Help with integral

[SomeRandomGuy]

Can anyone help me through the following integral? I already know the solution, the process of finding it is what I am concerned with.
(Not sure what the proper notation is for writing this on a forum)

Int(sqrt(a^2sin^2(t)+b^2cos^2(t))dt from t = 0, t = 2pi

I tried simplifying by dividing through with a bcos(t) so the sqrt would be tan^2+1 and so forth. This really didn't get me anywhere, however. All help is appreciated.

[Galileo]

Just split the integral in two and use the identities:
sin^2(x)=1/2(1-cos(2x))
cos^2(x)=1/2(1+cos(2x))

or notice that \int_0^{2\pi}sin^2xdx=\int_0^{2\pi}cos^2xdx and use cos^2x+sin^2x=1 for a quick evaluation.

[SomeRandomGuy]

Thanks for your reply, however, I think your incorrect. First, I didn't mention that a and b are some arbitrary constants. Secondly, the integral isn't a^2sin^2(t)+ b^2cos^2(t), it's the square root of that quantity, so splitting the integral doesn't apply here, I believe. I tried half angle formula's and that didn't seem to get me anywhere, either. By the way, incase anyone is interested, this integral is suppose to yield the circumference of an ellipse.

[arildno]

"By the way, incase anyone is interested, this integral is suppose to yield the circumference of an ellipse."

Sure enough, and no one today has found an analytical solution to that problem.

[SomeRandomGuy]

Isn't it suppose to be expressed as an infinite series? Like I said earlier, I was looking for the process of how to get that answer.

[Galileo]

The integral isn't a^2sin^2(t)+ b^2cos^2(t), it's the square root
Ah, I overlooked the 'sqrt' part.

The integral represents the arc lenght of the curve with parametric equations:
x=acos(t)
y=bsint(t)

Which is, as you said, an ellipse. The parametric equation can be written as an equation in x and y alone:
\left(\frac{x}{a}\right)^2+\left(\frac{y}{b}\right )^2=1
which is the general equation of an ellipse with eccentricity e=\sqrt{1-\frac{b^2}{a^2}}.

Exact expressions exists. This one is by MacLaurin (in 1742):
P=2a\pi\sum_{n=0}^{\infty}\left(\frac{-1}{(2n-1)}\right)\left(\frac{(2n)!}{(2^n n!)^2}\right)^2e^{2n}
where e is the eccentricity.

[SomeRandomGuy]

Thanks for your reply. That seems to make some sense.