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Integration with a square root

  • Thread starter Molecular
  • Start date
  • #1
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Homework Statement


The problem is really about finding the length of a parametrically defined curve, this in itself seems easy enough, but it's the integration part that's hard.
The problem is as follows:
x = cost
y = t + sint
t = 0.....pi


The Attempt at a Solution



So in order to find the length of this curve I need to find the integral from 0 to pi of:
sqrt((dy/dt)^2 + (dx/dt)^2)
dy/dt = 1 + cost
dx/dt = -sint

So I get the integral of:
sqrt((-sint)^2 + (1+cost)^2)) =
sqrt((sint)^2 + (cost)^2 + 2cost + 1)
(sint)^2 + (cost)^2 = 1, so I get the integral of:
sqrt(2+2cost).

And to be quite frank this has left me completely stumped. Googling around the internet I've read some stuff about trigonometric integration but there's nothing in the whole book about this, so I doubt that's how I'm supposed to solve it. I'm thinking I might be supposed to find a way to rewrite 2+2cost so that I can integrate but it has been to no avail.

A push in the right direction would definately be appreciated.
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,258
618
1+cos(t) has a nice expression in terms of a half angle formula.
 
  • #3
492
1
you can use the double-angle formula: [tex]\cos^2 t=\frac{1+\cos 2t}{2}[/tex]

from there the integral is just a |cos (some multiple of t)| and you will need to split up the integral into 2 I think to get rid of the absolute value.
 

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