Question

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

Answer 1

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`