I hope it is only because it's late and I've been studying for a calculus exam for several hours, but I cannot see how to solve this integral.

The problem states:

Find the exact length of the curve. Use a graph to determine the parameter interval. $$r=\cos^2\left(\frac{\theta}{2}\right)$$

Based on a Desmos graph, I determined the limits of integration to be $0$ and $2\pi$.

So, based on the formula for arc length of curves in the polar plane, $L = \int_a^b \sqrt{ r^2 + \left( \frac{dr}{d\theta} \right)^2 } d\theta$, I have constructed the following integral:

$$L = \int_0^{2\pi} \sqrt{ \cos^4\left(\frac{\theta}{2}\right) + \left(-\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right)\right)^2} d\theta$$

This is, in part, based on the derivative I found for $r$:

$$\frac{d}{d\theta}\left[ \cos^2\left(\frac{\theta}{2}\right) \right] = 2\cdot\frac{d}{d\theta}\left[\cos\left(\frac{\theta}{2}\right)\right]\cdot\frac{d}{d\theta}\left[\frac{1}{2}\theta\right] = 2\cdot\left[ -\sin\left(\frac{\theta}{2}\right)\right]\left(\frac{1}{2}\right)$$

$$\frac{dr}{d\theta} = -\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right)$$

I considered the Substitution Rule, but there is no part of the derivative of the expression under the root outside of the root. Then, I considered Trigonometric Substitution, but I don't see a term I could use in the place of $a$, so that I end up with $x = a^2-a^2\sin(\theta)$ or any of the other trigonometric substitutions.

Wolfram Alpha was able to solve the integral, but it wasn't able to give me an exact answer or to show me the steps to solve the problem. I have the Wolfram Alpha Calculus Course Assistant app for iOS, and it is normally very good about showing specific steps in solving integration problems.

I appreciate any answers. I just hope the answer isn't a simple solution which I overlooked. ;) Thanks.

  • $\begingroup$ Careful. $dr/d\theta = -\sin(\theta/2)\cos(\theta/2)$. And hence ... $\endgroup$ – Simon S Dec 5 '15 at 6:06
  • $\begingroup$ D'oh! You're right. I missed that last night! Thank you. Now, I've fixed it, but that makes the integral even uglier. And I'm afraid I'm still stumped as to how to integrate it. $\endgroup$ – tommytwoeyes Dec 5 '15 at 22:50
  • $\begingroup$ The integrand equals $\sqrt{ \cos^4\left(\frac{\theta}{2}\right) + \sin^2\left(\frac{\theta}{2}\right)\cos^2\left(\frac{\theta}{2}\right)}$ Do you see anything to do next? $\endgroup$ – Steve Kass Dec 5 '15 at 23:05
  • $\begingroup$ What @SteveKass said. This simplifies significantly $\endgroup$ – Simon S Dec 6 '15 at 0:36
  • $\begingroup$ Oh, wow, I can't believe I missed that. It simplifies to: $$\sqrt{\cos^2\left(\frac{\theta}{2}\right) \left[ \sin^2\left(\frac{\theta}{2}\right) + \cos^2\left(\frac{\theta}{2}\right) \right]} \implies \sqrt{\cos^2\left(\frac{\theta}{2}\right)(1)} \implies \cos\left(\frac{\theta}{2}\right)$$ Now I can see how to solve it. Thanks. $\endgroup$ – tommytwoeyes Dec 15 '15 at 3:06

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