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I'm learning integrals and I'm having trouble with evaluating this one:

$$ \int_0^{8 \pi} \sqrt{ 4 - 4 \cos(t)} \mathrm{d} t $$

The end result is $32 \sqrt{2}$.

I tried this by changing the form of the integral to

$\sqrt{8 \cdot \sin^2(t/2)}$

but I can't get the end result, which I specified above.

Any help is greatly appreciated, especially if explained in steps.

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up vote 4 down vote accepted

So first rewrite your integral as you did as \[ \int_0^{8\pi} \sqrt{4 - 4\cos t} \, dt = \int_0^{8\pi} \sqrt{8\sin^2 \frac t2}\, dt \] Now we get rid of the $\frac 12$ letting $s = \frac t2$ and obtain \[ \int_0^{8\pi} \sqrt{8\sin^2 \frac t2}\, dt = 2\int_0^{4\pi} \sqrt{8\sin^2 s }\, ds \] Now $\sqrt{\sin^2 s} = |\sin s|$ and $\sqrt{8} = 2\sqrt 2$. More over the function $s \mapsto |\sin s|$ is periodic with period $\pi$, so the integral over $[0,4\pi]$ equals four times the integral over $[0,\pi]$. We obtain \[ 2\int_0^{4\pi} \sqrt{8\sin^2 s }\, ds = 4\sqrt 2 \int_0^{4\pi}|\sin s|\, ds = 16\sqrt 2 \int_0^\pi \sin s \, ds \] where in the last step we also used $\sin s \ge 0$ for $s \in [0,\pi]$. Now the last integral is 2, so altogether we obtained \[ \int_0^{8\pi} \sqrt{4 - 4\cos t} \, dt = 32\sqrt 2.\]

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$\sqrt{4-4\cos(t)} = 2 \sqrt{1-\cos(t)}$


$1-\cos(t) = 2\sin^2(\frac{t}{2})$

Also you have to remember to integrate over absolute values.

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Perhaps you mean $2\sin^2(t/2)$? – Alex Becker May 22 '12 at 14:28
The OP already mentioned that they were able to get that far. They need help on the latter steps. – Ragib Zaman May 22 '12 at 14:30
Whoa sorry for all the typos. Also I gave a hint for the latter steps. – Eugene May 22 '12 at 14:31
Also, could you include the backslash in \sin and \cos? If you write 3sin x in $\TeX$, it looks like this: $3sin x$. But if you write 3\sin x, it looks like this: $3\sin x$. Notice that the latter has proper spacing before and after "$\sin$". Also, "$\sin$" is not italicized and $x$ is. That is standard $\TeX$ usage. – Michael Hardy May 22 '12 at 14:33
Sorry I didn't know that before now. Thanks for the information. – Eugene May 22 '12 at 14:35

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