Question

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

Answer 1

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`