Question

Evaluate the integral `int \ 1/(x^2sqrt(x^2-9)) \ dx`?

Answer 1

`sqrt(x^2-9)/(9|x|)+C.`

Explanation:

We substitute, `x=3secy rArr dx=3secytanydy.`

`:. I=int1/(x^2sqrt(x^2-9))dx,`

`=int{(3secy tany)dy}/{9sec^2ysqrt(9sec^2y-9),`

`=int{(cancel(3)secycancel(tany))dy}/{9sec^2y*cancel(3tany)},`

`=1/9int1/secydy=1/9intcosydy,`

`=1/9siny,`

`=1/9sqrt(1-cos^2y),`

`=1/9sqrt(1-1/sec^2y),`

`=1/9sqrt(1-(3/x)^2).......................[because, x=3secy],`

`=1/9sqrt(x^2-9)/|x|.`

`rArr I=sqrt(x^2-9)/(9|x|)+C.`

Enjoy Maths.!

Answer 2

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

Explanation:

We seek:

`I = int \ 1/(x^2sqrt(x^2-9)) \ dx`

We can write the integral as follows:

`I = int \ 1/(x^2sqrt(9(x^2/9-1))) \ dx`
`\ \ = int \ 1/(x^2sqrt(9)sqrt((x/3)^2-1)) \ dx`
`\ \ = 1/3 \ int \ 1/(x^2 sqrt((x/3)^2-1)) \ dx`

Let us attempt a substitution of the form:

`3sec theta = x => 3sectheta tantheta (d theta)/dx = 1`

Then substituting into the integral, we get:

`I = 1/3 \ int \ 1/((3sec theta)^2 sqrt((sec theta)^2-1)) \ 3sectheta tantheta \ d theta`

`\ \ = int \ (sectheta tantheta)/(9sec^2 theta sqrt(sec^2theta-1)) \ d theta`

`\ \ = 1/9 \ int \ (tantheta)/(sec theta sqrt(tan^2 theta)) \ d theta`

`\ \ = 1/9 \ int \ (tantheta)/(sec theta tan theta) \ d theta`

`\ \ = 1/9 \ int \ (1)/(sec theta) \ d theta`

`\ \ = 1/9 \ int \ cos theta \ d theta`

`\ \ = 1/9 sin theta + C`

`\ \ = 1/9 sqrt(1-cos^2 theta) + C`

`\ \ = 1/9 sqrt(1-(1/sec^2 theta)) + C`

And we can now restore the earlier substitution:

`I = 1/9 sqrt(1-(1/(x/3)^2)) + C`
`\ \ = 1/9 sqrt(1-(1/(x^2/9)) + C`
`\ \ = 1/9 sqrt(1-9/x^2) + C`

Answer 3

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

Explanation:

Restricting to the interval `x in (3,+oo)`, substitute:

`x=3sect`

`dx = 3 sect tant dt`

with `t in (0,pi/2)` and get:

`int dx/(x^2sqrt(x^2-9)) = 3 int (sect tant dt)/(9sec^2t sqrt(9sec^2t -9))`

`int dx/(x^2sqrt(x^2-9)) = 1/9 int ( tant dt)/(sect sqrt(sec^2t -1))`

use now the trigonometric identity:

`sec^2t -1 = 1/cos^2-1 = (1-cos^2t)/cos^2t =sin^2t/cos^2t = tan^2t`

so that:

`int dx/(x^2sqrt(x^2-9)) = 1/9 int ( tant dt)/(sect sqrt(tan^2t))`

and as in the selected interval the tangent is positive:

`int dx/(x^2sqrt(x^2-9)) = 1/9 int ( tant dt)/(sect tant)`

`int dx/(x^2sqrt(x^2-9)) = 1/9 int dt/sect`

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

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

To undo the substitution note that:

`x= 3sect = 3/cost = 3/sqrt(1-sin^2t)`

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

`1-sin^2t = 9/x^2`

`sin^2t = 1-9/x^2`

and as in the interval the sine is positive:

`sint = sqrt(x^2-9)/x`

So:

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

And using the substitution `x=-sect` for `x in (-oo,-3)` we obtain the same result.