# Help me find the arc length of a parametric equation...

1. May 1, 2017

1. The problem statement, all variables and given/k
nown data

$x = (sin(t))^2$ $y = (cos(t))^2$ t goes from 0 to 3 pi

2. Relevant equations
∫$\sqrt{ {(dx/dt)}^2 + {(dy/dt)}^2 } dt$

3. The attempt at a solution
∫$\sqrt{ {(2sin(t)cost)}^2 + {(-2cos(t)sin(t))}^2 } dt$

∫$\sqrt{ 4(sin(t)cos(t))^2 + {(-2cos(t)sin(t))}^2 } dt$

∫$\sqrt{ 4(sin(t)cos(t))^2 + 4(cos(t)sin(t))^2 } dt$

∫$\sqrt{ 8(sin(t)cos(t))^2 } dt$

$2\sqrt 2$∫$\sqrt{ (sin(t)cos(t))^2 } dt$

$2\sqrt 2$∫${ (sin(t)cos(t)) } dt$

keep in mind my bounds are from 0 to 3 pi right now

so I integrate this using u = sin (t) du = cos(t) dt
and get
(root 2)(sin(t))^2 from 0 to pi/3

now when I evaluate I get 0-0

where did my integration go wrong?

2. May 1, 2017

### Ray Vickson

Never, never just write $\sqrt{X^2} = X$ because half the time it is wrong. What IS always correct is $\sqrt{X^2} = |X|.$

3. May 1, 2017

so it should be absolute value sin(t)cos(t) ??

4. May 1, 2017

### Ray Vickson

I thought that was what I said.

5. May 2, 2017

So where do I go from here? can I just call that the absolute value of 2sin since sin and cos have the same range?

6. May 2, 2017

### Ray Vickson

I have already offered all the help I feel I can according to PF rules.

7. May 2, 2017