graph the curve and find its length, $r=\cos^2(\frac {\theta}{2}) $ graph the curve and find it's length, $r=\cos^2(\frac {\theta}{2}) $
I graphed it and found that it was a cardioid (or a sideways heart). I am getting stuck on the arc length.
this is what I have:
$$r=(\frac 12(1+\cos\theta) $$
$$\frac {dr}{d\theta}= -\frac 12\sin\theta  $$
what I get for under the square root is :
$$\frac 14+\frac 12\cos\theta+\frac 14\cos^2\theta+ \frac 14\sin^2\theta  $$
I ended up getting stuck with $\frac12+\frac12\cos\theta  $
But I don't think this is right, where did I go wrong?
 A: Things look fine. By the double-angle identity you have already used, the square root is $|\cos(\theta/2)|$. Integrate. Note the absolute value sign. 
A: $\newcommand{\+}{^{\dagger}}
 \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\down}{\downarrow}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\isdiv}{\,\left.\right\vert\,}
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}
 \newcommand{\wt}[1]{\widetilde{#1}}$


\begin{align}
\color{#66f}{\Large L}&=2\int_{0}^{\pi}\root{\pars{\dd r}^{2} + r^{2}\pars{\dd\theta}^{2}}
=2\int_{0}^{\pi}\root{\bracks{\totald{{\rm r}\pars{\theta}}{\theta}}^{2} + {\rm r}^{2}\pars{\theta}}\,\dd\theta
\\[3mm]&=2\int_{0}^{\pi}
\root{{1 \over 4}\,\sin^{2}\pars{\theta} + \cos^{4}\pars{\theta \over 2}}
\,\dd\theta
\\[3mm]&=2\int_{0}^{\pi}\root{{1 \over 4}\,\sin^{2}\pars{\theta}
+ \bracks{1 + \cos\pars{\theta} \over 2}^{2}}\,\dd\theta
=\int_{0}^{\pi}\root{2\bracks{1 + \cos\pars{\theta}}}\,\dd\theta
\\[3mm]&=\int_{0}^{\pi}\root{4\cos^{2}\pars{\theta \over 2}}\,\dd\theta
=2\int_{0}^{\pi}\verts{\cos\pars{\theta \over 2}}\,\dd\theta
=4\int_{0}^{\pi/2}\cos\pars{\theta}\,\dd\theta = \color{#66f}{\Large 4}
\end{align}

A: $L=\int_{0}^{2\pi}\sqrt{r^2+(\frac{dr}{d\theta})^2}d\theta = \int_{0}^{2\pi}\sqrt{\frac{1}{4}(1+2cos\theta+cos^2\theta)+\frac{1}{4}sin^2\theta}d\theta = \int_{0}^{2\pi}\sqrt{\frac{1}{4}(2+2cos\theta)}d\theta = \int_{0}^{2\pi}\sqrt{\frac{1}{4}(2+2(2cos^2\frac{\theta}{2}-1))}d\theta = \int_{0}^{2\pi}\sqrt{cos^2\frac{\theta}{2}}d\theta = \int_{0}^{2\pi}|\cos\frac{\theta}{2}|d\theta = \int_{0}^{\pi}\cos\frac{\theta}{2}d\theta - \int_{\pi}^{2\pi}\cos\frac{\theta}{2}d\theta = [2sin\frac{\theta}{2}]_{0}^{\pi} - [2sin\frac{\theta}{2}]_{\pi}^{2\pi} = (2-0)-(0-2)=4$
