Integration with hyperbolic secant

[spaghetti3451]

Homework Statement

Solve $\displaystyle{d\sigma = \frac{d\rho}{\cosh\rho}.}$

Homework Equations

The Attempt at a Solution

The answer is $\displaystyle{\sigma = 2 \tan^{-1}\text{sinh}(\rho/2)}$. See equation (10.2) in page 102 of the lecture notes in http://www.hartmanhep.net/topics2015/gravity-lectures.pdf. There is a typo in the equation.

Let us first try to check by differentiation. Using $\displaystyle{d\tan^{-1}(x) = \frac{dx}{1+x^{2}}}$, we have

$\displaystyle{d\sigma = \frac{2\sinh'(\rho/2)d\rho}{1+\sinh^{2}(\rho/2)}}$

$\displaystyle{d\sigma = \frac{\cosh(\rho/2)d\rho}{\cosh^{2}(\rho/2)}}$

$\displaystyle{d\sigma = \frac{d\rho}{\cosh(\rho/2)}}$

Is this correct?

 

[Ssnow]

Hi, I think the error is in your derivation:

$d\sigma=\frac{2\sinh'(\rho/2)d\rho/2}{1+\sinh^2{(\rho/2)}}=\frac{d\rho/2}{\cosh{(\rho/2)}}$

bercause you work with differentials you have $1/2$ in addiction ...
Ssnow