# 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
