FluxCapacitator said:
Quite frankly, I'm feeling waaaaaaaaay out of depth with this problem. I can understand that some mathematician found a solution to a similar problem as mine, but it was so complex that he just called it EllipticE.
Hey Flux, I was kinda' funnin' about the report. Suppose we have this:
[tex]\int_a^b \sqrt{1+4Cos^2(2x)}dx[/tex]
As per Tom, first let [itex]\theta=2x[/itex]. This gives:
[tex]\frac{1}{2}\int_{2a}^{2b}\sqrt{1+4Cos^2(\theta)}d\theta[/tex]
Letting:
[tex]Cos^2(\theta)=1-Sin^2(\theta)[/tex]
[tex]\frac{1}{2}\int_{2a}^{2b}\sqrt{1+4(1-Sin^2(\theta))}d\theta[/tex]
or:
[tex]\frac{1}{2}\int_{2a}^{2b}\sqrt{5-4Sin^2(\theta)}d\theta[/tex]
Pulling the 5 out of the radical gives us:
[tex]\frac{\sqrt{5}}{2}\int_{2a}^{2b}\sqrt{1-\frac{4}{5}Sin^2(\theta)}d\theta[/tex]
But by definition:
[tex]\int_0^u \sqrt{1-mSin^2(\theta)}d\theta=EllipticE[u,m][/tex]
we can write the definite integral as:
[tex]
\begin{align*}<br />
\frac{\sqrt{5}}{2}\int_{2a}^{2b}\sqrt{1-\frac{4}{5}Sin^2(\theta)}d\theta &= \frac{\sqrt{5}}{2}\left(<br />
\int_{0}^{2b}\sqrt{1-\frac{4}{5}Sin^2(\theta)}d\theta-\int_{0}^{2a}\sqrt{1-\frac{4}{5}Sin^2(\theta)}d\theta<br />
\right) \\<br />
&= \frac{\sqrt{5}}{2}\left(\text{EllipticE}[2b,4/5]-\text{EllipticE}[2a,4/5]\right)<br />
\end{align}[/tex]
And so we can say:
[tex]\int \sqrt{1+4Cos^2(2x)}dx=\frac{\sqrt{5}}{2}\text{EllipticE}[2x,4/5][/tex]
And just cause I like some concrete evidence of all this I calculated:
[tex]\int_2^5 \sqrt{1+Cos^2(2x)}dx[/tex]
both using EllipticE functions and numerically. Both return 5.01464.