How do you find the integral of #int (dx/sqrt(t^2-1))# from negative infinity to -2?

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Answer 1 · 2016-09-24T11:31:40

Does not converge

#int_(-oo)^(-2) 1/sqrt(x^2-1) dx#

#x = cosh u, dx/(du) = sinh u#

#implies int_(u_1)^(u_2) 1/sqrt(cosh^2 u -1) dx/(du) du#

#= int_(u_1)^(u_2) 1/(sinh u) sinh u du#

#= [u]_(u_1)^(u_2)#

#= lim_(t to -oo) [cosh^(-1) x]_(t)^(-2)#

Does not converge