How do you integrate #int 1/sqrt(4x+8sqrtx-15) # using trigonometric substitution?
1 Answer
May 17, 2018
Use the substitution
Explanation:
Let
#I=int1/sqrt(4x+8sqrtx-15)dx#
Complete the square in the denominator:
#I=int1/sqrt(4(sqrtx+1)^2-19)dx#
Apply the substitution
#I=1/2int(sqrt19sec^2theta-2sectheta)d theta#
Integrate directly:
#I=1/2(sqrt19tantheta-2ln|sectheta+tantheta|)+C#
Reverse the substitution:
#I=1/2sqrt(4x+8sqrtx-15)-ln|2(sqrtx+1)+sqrt(4x+8sqrtx-15)|+C#