}\int\sqrt{\sec z}~ dz.$
That simplified a lot. But the last integral can't be expressed by elementary functions. Set $$z=2\theta$$ so that $$dz=2d\theta$$. Then we do some manipulations:
$\dfrac 1 2\int\sqrt{\sec z}~dz=\int\dfrac{d\theta}{\sqrt{\cos 2\theta}}=\int\dfrac{d\theta}{\sqrt{1-2\sin^2\theta}}=\int_0^{\theta} \dfrac{dn}{\sqrt{1-2\sin^2 n}}+\text{C}=F\left(\theta\mid 2\right)+\text{C}.$
Here $$F\left(\phi \mid k^2\right)=\displaystyle\int_0^\phi \dfrac{d\theta}{\sqrt{1-k^2\sin^2\theta}}$$ is the incomplete elliptic integral of the first kind. Now just plug everything back:
$F\left(\theta\mid 2\right)+\text{C}=F\left(\frac{z}{2}\mid 2\right)+\text{C}=\boxed{F\left(\frac{1}{2}\sec^{-1} x^2\mid 2\right)+\text{C}.}$
Read here about the elliptic integrals. (http://en.wikipedia.org/wiki/Elliptic_integral)
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