I am sked to find the length of the polar curve $r=\frac{6}{1+\cos(\theta))}$, where $0 \leq \theta \leq \frac{\pi}{2} $ .

So the formula is essentially:

$L = \int_{a}^{b} \sqrt {r^2+({\frac{dr}{d \theta}})^2} d\theta$

I calculated ${r^2+({\frac{dr}{d \theta}}})^2$ which turns out to be $\frac{72}{(1+\cos(\theta))^3}$ according to this

But I have a problem now.

I get:

$\int_{0}^{\frac{\pi}{2}} \sqrt {\frac{72}{(1+\cos(\theta))^3}} d\theta$

I'm not even sure how I would begin to integrate this. Is there another way or it there some sort of trick to simplify the integral?


EDIT: Weierstrass Substitution is not allowed otherwise I would have done that.


One option is to say:

$r(1+\cos\theta) = 6\\ r = 6-x\\ x^2 + y^2 = (6-x)^2\\ x = 3 - \frac {1}{12} y^2$

And do this in Cartesian.

Supposing you don't want to do that.

$\sqrt {\frac {1+\cos\theta}{2}} = \cos \frac 12 \theta\\ 1+\cos\theta = 2 \cos^2 \frac 12\theta$

Is this too close to the Weirstrass substitution? There is no substitution that is actually happening though.

$\int_0^{\frac \pi2} 3 \sec^3 \frac 12\theta \ d\theta$

$\frac {3}{2} (\sec \frac 12 \theta \tan \frac 12 \theta + \ln |\sec \frac 12 \theta + \tan \frac 12 \theta|) |_0^\frac{\pi}{2}\\ \frac {3\sqrt2}{2} + \frac 32\ln(\sqrt 2 + 1)) $

  • $\begingroup$ I got it now. Thank you so much! $\endgroup$ – Future Math person Mar 23 '17 at 1:35

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