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spaghetti3451
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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?
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