Question

Using `int sec x dx = ln|sec x + tan x | + c` , find `int 1/ sqrt (2x^2 - 4) dx` using suitable trigonometric substitution?

Answer 1

`int \ 1/sqrt(2x^2-4) \ dx = (sqrt2)/2 \ ln | x+sqrt( (x^2-2) ) | + C`

Explanation:

We seek:

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

Comparing with the trig identity `tan^2 A -= sec^2A-1`, Consider a substitution of the form:

`x/sqrt(2) = sec theta => dx/(d theta) = sqrt(2) sec theta tan theta`

If we substitute this into our integral we have:

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

`\ \ = (sqrt2)/2 \ int \ (sec theta tan theta) / (sqrt( tan^2 theta ) ) \ d theta`

`\ \ = (sqrt2)/2 \ int \ (sec theta tan theta) / (tan theta ) \ d theta`

`\ \ = (sqrt2)/2 \ int \ sec theta \ d theta`

`\ \ = (sqrt2)/2 \ ln|sectheta+tantheta| + A`

Again Using `tan^2 A -= sec^2A-1` we have:

`tan^2 theta -= x^2/2 - 1 = (x^2-2)/2`
`=> tan theta = sqrt( (x^2-2)/2 )`

And using the above results to restore the substitution we have:

`I = (sqrt2)/2 \ ln |x/sqrt(2)+sqrt( (x^2-2)/2 )| + A`
`\ \ = (sqrt2)/2 \ ln | 1/sqrt(2) ( x+sqrt( (x^2-2) ) ) | + A`
`\ \ = (sqrt2)/2 ln(1/sqrt(2)) + (sqrt2)/2 \ ln | x+sqrt( (x^2-2) ) | + A`
`\ \ = (sqrt2)/2 \ ln | x+sqrt( (x^2-2) ) | + C`