Question

How do you integrate `int dx/sqrt(x^2-64)` using trig substitutions?

Answer 1

`I = ln | sqrt(x^2/64 -1 ) + x/8 | + C`

Explanation:

We seek:

`I = int \ 1/sqrt(x^2-64) \ dx`

We can perform the substitution:

`x = 8sec theta => sec theta = x/8`

And differentiating wrt `theta`:

`dx/(d theta) = 8sec theta tan theta`

And Substituting into the integral, it becomes:

`I = int \ 1/sqrt((8sec theta)^2-64) \ 8sec theta tan theta \ d theta`

`\ \ = int \ (8sec theta tan theta)/sqrt(64sec^2 theta-64) \ d theta`

`\ \ = int \ (sec theta tan theta)/sqrt(sec^2 theta-1) \ d theta`

`\ \ = int \ (sec theta tan theta)/sqrt(tan^2 theta) \ d theta`

`\ \ = int \ sec theta \ d theta`

`\ \ = ln | tan theta + sec theta | + C`

And using the identity:

`1 + tan^2 A -= sec^2 A => tan theta = sqrt(sec^2 theta -1 )`

Allowing us to restore the earlier substitution:

`I = ln | sqrt(sec^2 theta -1 ) + sec theta | + C`

`\ \ = ln | sqrt((x/8)^2 -1 ) + x/8 | + C`

`\ \ = ln | sqrt(x^2/64 -1 ) + x/8 | + C`