How do you integrate #int 1/(xsqrt(x^2-1) )dx# using trigonometric substitution?
1 Answer
Nov 5, 2017
# int \ 1/(xsqrt(x^2-1)) \ dx = arcsecx + C #
Explanation:
We seek:
# I = int \ 1/(xsqrt(x^2-1)) \ dx #
Let us attempt a substitution of the form:
# sectheta=x #
Then differentiating wrt
# sectheta tan theta (d theta)/dx = 1 #
Substituting into the integral we have:
# I = int \ 1/(sec theta sqrt(sec^2theta-1)) \ sectheta tan theta \ d theta#
# \ \ = int \ 1/(sec theta sqrt(tan^2theta)) \ sectheta tan theta \ d theta#
# \ \ = int \ 1/(sec theta tan theta) \ sectheta tan theta \ d theta#
# \ \ = int \ d theta#
# \ \ = theta + C#
Restoring the substitution gives:
# I = arcsecx + C #