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So I was given the following prompt:

What is the length of the curve $y=1-\cos(x)$ from $x=0$ to $x=4\pi$?

I was able to set up an integral and find a numerical answer for this problem, but I just wanted some clarification as to whether or not this is correct. I set up the integral of: $$L=\int_0^{4\pi}\sqrt{1+[\sin(x)]^2}\,dx$$ and I found the answer of $15.281$ from this integral, any clarification about whether or not this is correct would be appreciated!

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    $\begingroup$ Looks correct.. $\endgroup$ – Chrystomath Jan 19 at 16:44
  • $\begingroup$ Two comments about Mathjax: Use \sin(x) instead of sin(x) so that you display $\sin(x)$ rather than $sin(x)$, and second, use \sqrt{} rather than \sqrt() to display $\sqrt{1 + [\sin(x)]^2}$ rather than $\sqrt(1+[\sin(x)]^2)$. $\endgroup$ – DMcMor Jan 19 at 16:49
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Yes, you have applied the arc length formula correctly: $$ \int_a^b\sqrt{1+(f'(x))^2}\,dx $$

You then got an elliptic integral and you have the correct approximate value.

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