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## Homework Statement

My textbook sets up the integral, but does not solve, claiming that it's "trivial to solve manually or by using a CAS". I put the integral into my TI-89, and sure enough, there is a solution, and that solution happens to be "8". However...

## Homework Equations

The actual problem is to find the total arclength of r = 1+sin(x)

(It's polar, I just let theta = x, since theta is a pain to LaTeX)

## The Attempt at a Solution

Here is my attempt at solving this integral manually:

[tex]s = \int_{0}^{2π}\sqrt{(1+sin^{2}(x))+cos(x)^{2}}dx[/tex]

[tex]\int_{0}^{2π}\sqrt{2+2sin(x)}dx[/tex]

[tex]\int_{0}^{2π}\sqrt{2+2sin(x)} \frac{\sqrt{2-2sin(x)}}{\sqrt{2-2sin(x)}}dx[/tex]

[tex]\int_{0}^{2π}\sqrt{\frac{4-4sin^{2}(x)}{2-2sin(x)}}dx[/tex]

[tex]\int_{0}^{2π}\sqrt{\frac{4cos^{2}(x)}{2(1-sin(x))}}dx[/tex]

[tex]\sqrt{2}\int_{0}^{2π}\frac{cos(x)}{\sqrt{1-sin(x)}}dx[/tex]

[itex]u = 1-sin(x)[/itex]

[itex]du = -cos(x)dx[/itex]

[tex]-\sqrt{2}\int_{1}^{1}\frac{1}{\sqrt{u}}du[/tex]

I stopped here, because it's clear to me that the integral from one to one will be zero...

...and now the whole thing equals zero... I have tried doing the integral with substitutions other than [itex]u=1-sin(x)[/itex] to no avail. I tried to devise a way so I could take a limit (much like some improper integral problems are done), no luck. I can't seem to figure this out without using a CAS! It is one thing if I just got to a point where I didn't know how to proceed I could deal with, but I come up with zero every time!

This is my best attempt, I have done it 11 different ways now..

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