How do you integrate #int 1/sqrt(9x^2-6x-3) # using trigonometric substitution?

1 Answer
May 17, 2018

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

Explanation:

Complete the square at the denominator:

#int dx/sqrt(9x^2-6x-3) = int dx/sqrt( (3x-1)^2 -4) = 1/2 int dx/sqrt( ((3x-1)/2)^2 -1)#

Note that the integrand is defined only for:

#abs (3x-1) > 2#

that is for #x > 1# or #x <-1/3#.

Let #x > 1# and substitute:

#(3x-1)/2 = sect#

#dx =2/3 sect tant dt#

with #t in (0,pi/2)#. Then

#int dx/sqrt(9x^2-6x-3) = 1/3 int (sect tantdt)/sqrt(sec^2t-1)#

Using the trigonometric identity:

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

and considering that for #t in (0,pi/2)# the tangent is positive:

#int dx/sqrt(9x^2-6x-3) = 1/3 int (sect tantdt)/sqrt(tan^2t)#

#int dx/sqrt(9x^2-6x-3) = 1/3 int (sect tantdt)/tant#

#int dx/sqrt(9x^2-6x-3) = 1/3 int sect dt#

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

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

Multiplying the argument of the logarithm by #2# which is equivalent to adding a constant:

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

and by direct differentiation we can see that this is valid also for #x < -1/3#