Question

How do you integrate `int 1/sqrt(x^2-16x+3)` using trigonometric substitution?

Answer 1

Use the substitution `x-8=sqrt61sectheta`.

Explanation:

Let

`I=int1/sqrt(x^2-16x+3)dx`

Complete the square in the square root:

`I=int1/sqrt((x-8)^2-61)dx`

Apply the substitution `x-8=sqrt61sectheta`:

`I=int1/(sqrt61tantheta)(sqrt61secthetatanthetad theta)`

Simplify:

`I=intsecthetad theta`

Integrate directly:

`I=ln|sectheta+tantheta|+C`

Rescale `C`:

`I=ln|sqrt61sectheta+sqrt61tantheta|+C`

Reverse the substitution:

`I=ln|x-8+sqrt(x^2-16x+3)|+C`

Answer 2

`I=log|(x-8)+sqrt(x^2-16x+3)|+C`, where,
`C=-logsqrt(61)+c`.
NOTE : -It is better to use `int1/sqrt(X^2+k)dX=ln|x+sqrt(x^2+k)|+c`, from (A)

Explanation:

#I=int1/sqrt(x^2-16x+3)dx=int1/sqrt(x^2-16x+64-61)dx =int1/sqrt((x-8)^2-(sqrt61)^2)dx,.(A)#
We take,`x-8=sqrt(61)sectheta=>x=8+sqrt(61)sectheta=>dx=sqrt(61)secthetatantheta*d(theta)`
`I=int((sqrt(61)secthetatantheta))/sqrt(61sec^2theta-61)d(theta)`
`I=int((sqrt(61)secthetatantheta))/(sqrt(61)sqrt(sec^2theta-1))d(theta)` `=int(secthetatantheta)/(tantheta)d(theta)=intsecthetad(theta)` `=log|sectheta+tantheta|+c`
`I=log|sectheta+sqrt(sec^2theta-1)|+c`
`I=log|(x-8)/sqrt(61)+sqrt(((x-8)/sqrt61)^2-1)|+c`
`I=log|(x-8)/sqrt(61)+sqrt((x-8)^2-61)/sqrt(61)|+c`
`I=log|(x-8)+sqrt(x^2-16x+3)|-log|sqrt(61)|+c`