How do you integrate int 1/sqrt(3x-12sqrtx) using trigonometric substitution?
1 Answer
May 21, 2018
Use the substitution
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
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