help rewriting an integral

**minivan15** · Sep 15th 2008, 08:53 PM

hey...so my problem is to rewrite the [integral from 0 to pi/2] of $SQRT(1 + (asinx)^2)dx$ in terms of the elliptic integral E(k, phi) = [integral from 0 to phi] $SQRT(1-(ksinu)^2)du$ .

please note i am aware elliptic integrals are unsolvable in closed form...i am not looking for a solution, I am just attempting to rewrite the integral to make it look like the form of the elliptic integral E

using the substitution tanu = asinx, I have worked it out to be: [integral from 0 to arctan(asinx)] $du/((cosu)^2*SQRT(a^2 - (a^2 + 1)(sinu)^2))$

but I can't figure out where to go from here. Please help!

**CaptainBlack** · Sep 15th 2008, 11:21 PM

Originally Posted by minivan15

hey...so my problem is to rewrite the [integral from 0 to pi/2] of $SQRT(1 + (asinx)^2)dx$ in terms of the elliptic integral E(k, phi) = [integral from 0 to phi] $SQRT(1-(ksinu)^2)du$ .

Assuming no typos:

$E(a{\bf{i}},\pi/2)=\int_0^{\pi/2}\sqrt{1 + (a\sin(x))^2}dx$

an alterative approach if you don't like the imaginary argument is to replace $\sin(x)$ by $\cos(x-\pi/2)$ and use the identity $\cos^2(x-\pi/2)=1-\sin^2(x-\pi/2)$

RonL

**minivan15** · Sep 16th 2008, 01:45 PM

thanks, but unfortunately I need to solve it a specific way, because I'll use the results in later problems. I've worked it down to:

subst. $tanu = asinx$

so $x = arcsin((tanu)/a)$

and dx = $1/ ROOT(1-((tanu)/a)^2) * 1/a(cosu)^2 * du$

the integral becomes: [integral from 0 to arctan(a)] $du / ((cosu)^2 * ROOT((acosu)^2 - (sinu)^2))$

(after a few simple steps of manipulation)

I need to finish it with this method to solve the problem, but am stuck here. Can someone please help? I have been stuck for 3 days and it's driving me crazy!!

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