How do you integrate int dx/sqrt(x^2-64)∫dx√x2−64 using trig substitutions?
1 Answer
I = ln | sqrt(x^2/64 -1 ) + x/8 | + C I=ln∣∣∣√x264−1+x8∣∣∣+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
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
