Finding the Arc Length of a Polar Function

[jgiarrusso]

Hi, I've been having some issues in solving this problem.

Homework Statement

Find the arc length of r=2/(1-cosθ) from π/2 to π

Homework Equations

L =(integrate) sqrt(r²+(dr/dθ)²)dθ

The Attempt at a Solution

I found (dr/dθ) = (-2sinθ)/(1-cosθ)²

so (dr/dθ)² = (4sin²θ)/(1-cosθ)⁴

Then r² = 4/(1-cosθ)²

both have a factor of 4/(1-cosθ)², so I pulled that outside the sqrt to get

L=(integrate) 2/(1-cosθ) * sqrt(1+(sin²θ)/(1-cosθ)²)

then I multiplied the 1 by (1-cosθ)²/(1-cosθ)² to give common denominators.

After multiplying it out, the numerator of the fraction was 1-2cosθ+cos²θ+sin²θ, so I got rid of the sin and cos and added a 1 to get 2 - 2cosθ

I pulled out a factor of sqrt2 and ended up with:

L = (integrate) 2sqrt2/(1-cosθ) * sqrt(1/(1-cosθ))
or
L= (integrate) 2sqrt2*(1-cosθ)^-3/2

This is where I got stuck. I can't think of any way to integrate that problem using any of the means we have gone over so far.

Some direction in which way to go would be extremely helpful, thanks in advance.

 

[Dick]

Can you think of a way to express 1-cos(theta) as the square of something using a half-angle formula?

 

[jgiarrusso]

Aha, thank you so much. That was just the push I needed!

 

[jgiarrusso]

Hmm, that did allow me to solve the integration (at least, I think I did it properly). But now, when I go to plug in my evaluation, I'm getting sec(π/2) which is doesn't work out.

Using the half angle, I got (1-cosθ)=2cos²(θ/2).

Plugging that in canceled out my 2sqrt2, and made the integration into sec³(θ/2)

Solving that gave me: sec(θ/2)tan(θ/2) + ln|sec(θ/2)+tan(θ/2)|

Now in evaluating from π to π/2, plugging in for π is not a finite number. Did I go wrong in my integration of sec³?

 

[ideasrule]

1-cosθ is not equal to 2cos²(θ/2); it's equal to 2sin²(θ/2).

 

[jgiarrusso]

Thanks! That's what I get for trying to do my homework far too late at night and without my list of identities nearby ~_~