# Integral

• Nov 28th 2012, 03:01 PM
Chipset3600
Integral
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?
• Nov 28th 2012, 03:22 PM
Plato
Re: Integral
Quote:

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
• Nov 28th 2012, 03:28 PM
kjchauhan
Re: Integral
Note that,

$\int\sin(ax) dx=-\frac{cos(ax)}{a}, a \ne 0$
• Nov 28th 2012, 03:29 PM
MarkFL
Re: Integral
Your substitution looks good, and observing that $0\le1-\cos(x)$ for all real $x$, we may write the integral as:

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