# Integration and trig

1. Dec 25, 2005

### 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:

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

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

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

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?

Last edited: Dec 25, 2005
2. Dec 25, 2005

### d_leet

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

3. Dec 25, 2005

### 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.

Last edited: Dec 26, 2005
4. Dec 25, 2005

### Alkatran

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