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

1 Answer
May 21, 2018

Use the substitution sqrtx+2=2sectheta.

Explanation:

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