How do you integrate #int 1/sqrt(4x+8sqrtx-15) # using trigonometric substitution?

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Answer 1 · 2018-05-17T04:26:25

Use the substitution #2(sqrtx+1)=sqrt19sectheta#.

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 #2(sqrtx+1)=sqrt19sectheta#:

#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#