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jgiarrusso

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

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

L =(integrate) sqrt(r

I found (dr/dθ) = (-2sinθ)/(1-cosθ)

so (dr/dθ)

Then r

both have a factor of 4/(1-cosθ)

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

then I multiplied the 1 by (1-cosθ)

After multiplying it out, the numerator of the fraction was 1-2cosθ+cos

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)

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.

## Homework Statement

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

## Homework Equations

L =(integrate) sqrt(r

^{2}+(dr/dθ)^{2})dθ## The Attempt at a Solution

I found (dr/dθ) = (-2sinθ)/(1-cosθ)

^{2}so (dr/dθ)

^{2}= (4sin^{2}θ)/(1-cosθ)^{4}Then r

^{2}= 4/(1-cosθ)^{2}both have a factor of 4/(1-cosθ)

^{2}, so I pulled that outside the sqrt to getL=(integrate) 2/(1-cosθ) * sqrt(1+(sin

^{2}θ)/(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+cos

^{2}θ+sin^{2}θ, so I got rid of the sin and cos and added a 1 to get 2 - 2cosθI 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.

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