How do you integrate #int 1/sqrt(3x^2-12x+29)dx# using trigonometric substitution?
1 Answer
Mar 16, 2018
Use the substitution
Explanation:
Let
#I=int1/sqrt(3x^2-12x+29)dx#
Complete the square in the square root:
#I=sqrt3int1/sqrt((3x-6)^2+51)dx#
Apply the substitution
#I=sqrt17intsecthetad theta#
Integrate directly:
#I=sqrt17ln|sectheta+tantheta|+C#
Rescale
#I=sqrt17ln|3x-6+sqrt(3x^2-12x+29)|+C#