How do you integrate int 1/sqrt(4x+8sqrtx-15) using trigonometric substitution?

1 Answer
May 17, 2018

Use the substitution 2(sqrtx+1)=sqrt19sectheta.

Explanation:

Let

I=int1/sqrt(4x+8sqrtx-15)dx

Complete the square in the denominator:

I=int1/sqrt(4(sqrtx+1)^2-19)dx

Apply the substitution 2(sqrtx+1)=sqrt19sectheta:

I=1/2int(sqrt19sec^2theta-2sectheta)d theta

Integrate directly:

I=1/2(sqrt19tantheta-2ln|sectheta+tantheta|)+C

Reverse the substitution:

I=1/2sqrt(4x+8sqrtx-15)-ln|2(sqrtx+1)+sqrt(4x+8sqrtx-15)|+C