Question

How do you integrate `int 1/sqrt(4x^2-12x+34)` using trigonometric substitution?

Answer 1

`=1/2ln|sqrt(4x^2-12x+34)+(2x-3)|+C`

Explanation:

Here,

`I=int1/sqrt(4x^2-12x+34)dx`

`=int1/sqrt(4x^2-12x+9+25)dx`

`=int1/sqrt((2x-3)^2+5^2)dx`

Subst, `2x-3=5tanu=>2dx=5sec^2udu`

`=>dx=5/2sec^2udu and color(blue)(tanu=(2x-3)/5`

So

`I=int1/sqrt(5^2tan^2u+5^2)xx5/2sec^2udu`

`=5/2intsec^2u/(5secu)du`

`=1/2intsecudu`

`=1/2ln|secu+tanu|+c`

`=1/2ln|sqrt(tan^2u+1)+tanu|+c`

Subst. back , `color(blue)(tanu=(2x-3)/5` we get

`I=1/2ln|sqrt(((2x-3)/5)^2+1)+((2x-3)/5)|+c`

`=1/2ln|(sqrt(4x^2-12x+34))/5+(2x-3)/5|+c`

`=1/2ln|((sqrt(4x^2-12x+34))+(2x-3))/5|+c`

`=1/2ln|sqrt(4x^2-12x+34)+(2x-3)|-1/2ln5+c`

`=1/2ln|sqrt(4x^2-12x+34)+(2x-3)|+C`

`where,C=-1/2ln5+c`

Answer 2

Integration without trigonometric substitution.
`I=1/2ln|(2x-3)+sqrt(4x^2-12x+34)|+c`

Explanation:

Here,

`I=int1/sqrt(4x^2-12x+34)dx`

`=int1/sqrt(4x^2-12x+9+25)dx`

`=int1/sqrt((2x-3)^2+5^2)dx`

Subst, `color(blue)(2x-3=u)=>2dx=du=>dx=1/2du`

So,

`I=int1/sqrt(u^2+5^2)xx1/2du`

`=1/2color(red)(int1/sqrt(u^2+5^2)du`

`=1/2color(red)(ln|u+sqrt(u^2+5^2)|+c`

Subst. back ,`color(blue)(u=2x-3` ,we get

`I=1/2ln|(2x-3)+sqrt((2x-3)^2+5^2)|+c`

`=1/2ln|(2x-3)+sqrt(4x^2-12x+34)|+c`
.........................................................................................

Note :

`color(red)((1)int1/sqrt(x^2+a^2)dx=ln|x+sqrt(x^2+a^2)|+c`