Question

How do you integrate `int 1/sqrt(x^2-6x+4)` using trigonometric substitution?

Answer 1

Hint:
`I=int1/sqrt((x-3)^2-(sqrt5^2))dx`
Use:
`color(red)(int1/sqrt(X^2-A^2)dX=ln|X+sqrt(X^2-A^2)|+c`
`I=ln|(x-3)+sqrt((x-3)^2-(sqrt5)^2)|+c`
`I=ln|(x-3)+sqrt(x^2-6x+4)|+c`

Explanation:

Here,

`I=int1/sqrt(x^2-6x+4)dx`

`I=int1/sqrt(x^2-6x+9-5)dx`

`I=int1/sqrt((x-3)^2-(sqrt5^2))dx`

Taking ,`x-3=sqrt5secu=>x=3+sqrt5secu`

`:.dx=sqrt5secutanudu`

So,

`I=int(sqrt5secutanu)/sqrt((sqrt5secu)^2-(sqrt5)^2)du`

`I=int(sqrt5secutanu)/(sqrt5sqrt(sec^2u-1))du`

`=int((secutanu)/tanu)du`

`=intsecudu`

`=ln|secu+tanu|+c`

`=ln|secu+sqrt(sec^2u-1)|+c`

`=ln|(x-3)/sqrt5+sqrt(((x-3)/sqrt5)^2-1)|+c`

`=ln|(x-3)/sqrt5+sqrt((x-3)^2-5)/sqrt5|+c`

`=ln|((x-3)+sqrt(x^2-6x+9-5))/sqrt5|+c`

`=ln|(x-3)+sqrt(x^2-6x+4)|-lnsqrt5+c`

`=ln|(x-3)+sqrt(x^2-6x+4)|+C,where,C=c-lnsqrt5`