Question

How do you integrate `int 1/sqrt(9x^2-6x+5)` using trigonometric substitution?

Answer 1

`int dx/sqrt(9x^2-6x+5 ) = 1/3 ln abs(3x-1+sqrt(9x^2-6x+5 )) +C`

Explanation:

Complete the square at the denominator:

`9x^2-6x+5 = 9x^2-6x+1+4 = (3x-1)^2+4`

then write the integrand as:

`1/sqrt(9x^2-6x+5 ) = 1/sqrt( (3x-1)^2+4) = 1/2 1/sqrt(((3x-1)/2)^2+1)`

Substitute now:

`(3x-1)/2 = tant`

with `t in (-pi/2,pi/2)`

`dx = 2/3sec^2t dt`

so that:

`int dx/sqrt(9x^2-6x+5 ) = 1/3 int (sec^2tdt)/sqrt(tan^2t+1)`

Use now the trigonometric identity:

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

and as for `t in (-pi/2,pi/2)` the secant is positive:

`sqrt(tan^2t+1) = sec t`

So:

`int dx/sqrt(9x^2-6x+5 ) = 1/3 int (sec^2tdt)/sect`

`int dx/sqrt(9x^2-6x+5 ) = 1/3 int sectdt`

`int dx/sqrt(9x^2-6x+5 ) = 1/3 ln abs(sect + tant) +C`

Undo the substitution, considering that:

`sect = sqrt(tan^2t +1) = sqrt(((3x-1)/2)^2+1) = 1/2sqrt(9x^2-6x+5 )`

`int dx/sqrt(9x^2-6x+5 ) = 1/3 ln abs(1/2sqrt(9x^2-6x+5 )+(3x-1)/2) +C`

and as multiplying the argument of the logarithm by a constant is equivalent to adding a constant to the expression:

`int dx/sqrt(9x^2-6x+5 ) = 1/3 ln abs(3x-1+sqrt(9x^2-6x+5 )) +C`