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

1 Answer
Jun 28, 2018

# 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 #