How to calculate this integral?

[Robin04]

Homework Statement

I have a problem with this integral:
$\int_{x_0}^x \frac{dx}{\sqrt{2-2\cos{x}}}$

Homework Equations

The Attempt at a Solution

I came across this while reading a book and the author says that this can be written in the form $\int_{x_0}^x \frac{dx}{2\sin{\frac{x}{2}}}$. I don't see how this can be achieved. What identity is used here for the trigonometric functions? Or this is an integral trick?

 

[BvU]

Badly typeset book (what book ?) this seems to me. Or did you make an error copying the equations ? Numerator is non-negative in 1. and not in 3.

Trig function to look at is formula for cosine of double angle

 

[Robin04]

Badly typeset book (what book ?) this seems to me. Or did you make an error copying the equations ? Numerator is non-negative in 1. and not in 3.

Trig function to look at is formula for cosine of double angle

It's a book called Solitons and Instantons by R. Rajaraman. Page 36-37, eq. (2.71), you can download it here: <link redacted>
Moderator's note: do not post links to resources that violate copyright.

 

[BvU]

So post the correct expressions using $\LaTeX$

Did you understand the double angle hint ?

 

[Robin04]

So post the correct expressions using LATEX

This is the correct expression as written in the book.

Did you understand the double angle hint ?

Yes, but if I understand well these two expressions are not equivalent, so there's not much I can do here with it.

 

[BvU]

but if I understand well these two expressions are not equivalent, so there's not much I can do here with it.

Wrong. Write it out and post.

 

[DrClaude]

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

 

[BvU]

$$\cos 2\alpha = \cos^2 \alpha - \sin^2\alpha=1-2\sin^2\alpha \quad\Rightarrow\quad 1-\cos x = 2\,\sin^2{x\over 2}$$ meaning $$
{1\over \sqrt{2-2\cos x} } = {1\over \sqrt{4\sin^2 {x\over 2} }} = {1\over 2\left | \sin {x\over 2}\right | } $$ So, depending on the constraints on the bounds (any described?), the book may be right.

 

[Robin04]

It's not as hard as I thought. Thank you for your help!

So, depending on the constraints on the bounds (any described?), the book may be right.

Actually, there's a $\pm$ in front of the integral in the book, so the absolute value doesn't change much on that.