Polar Arc Length: Solve Integral of r=6cos6θ

[spinnaker]

Homework Statement

Find the arc length of one of the leaves of the polar curve r= 6 cos 6θ.

Homework Equations

L = ∫sqrt(r^2 + (dr/dθ)^2) dθ
(I use twice that since the length from 0 to π/12 is only half the petal)

The Attempt at a Solution

I seem to get an integral that can't be solved:

L = 2∫sqrt((6 cos 6θ)^2 + (-36 sin 6θ)^2) dθ
= 2∫sqrt(36 cos^2 6θ + 1296 sin^2 6θ) dθ

I simplify the cos^2 and sin^2 to get

L = 2∫sqrt(36 + 1260 sin^6θ) dθ
= 12∫sqrt(1+35 sin^2 6θ) dθ

but that's where I'm stuck. I have no idea how to do that integral. Any help would be sincerely appreciated!

 

[ShayanJ]

You can't do that analytically.
Here's what www.wolframalpha.com says:
<br /> \int_0^{\frac{\pi}{12}} \sqrt{1+35 \sin^2{6\theta}}d\theta=\frac{E(-35)}{6}\approx1.0375<br />
Where E(m) is the Complete Elliptic Integral of the Second Kind.

 

[spinnaker]

But there's got to be some sort of trick to make this easy to work with - nothing in the notes or textbook has this Elliptic Integral thing in it.

Any ideas?