How do you integrate #int 1/sqrt(4x+8sqrtx-24) # using trigonometric substitution?

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Answer 1 · 2018-03-09T12:31:19

Use the substitution #sqrtx+1=sqrt7sectheta#.

Let

#I=int1/sqrt(4x+8sqrtx-24)dx#

Complete the square in the square root:

#I=1/2int1/sqrt((sqrtx+1)^2-7)dx#

Apply the substitution #sqrtx+1=sqrt7u#:

#I=int(sqrt7u-1)/sqrt(u^2-1)du#

Rearrange:

#I=sqrt7intu/sqrt(u^2-1)du-int1/sqrt(u^2-1)du#

Apply the substitution #u=sectheta#:

#I=sqrt7sqrt(u^2-1)-intsecthetad theta#

Hence

#I=sqrt(7u^2-7)-ln|sectheta+tantheta|+C#

Reverse the last substitution:

#I=sqrt(7u^2-7)-ln|u+sqrt(u^2-1)|+C#

Reverse the first substitution:

#I=sqrt((sqrtx+1)^2-7)-ln|(sqrtx+1)+sqrt((sqrtx+1)^2-7)|+C#

Simplify:

#I=sqrt(x^2+2sqrtx-6)-ln|(sqrtx+1)+sqrt(x^2+2sqrtx-6)|+C#