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What is #int 1/ sqrt(x^2 - 8^2) dx#?

1 Answer

Apr 23, 2018

See below

Explanation:

Try the change #x=8sectheta#

Then #dx=8secthetatanthetad theta#

#int1/(sqrt(x^2-8))dx=int(sqrt8secthetatanthetad theta)/sqrt(8^2(sec^2theta-1))=#

#=int(cancel8secthetacanceltanthetad theta)/(cancel8canceltantheta))=intsecthetad theta=ln(sectheta+tantheta)+C=ln(x/8+sqrt(x^2-8^2)/8)+C#

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