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

Question

2392 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-04T05:33:48

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

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 · 2018-04-04T06:00:05

#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.#

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#