Solving a Length of Curve with 0<t<2π: Find the Answer

[manenbu]

Homework Statement

Find the length of the curve:
x = 2a cos t - a cos 2t
y = 2a sin t - a sin 2t

0<t<2π

Homework Equations

$$L = \int_{C} dl$$

The Attempt at a Solution

Well, after a lot of manipulating and using trig identities, I get this:

$$L= 4a \int_{0}^{2\pi} \sin{t} dt$$
When I solve, I get 0.
But - if I change the range to be not 0<t<2π but 0<t<π, and then multiply by 2 - I get 16a as the answer, which is correct.
My question is why? I can see why it is like that "numerically" but not "philosophically".
How do I identify a situation where I need to do it, without looking at the answers, next time?
And was I doing everything correct? Or maybe it just turned out to be the correct answer by accident?
Thanks!

 

[LCKurtz]

Likely you have a mistake somewhere. I get:

$$ds = \sqrt{x'(t)^2 +y'(t)^2}\ dt = 2\sqrt2 a \sqrt{1 - cos(t)}\ dt$$

which gives the correct answer.

 

[manenbu]

I had that too.
But:
$$2\sqrt2 a \sqrt{1 - cos(t)}\ dt = 2\sqrt2 a \sqrt{2 sin^2(\frac{t}{2}} dt = 4a sin(\frac{t}{2})$$

Assuming that the trig identity is correct, what went wrong here?
The latter form is much easier to integrate.

 

[manenbu]

oh. in the original integral I missed the $\frac{t}{2}$.
nevermind, problem solved.