How do you integrate #int 3/(xsqrt(x^2-9))# by trigonometric substitution?
1 Answer
Sep 11, 2016
Explanation:
We have:
#int3/(xsqrt(x^2-9))dx#
We will use the substitution
#=int3/(3secthetasqrt(9sec^2theta-9))(3secthetatanthetad theta)#
#=int1/(secthetasqrt(sec^2theta-1))(secthetatanthetad theta)#
Note that
#=int1/(secthetatantheta)(secthetatanthetad theta)#
#=intd theta#
#=theta+C#
From
#="arcsec"(x/3)+C#