How to Simplify the Integral in an Arc Length Problem with Parametric Equations?

[dark_omen]

Okay, so I was given the parametric equations of x = (cos(t))^2 and y = cos(t). So I found dy/dt = -sin(t) and dx/dt = -2sin(t)cos(t). This is where I am getting stuck, so I have the L = integral from 0 to 4pi (sqrt((dx/dt)^2+(dy/dt)^2)) , but I don't know how to simplify this to get the answer to the problem. Can anyone help, thanks!

 

[George Jones]

Take the square root of the common factor of sin^t, and then integrate either by substitution or by inspection.

Regards,
George

 

[George Jones]

Sorry, forget the "by inspection" part of my previous post.

You end up with

\int_{0}^{4\pi} \sqrt{1 + 4 \cos^2 t} \sin t dt,

right?

I don't see a fast way to do this integral, but I can see how to do it using a couple of substitutions.

What might be a good first substitution?

Regards,
George

 

[StatusX]

Also, make sure to take the absolute value of the sin(t) you pull out of the square root.

 

[dark_omen]

I took 1 + 4cos(x) as u, and I could integrate that, but when I solved for the length it was 0.
This is what I did:
- 1/4 * integral(sqrt(u^2)du)
- 1/4 * ((u)^2/(2))evaluated @ 0 to 4pi

 

[nrqed]

I took 1 + 4cos(x) as u, and I could integrate that, but when I solved for the length it was 0.
This is what I did:
- 1/4 * integral(sqrt(u^2)du)
- 1/4 * ((u)^2/(2))evaluated @ 0 to 4pi

If u = 1+ 4cos(t), then u^2 would be 1 + 8 cos(t) + 16 cos^2(t) !
So wrong u!

Now, you should take u= cos(t) and you will have something like the integral of sqrt(1+4 u^2)