Why does the Pythagorean identity hide parametrization errors in arc length?

[Alkatran]

Let's say I wanted to find the length of the perimeter of a semi-circle using integration with x(t) = cos(t) and y(t) = sin(t). I would do like so:

$$\int_{0}^{\pi} \sqrt{sin(t)^2 + cos(t)^2} dt<br /> = \int_{0}^{\pi} \sqrt{1} dt<br /> = \pi$$

now if I wanted to do it for the whole circle I would do it like so:

$$\int_{0}^{\pi} \sqrt{sin(2t)^2 + cos(2t)^2} dt<br /> = \int_{0}^{\pi} \sqrt{1} dt<br /> = \pi$$

Now obviously I made a mistake: I didn't replace 2x with u (etc...). But my question is: why can't I do this? What rule stops me from eliminating the cos^2 + sin^2 and getting the wrong answer?

 

[d_leet]

Well even if you make the substitution u=2t you would still arrive at an answer of just pi, because you have to account for du = 2dt.

 

[Tide]

If $x = \cos(2t)$ then $dx^2 = 4 \sin^2(2t) dt^2$ and similarly for y so your distance/perimeter formula is incorrect.

 

[Alkatran]

If $x = \cos(2t)$ then $dx^2 = 4 \sin^2(2t) dt^2$ and similarly for y so your distance/perimeter forumula is incorrect.

No wonder I was having trouble doing the problem, I forgot to differentiate x(t) and y(t)!