How do you integrate #int 1/sqrt(x^2-16x+3) # using trigonometric substitution?
2 Answers
Use the substitution
Explanation:
Let
#I=int1/sqrt(x^2-16x+3)dx#
Complete the square in the square root:
#I=int1/sqrt((x-8)^2-61)dx#
Apply the substitution
#I=int1/(sqrt61tantheta)(sqrt61secthetatanthetad theta)#
Simplify:
#I=intsecthetad theta#
Integrate directly:
#I=ln|sectheta+tantheta|+C#
Rescale
#I=ln|sqrt61sectheta+sqrt61tantheta|+C#
Reverse the substitution:
#I=ln|x-8+sqrt(x^2-16x+3)|+C#
NOTE : -It is better to use
Explanation:
We take,