# Simple way to do an integral

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• Jan 12th 2011, 06:22 PM
billa
Simple way to do an integral
I am trying to do ...

$\displaystyle \int_0^{2\pi}\sqrt{1-cos(t)}dt$

Here is what I did. Can you either comment on my work or post a better solution?

Assume the domain of all functions is $\displaystyle (0,\pi)$. Let $\displaystyle u=1-cos(t)$

$\displaystyle \displaystyle\int\sqrt{1-cos(x)}dx=\int\frac{sin(x)}{\sqrt{1+cos(x)}}dx=\in t\frac{1}{\sqrt{2-u}}du=-2\sqrt{1+cos(x)}$

Therefore, $\displaystyle \int_0^\pi\sqrt{1-cos(x)}dx=2\sqrt{2}$

Since the integrand is symmetric about $\displaystyle y=\pi$,

$\displaystyle \int_0^{2\pi}\sqrt{1-cos(t)}dt=2*2\sqrt{2}$
• Jan 12th 2011, 06:48 PM
Krizalid
well the answer is right, but your procedure seems to be a little bit messy.

we use the fact that $\displaystyle \cos t=1-2{{\sin }^{2}}\dfrac{t}{2},$ hence your integral becomes $\displaystyle \displaystyle2\sqrt{2}\int_{0}^{\pi }{\left| \sin t \right|\,dt}=2\sqrt{2}\int_{0}^{\pi }{\sin t\,dt}=4\sqrt{2}.$
• Jan 12th 2011, 07:02 PM
billa
Awesome! Many thanks, Krizalid.