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52.$x=\cos^2 t, y=\cos t, 0\leq t\leq 4\pi$. $(\frac{dx}{dt})^2+(\frac{dy}{dt})^2=(-2\cos(t)\sin(t))^2+(-\sin t)^2=\sin^2 t(4\cos^2 t+1)$ \begin{align} Distance &=\int_0^{4\pi}|\sin t|\sqrt{4\cos^2 t+1}dt=4\int_0^\pi \sin t\sqrt{4\cos^2 t+1}dt\\ &=-4\int_1^{-1}\sqrt{4u^2+1}du [u=\cos t, du=-\sin tdt] = 4\int_{-1}^1\sqrt{4u^2+1}du\\ &=8\int_0^1\sqrt{4u^2+1}du=8\int_0^{\tan^{-1} 2}\sec\theta\cdot \frac{1}{2}\sec^2\theta d\theta [2u=\tan\theta, 2du=\sec^2\theta d\theta ]\\ &=4\int_0^{\tan^{-1} 2}\sec^3\theta d\theta = [2\sec \theta\tan\theta+2\ln |\sec\theta+\tan\theta |]_0^{\tan^{-1} 2}=4\sqrt{5}+2\ln(\sqrt{5}+2) \end{align} Thus, $L=\int_0^\pi |\sin t|\sqrt{4\cos^2 t+1}dt=\sqrt{5}+\frac{1}{2}\ln(\sqrt{5}+2)$.

I decided to factor $4$ out of the square root and replaced $u^2$ by $u^2$ and $1/4$, and then used an integration identity. I don't remember how to integrate by substituting variables by tan and sec, so I was wondering if it was absolutely crucial that I remember it for my mid-term.

I got $4(-\sqrt{(5)} -1/4 \ln(-1\sqrt{(5/4)}/1\sqrt{(5/4)})$ in the end. Is there a simpler way? And did I make any mistake?

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Geez, that ad-ridden page was a surprise! You can post images directly into MSE questions via the button above the question box that looks like a pretty landscape painting. –  Kevin Carlson Oct 11 '12 at 20:57
    
there's no ad. what are you talking about? –  Gladstone Asder Oct 11 '12 at 21:59
    
I get a page full of ads at your link, though perhaps this isn't the case for you. –  Kevin Carlson Oct 11 '12 at 22:51
    
The link worked for me (I got a webpage with mathematical text). –  Christopher A. Wong Oct 11 '12 at 23:20
    
I can't post the pic here, because I am new. Why don't you help me? –  Gladstone Asder Oct 11 '12 at 23:20

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