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:
\int_a^b \sqrt{1+4Cos^2(2x)}dx
As per Tom, first let \theta=2x. This gives:
\frac{1}{2}\int_{2a}^{2b}\sqrt{1+4Cos^2(\theta)}d\theta
Letting:
Cos^2(\theta)=1-Sin^2(\theta)
\frac{1}{2}\int_{2a}^{2b}\sqrt{1+4(1-Sin^2(\theta))}d\theta
or:
\frac{1}{2}\int_{2a}^{2b}\sqrt{5-4Sin^2(\theta)}d\theta
Pulling the 5 out of the radical gives us:
\frac{\sqrt{5}}{2}\int_{2a}^{2b}\sqrt{1-\frac{4}{5}Sin^2(\theta)}d\theta
But by definition:
\int_0^u \sqrt{1-mSin^2(\theta)}d\theta=EllipticE[u,m]
we can write the definite integral as:
<br />
\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}<br />
And so we can say:
\int \sqrt{1+4Cos^2(2x)}dx=\frac{\sqrt{5}}{2}\text{EllipticE}[2x,4/5]
And just cause I like some concrete evidence of all this I calculated:
\int_2^5 \sqrt{1+Cos^2(2x)}dx
both using EllipticE functions and numerically. Both return 5.01464.
See Flux, I have trouble with this stuf too you know. 
