Simple way to do an integral

**billa** · January 12th 2011, 06:22 PM

I am trying to do ...

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

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

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

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

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

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

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

**Krizalid** · January 12th 2011, 06:48 PM

well the answer is right, but your procedure seems to be a little bit messy.

we use the fact that $\cos t=1-2{{\sin }^{2}}\dfrac{t}{2},$ hence your integral becomes $\displaystyle2\sqrt{2}\int_{0}^{\pi }{\left| \sin t \right|\,dt}=2\sqrt{2}\int_{0}^{\pi }{\sin t\,dt}=4\sqrt{2}.$

**billa** · January 12th 2011, 07:02 PM

Awesome! Many thanks, Krizalid.

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