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How do I find the exact length of the polar curve $$r = 1+sin(\theta)$$ from $$\frac{\pi}{3} \leq \theta \leq \pi $$?

I had originally setup my equation as $$\int_{\frac{\pi}{3}}^{\pi} \sqrt{(1+\sin(\theta) )^2 + (\cos\theta)^2} * d\theta$$ but that got me the incorrect answer. What am I doing wrong?

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    $\begingroup$ Are you sure of the definition of the length of a polar curve? $\endgroup$ – Mhenni Benghorbal Apr 11 '16 at 3:07
  • $\begingroup$ @MhenniBenghorbal Yes $\endgroup$ – Greencat Apr 11 '16 at 3:22
  • $\begingroup$ Your expression looks right, but the parentheses are hard to read. We want to integrate the square root of $(1+\sin\theta)^2+\cos^2\theta$, so we want to integrate $\sqrt{2+2\sin\theta}$. $\endgroup$ – André Nicolas Apr 11 '16 at 3:24
  • $\begingroup$ @AndréNicolas sorry. I fixed the parenthesis. Could you explain that a bit more? $\endgroup$ – Greencat Apr 11 '16 at 3:28
  • $\begingroup$ @imranfat that is the derivative of r (necessary for the equation being used) $\endgroup$ – Greencat Apr 11 '16 at 3:28
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As I mentioned in comments, we want $$\int_{\pi/3}^{\pi}\sqrt{2+2\sin\theta}\,d\theta.$$ Let $\theta=\frac{\pi}{2}-t$. Then $\sin\theta=\cos t$, and after the substitution we want $$\int_{-\pi/2}^{\pi/6}\sqrt{2+2\cos t}\,dt.$$ But by the identity $\cos(2y)=2\cos^2y-1$, we have $2+2\cos t=4\cos^2(t/2)$. So we want $$\int_{-\pi/2}^{\pi/6}2\cos(t/2)\,dt.$$ This integral is straightforward. Note that the $\sin(\pi/12)$ we get as a component of the answer can be expressed exactly in terms of square roots, so the final answer has a nice simple form.

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