Solving a polar integral

Hey ya'll thank you in advance for taking the time to answer some math questions... so let's get into it, I was helping a student trying to work an example from the Stewart Calculus book and it all made sense up until we hit this definite integral, the book says that the answer is a length of 8, since we are dealing with arc length of a polar function.

I left out a few earlier steps, but like I said the book narrows it down to the following definite integral:

$$L = \int_{0}^{2\pi} \,\sqrt{\, 2 + 2\sin\left(\theta\right)\,}\,\mathrm{d}\theta$$

Please try to show as many steps and their justifications, thank.

Apologies for the syntax, but the limits of integration should be from $$0$$-$$2\pi$$.

P.S. I got zero, and I also used photomath which also got zero, so I don't know what I'm missing. The book does say to multiply top and bottom by the conjugate, which is what I did, to get started. It seems everything goes ok until you have to convert your limits of integration for a $$u$$-substitution. Any feedback is appreciated. Have a great day !! :)

• It is suggested that you use $\LaTeX$ to improve your math typesetting. – Trebor Apr 16 at 23:48

Firstly use that $$2+2\sin{(\theta)}=4\cos^2{\left(\frac\pi4-\frac\theta2\right)}$$

This follows from $$\sin{(\theta)}=\cos{\left(\frac\pi2-\theta\right)}$$ $$\cos{(\theta)}=2\cos^2{\left(\frac\theta2\right)}-1$$

Hence the integral is \begin{align} \int_0^{2\pi}\sqrt{4\cos^2{\left(\frac\pi4-\frac\theta2\right)}}d\theta &=\int_0^{2\pi}2\left|\cos{\left(\frac\pi4-\frac\theta2\right)}\right|d\theta\\ &=\int_0^\frac{3\pi}22\cos{\left(\frac\pi4-\frac\theta2\right)}d\theta-\int_\frac{3\pi}2^{2\pi}2\cos{\left(\frac\pi4-\frac\theta2\right)}d\theta\\ &=\left[-4\sin{\left(\frac\pi4-\frac\theta2\right)}\right]_0^\frac{3\pi}2-\left[-4\sin{\left(\frac\pi4-\frac\theta2\right)}\right]_\frac{3\pi}2^{2\pi}\\ &=4-(-2\sqrt{2})-(2\sqrt{2}-4)\\ &=8\\ \end{align}

• This honestly doesn't help much, I need the process. Thank you though, I'm trying it out with your hints. – Jason Gabriel Apr 16 at 23:54
• Where does the last equation come from? your own or is it an identity? – Jason Gabriel Apr 16 at 23:55
• @Jason Gabriel I have written the full process now. – Peter Foreman Apr 17 at 0:13
• Thank you so much, I was able to explain most of it to my tutee! But can you briefly explain why you split up the limits of integration as you did...?(0 - 3π/2 and 3π/2 - 2π)? What's the reasoning? – Jason Gabriel Apr 17 at 1:13
• I graphed the function in the integral (after square rooting) and found that from 0 to 0 - 3π/2, the area of the graph is positive and negative from 3π/2 - 2π? Still having minor complications for this reasoning, but is that why? – Jason Gabriel Apr 17 at 1:29

$$\newcommand{\bbx}{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}{\displaystyle{#1}} \newcommand{\expo}{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}{\mathcal{#1}} \newcommand{\mrm}{\mathrm{#1}} \newcommand{\pars}{\left(\,{#1}\,\right)} \newcommand{\partiald}[]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}{\left\vert\,{#1}\,\right\vert}$$ \begin{align} L & \equiv \int_{0}^{2\pi}\root{2 + 2\sin\pars{\theta}}\,\dd\theta = \root{2}\int_{-\pi}^{\pi}\root{1 - \sin\pars{\theta}}\,\dd\theta \\[5mm] & = \root{2}\sum_{\sigma = \pm 1}\int_{0}^{\pi} \root{1 + \sigma\sin\pars{\theta}}\,\dd\theta = \root{2}\sum_{\sigma = \pm 1}\int_{-\pi/2}^{\pi/2} \root{1 + \sigma\cos\pars{\theta}}\,\dd\theta \\[5mm] & = 2\root{2}\sum_{\sigma = \pm 1}\int_{0}^{\pi/2} \root{1 + \sigma\cos\pars{\theta}}\,\dd\theta = 2\root{2}\sum_{\sigma = \pm 1}\int_{0}^{\pi/2} {\sin\pars{\theta} \over \root{1 - \sigma\cos\pars{\theta}}}\,\dd\theta \\[5mm] & = 2\root{2}\sum_{\sigma = \pm 1}\int_{0}^{1} \pars{1 -\sigma x}^{-1/2}\,\dd x = 2\root{2}\sum_{\sigma = \pm 1} \left.\vphantom{\Large A}\pars{-2\sigma} \root{1 -\sigma x}\,\right\vert_{\ x\ =\ 0}^{\ x\ =\ 1} \\[5mm] & = 4\root{2}\sum_{\sigma = \pm 1} \bracks{-\sigma\root{1 - \sigma} + \sigma} = 4\root{2}\bracks{\pars{\root{2} - 1} + 1} = \bbx{8} \end{align}