Integration, Help needed!!

For students of class 11-12 (age 16+)
oro
Posts: 10
Joined: Thu Oct 24, 2013 11:36 pm

Integration, Help needed!!

Unread post by oro » 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?

Nirjhor
Posts: 136
Joined: Thu Aug 29, 2013 11:21 pm
Location: Varies.

Re: Integration, Help needed!!

Unread post by Nirjhor » 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)
- What is the value of the contour integral around Western Europe?

- Zero.

- Why?

- Because all the poles are in Eastern Europe.


Revive the IMO marathon.

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