Question

How do you integrate `int 1/sqrt(4x^2-12x-16)` using trigonometric substitution?

Answer 1

`int dx/sqrt(4x^2-12x-16) = 1/2ln abs ((2x-3)+sqrt(4x^2-12x-16))+C`

Explanation:

Complete the square under the root at the denominator:

`int dx/sqrt(4x^2-12x-16) = int dx/sqrt( (2x-3)^2 -25)`

Substitute now:

`2x -3 = 5sect`

`dx = 5/2sect tant dt`

then:

`int dx/sqrt(4x^2-12x-16) = 5/2 int (sect tantdt)/sqrt( 25sec^2t -25)`

`int dx/sqrt(4x^2-12x-16) = 1/2 int (sect tantdt)/sqrt( sec^2t -1)`

Use now the trigonometric identity:

`sec^2t-1 = tan^2t`

Note now that the function is defined for:

`abs(2x-3) > 5`

If we restrict to the interval:

`2x-3 > 5`

then `sect` is positive and:

`sqrt( sec^2t -1) = tant`

so that:

`int dx/sqrt(4x^2-12x-16) = 1/2 int (sect tantdt)/tant = 1/2 int sectdt`

To solve this last integral there is a well known trick:

`int sect dt = int sect (sect+tant)/(sect+tant) dt`

`int sect dt = int (sec^2t+sect tant)/(sect+tant) dt`

`int sect dt = int (d(sect+ tant))/(sect+tant) dt`

`int sect dt = ln abs (sect+tant) +C`

so:

`int dx/sqrt(4x^2-12x-16) = 1/2ln abs (sect+tant) +C`

and undoing the substitution:

`int dx/sqrt(4x^2-12x-16) = 1/2ln abs ((2x-3)/5 +sqrt(((2x-3)/5)^2-1))+C`

and simplifying:

`int dx/sqrt(4x^2-12x-16) = 1/2ln abs ((2x-3)+sqrt(4x^2-12x-16))+C`

Answer 2

`I=1/2ln|(2x-3)+sqrt(4x^2-12x-16)|+C`

Explanation:

Here,
`4x^2-12x-16=4x^2-12x+9-25=(2x-3)^2-(5)^2`
So,

`I=int1/sqrt(4x^2-12x-16)dx`
`=int1/sqrt((2x-3)^2-(5)^2)dx`

Take ,`color(red)(2x-3=5secu)=>secu=(2x-3)/5and`
`=>2dx=5secutanudu`
`=>dx=5/2secutanudu`
`I=int1/sqrt(25sec^2u-25)(5/2secutanu)du`
`I=5/2int(secutanu)/(5sqrt(sec^2u-1))du`
`=1/2int(secutanu)/(tanu)du`
`=1/2intsecudu`
`=1/2ln|secu+tanu|+c`
`=1/2ln|secu+sqrt(sec^2u-1)|+c`
`=1/2ln|(2x-3)/5+sqrt(((2x-3)/5)^2)-1|+c`
`=1/2ln|(2x-3)/5+sqrt((2x-3)^2-25)/5|+c`
`=1/2[ln|(2x-3)+sqrt(4x^2-12x-16)|-ln5]+c`
`I=1/2ln|(2x-3)+sqrt(4x^2-12x-16)|+C,`
where, `C=c-1/2ln5`
It is better to use `int1/sqrt(t^2-A^2)dt=ln|t+sqrt(t^2-A^2)|+c`
for ,`2x-3=t=>dx=1/2dt=>I=int1/sqrt(t^2-5^2)(1/2dt)=1/2ln|t+sqrt(t^2-25)|+c=1/2ln|(2x-3)+sqrt(4x^2-12x-16)|+c`

Answer 3

The answer is `=1/2ln(|sqrt((2x-3)^2/25-1)+(2x-3)/5|)+C`

Explanation:

The denominator is

`sqrt(4x^2-12x-16)=2sqrt(x^2-3x-4)`

`=2sqrt(x^2-3x+9/4-4-9/4)`

`=2sqrt((x-3/2)^2-25/4)`

`=5sqrt(((2x-3)/5)^2-1)`

`int(dx)/sqrt(4x^2-12x-16)=1/2int(dx)/(sqrt((x-3/2)^2-25/4))`

Let `u=(2x-3)/5`, `=>`, `du=2/5dx`

Therefore,

`int(dx)/sqrt(4x^2-12x-16)=1/2int(du)/(sqrt(u^2-1))`

Let `u=sectheta`, `=>`, `du=secthetatanthetad theta`

`1/2int(du)/(sqrt(u^2-1))=1/2int(secthetatanthetad theta)/(sqrt(sec^2theta-1))`

`=1/2intsecthetad theta`

`=1/2int(sectheta(sectheta+tantheta)d theta)/(sectheta+tantheta)`

`=1/2int((sec^2theta+sethetatantheta)d theta)/(sec theta+ tan theta)`

Let `v=sectheta+tantheta`

`=>`, `dv=(sec^2theta+secthetatantheta)d theta`

And finally,

`intsecthetad theta=int(dv)/v`

`=1/2ln(v)`

`=1/2ln(sqrt(u^2-1)+u)`

`=1/2ln(|sqrt((2x-3)^2/25-1)+(2x-3)/5|)+C`