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 #x# we have:

# 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 #