Question

Evaluate the integral? : `int 1/( x^2sqrt(x^2-4) ) dx`

Answer 1

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

Explanation:

We seek:

`I = int \ 1/( x^2sqrt(x^2-4) ) \ dx`

Because of the negative sign let us attempt the following substitution:

`x=2sec theta => x^2 = 4sec^2 theta`

And differentiating wrt `theta`, we get:

`(dx)/(d theta) =2sec theta tan theta`

Substituting into the integral we get:

`I = int \ 1/( (4sec^2theta)sqrt(4sec^2theta - 4) ) \ 2sec theta tan theta \ d theta`

`\ \ = 1/2 \ int \ 1/( secthetasqrt(4(sec^2theta - 1)) ) \ tan theta \ d theta`

`\ \ = 1/2 \ int \ 1/( secthetasqrt(4tan^2 theta) ) \ tan theta \ d theta`

`\ \ = 1/2 \ int \ 1/( sectheta(2tan theta) ) \ tan theta \ d theta`

`\ \ = 1/4 \ int \ 1/( sectheta ) \ d theta`

`\ \ = 1/4 \ int \ costheta \ d theta`

`\ \ = 1/4 \ sin theta + C` ..... [A]

We now have a simple solution, but we need to be able to restore the substitution; so we need to perform some further trigonometric manipulation. now:

`x=2sec theta => sec theta = x/2`
`:. 1/(cos theta) = x/2`
`:. cos theta = 2/x`

And using the pythagorean identity `sin^2A + cos^2A -= 1` then:

`sin^2theta + (2/x)^2 = 1`
`:. sin^2theta = 1 - 4/x^2 = (x^2-4)/x^2`
`:. sin theta = sqrt((x^2-4)/x^2) = sqrt(x^2-4)/x`

And now using this result we can restore the earlier substitution in [A], to give:

`I = 1/4 \ sqrt(x^2-4)/x + C`
`\ \ = sqrt(x^2-4)/(4x) + C`