Question

Help please?? :|

Answer 1

\[\int\limits_{0}^{2pi} (\sqrt{4\sin^2t + 9\cos^2t}) dt\]

Answer 2

can't figure this one out..
then i checked wolfram

http://www.wolframalpha.com/input/?i=integral+sqrt%284sin%28t%29^2+%2B9cos%28t%29^2%29+dt

Answer 3

How's that solved? :(
We're asked to evaluate the integral from 0 to 2pi. The answer's there but it didn't show how it was solved. :(

Answer 4

It says 12 E(5/9) ~ 15.8654

where E is the complete elliptic integral of the second kind....who knew?

Answer 5

I think we are asked to solve this using line integrals. How's that?

Answer 6

Arc length of an ellipse ?

Answer 7

yes! :)

Answer 8

The complete elliptic integral of the second kind is
int between o and pi/2 sqrt(1-k^2sin^2 theta) dtheta
http://en.wikipedia.org/wiki/Elliptic_integral#Complete_elliptic_integral_of_the_second_kind

Answer 9

@dumbcow @estudier Do you guys know how to use the elliptic integral step by step given that function? Thanks!

Answer 10

First put it in that form so becomes 9-5sin^2

Answer 11

or 1 - 5sin^2/9 where k^2 = 5/9

Answer 12

I have no idea what I am doing here, anyway follow Wikipedia.
So now we have int 0 to 1 of sqrt(1-k^2t^2)/sqrt(1-t^2) dt
Which looks like a doable integral...

Answer 13

So that's where the Wolfram E(5/9) comes from....

Answer 14

Am I talking to myself? :-)

Answer 15

Right so the only thing I am not seeing is that following Wiki we just get E(5/9) and if we feed it in directly we get 12*E(5/9) or 15.8654??