What is int 1/ sqrt(x^2 - 8^2) dx∫1√x2−82dx?

1 Answer
F. Javier B.
Apr 23, 2018

See below

Explanation:

Try the change x=8secthetax=8secθ

Then dx=8secthetatanthetad thetadx=8secθtanθdθ

int1/(sqrt(x^2-8))dx=int(sqrt8secthetatanthetad theta)/sqrt(8^2(sec^2theta-1))=∫1√x2−8dx=∫√8secθtanθdθ√82(sec2θ−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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