How do you integrate #int 1/sqrt(3x-12sqrtx) # using trigonometric substitution?

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Answer 1 · 2018-05-21T04:31:55

Use the substitution #sqrtx+2=2sectheta#.

Let

#I=int1/sqrt(3x-12sqrtx)dx#

Complete the square in the denominator:

#I=1/sqrt3int1/sqrt((sqrtx+2)^2-4)dx#

Apply the substitution #sqrtx+2=2sectheta#:

#I=4/sqrt3int(sec^2theta-sectheta)d theta#

Integrate term by term:

#I=4/sqrt3(tantheta-ln|sectheta+tantheta|)+C#

Reverse the substitution:

#I=2/sqrt3sqrt((sqrtx+2)^2-4)-4/sqrt3ln|sqrtx+2+sqrt((sqrtx+2)^2-4)|+C#

Simplify:

#I=2/sqrt3sqrt(x+4sqrtx)-4/sqrt3ln|sqrtx+2+sqrt(x+4sqrtx)|+C#