Integral

**Chipset3600** · November 28th 2012, 03:01 PM

Hi MHF!
I'm trying to solve this integral $\int \sqrt[]{1-cos(x)}$

i deduced that if: $sin^2(x)=\frac{1-cos(2x)}{2}$

so: $sin^2(\frac{x}{2})=\frac{1-cos(x)}{2}$

then: $2sin^2(\frac{x}{2})=1-cos(x))$

$\int \sqrt{2sin^2(\frac{x}{2})}dx= -\sqrt{2} cos(\frac{x}{2})+C$

Is that correct? what would be the best way to solve it?

**Plato** · November 28th 2012, 03:22 PM

Originally Posted by Chipset3600

Hi MHF!
I'm trying to solve this integral $\int \sqrt[]{1-cos(x)}$
i deduced that if: $sin^2(x)=\frac{1-cos(2x)}{2}$
so: $sin^2(\frac{x}{2})=\frac{1-cos(x)}{2}$
then: $2sin^2(\frac{x}{2})=1-cos(x))$
$\int \sqrt{2sin^2(\frac{x}{2})}dx= -\sqrt{2} cos(\frac{x}{2})+C$
Is that correct? what would be the best way to solve it?

You can use this resource. Look at this webpage.

Learn to use wolframalpha.com

**kjchauhan** · November 28th 2012, 03:28 PM

Note that,

$\int\sin(ax) dx=-\frac{cos(ax)}{a}, a \ne 0$

**MarkFL** · November 28th 2012, 03:29 PM

Your substitution looks good, and observing that $0\le1-\cos(x)$ for all real $x$ , we may write the integral as:

$\sqrt{2}\int\sin\left(\frac{x}{2} \right)\,dx=2\sqrt{2}\int\sin\left(\frac{x}{2} \right)\,\frac{1}{2}dx=-2\sqrt{2}\cos\left(\frac{x}{2} \right)+C$

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