Question

How do you find`int x/sqrt((x^2-4x+9) )dx` using trigonometric substitution?

Answer 1

`:. I=2ln|{(x-2)+sqrt(x^2-4x+9)}/sqrt5|+sqrt(x^2-4x+9)+C`.

Explanation:

Let, `I=intx/sqrt(x^2-4x+9)dx=intx/sqrt{(x-2)^2+(sqrt5)^2}dx`.

So, we use the subst. `(x-2)=sqrt5tany`.

`:. x=2+sqrt5tany, and, dx=sqrt5sec^2ydy`.

`:. I=int{(2+sqrt5tany)(sqrt5sec^2y)}/sqrt(5tan^2y+5)dy`,

`=int{(2+sqrt5tany)(sqrt5sec^2y)}/(sqrt5secy)dy`,

`=int{(2+sqrt5tany)secy}dy`,

`=2intsecydy+sqrt5intsecytanydy`,

`=2ln|(secy+tany)|+sqrt5secy`.

Note that, `(x-2)=sqrt5tany rArr tany=(x-2)/sqrt5, and,`

`sqrt(x^2-4x+9)=sqrt5secy`.

`:. I=2ln|{(x-2)+sqrt(x^2-4x+9)}/sqrt5|+sqrt(x^2-4x+9)+C`.

Feel & Spread The Joy of Maths.!

Answer 2

`I=sqrt(x^2-4x+9)+2ln|(x-2)+sqrt(x^2-4x+9)|+c`
Please see the answer given by Mr.Ratnaker Maheta for trig.substitution: `(x-2)=sqrt5tany.`

Explanation:

Here,

`I=intx/sqrt(x^2-4x+9)dx`

`=1/2int(2x)/sqrt(x^2-4x+9)dx`

`=1/2int(2x-4+4)/sqrt(x^2-4x+9)dx`

`=1/2int(2x-4)/sqrt(x^2-4x+9)dx+1/2int(4)/sqrt(x^2-4x+9)dx`

`=1/2int(d/(dx)(x^2-4x+9))/sqrt(x^2-4x+9)dx+4/2int(1)/sqrt(x^2- 4x+4+5)dx`

Applying `color(red)((1)` given below,

`=1/2*2sqrt(x^2-4x+9)+2int1/sqrt((x-2)^2+(sqrt5)^2)dx`

Applying `color(red)((2)` given below

`=sqrt(x^2-4x+9)+2ln|(x-2)+sqrt((x-2)^2+(sqrt5)^2)|+c`

`I=sqrt(x^2-4x+9)+2ln|(x-2)+sqrt(x^2-4x+9)|+c`

Hint:

`color(red)((1)int(d/(dx)(f(x)))/sqrt(f(x))dx=2sqrt(f(x))+c`

`color(red)((2)int1/sqrt(X^2+A^2)dx=ln|X+sqrt(X^2+A^2)|+c`