1. ## Integral

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

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

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

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

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

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

2. ## Re: Integral

Originally Posted by Chipset3600
Hi MHF!
I'm trying to solve this integral $\displaystyle \int \sqrt[]{1-cos(x)}$
i deduced that if: $\displaystyle sin^2(x)=\frac{1-cos(2x)}{2}$
so: $\displaystyle sin^2(\frac{x}{2})=\frac{1-cos(x)}{2}$
then: $\displaystyle 2sin^2(\frac{x}{2})=1-cos(x))$
$\displaystyle \int \sqrt{2sin^2(\frac{x}{2})}dx= -\sqrt{2} cos(\frac{x}{2})+C$
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

3. ## Re: Integral

Note that,

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

4. ## Re: Integral

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

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