Are you sure you've posted the correct integral? I get that this answer to this is 2 EllipticE[3/4] (I used pi/2 rather than 1.57)
where EllipticE(x) is the complete elliptic integral of the second kind.
If you can get this result w/o a table or a calculator you are more accomplished than I am.
Hey Romsek, this is an arc length problem for a polar equation r=3(sin (2x)) from 0 to 2times pi. I get integrand
(9sin^2 (2x))+(36cos^2(2x))^1/2 which gives 3 times integrand (sin^2(2x))+4cos^2(2x))1/2...(under square root sign)
The book says solve with a calculator but I was wondering if it can be done by just humans.
If they want a decimal answer then I'd suspect a calculator would be allowed. Especially with that probable pi/2 in the upper limit of the integration.
Hint: Use sin^2(2x) + cos^2(2x) = 1 under the square root. Then let y = 2x. Does the integrand look more familiar?
Hey Topsquark, do you mean factor out that expression or do integrand (sin^2(2x) + cos^2(2x) + 3cos^2(2x))^1/2 equals
integrand (1+ 3cos^2(2x))^1/2 ?
not this human... I can't speak for all humans.
It doesn't have a simpler form than the complete elliptic integral of the second kind. You'd have to be able to calculate that in your head. I suppose it's possible there's one of those numeric savants out there that could do this.
Technically the complete elliptic integral of the second kind is just a function like sine or cosine or exponential etc. We can't compute those in our heads either and I've been alive long enough to remember when we had to look sines and cosines up in tables. So despite it's fancy name it's just another function. It's just not frequently enough encountered to get commonly taught. You can bet Kepler probably had tables of it while he was trying to figure out orbits of the planets.