# Help with integral

1. Oct 29, 2004

### 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.

2. Oct 30, 2004

### 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.

3. Oct 30, 2004

### 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.

4. Oct 30, 2004

### 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.

5. Oct 30, 2004

### 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.

6. Oct 30, 2004

### Galileo

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.

Last edited: Oct 30, 2004
7. Oct 30, 2004