How do you integrate #int 1/sqrt(3x^2-12x-5)dx# using trigonometric substitution?

1 Answer
Jan 8, 2016

Assuming you know how to do substitution :

Complete the square, you want to have #(ax+-b)^2+-1# form

#int 1/sqrt(3x^2-12x-5)dx#

#3x^2-12x-5#

#a = sqrt(3)x#
#b = 2sqrt(3)#

#a^2 = 3x^2#
#b^2=12#

#3x^2-12x+12-17#

#(sqrt(3)x-2sqrt(3))^2-17#

So we have

#int 1/sqrt(3x^2-12x-5)dx = int 1/sqrt((sqrt(3)x-2sqrt(3))^2-17)dx#

#= int 1/sqrt(17(1/17(sqrt(3)x-2sqrt(3))^2-1))dx#

#= 1/sqrt17int 1/sqrt(1/17(sqrt(3)x-2sqrt(3))^2-1)dx#

In fact, at this moment you can do #u = 1/sqrt17(sqrt(3)x-2sqrt(3)) #

#du = sqrt(3/17)dx#
#dx = sqrt(17/3)du#

#u^2=1/17(sqrt(3)x-2sqrt(3))^2#

#= 1/sqrt3int 1/sqrt(u^2-1)du#

You have the perfect derivate of #1/sqrt(3)arccosh(u)#

One form of the integral of #int 1/sqrt(3x^2-12x-5)dx#

is #1/sqrt(3)arccosh(1/sqrt(17)(sqrt(3)x-2sqrt(3)))#

But i don't like this form, so coming back to :

#1/sqrt17int 1/sqrt(1/17(sqrt(3)x-2sqrt(3))^2-1)dx#

#sqrt(3)x-2sqrt(3) = sqrt(17)/cos(u)#

note that #u = arccos(sqrt(17)/(sqrt(3)x-2sqrt(3)))#

#sqrt(3)dx = sqrt(17)tan(u)/cos(u)du#

#dx = sqrt(17/3)tan(u)/cos(u)#

You get

#1/sqrt3int 1/sqrt(1/(cos^2(u))-1)tan(u)/cos(u)dx#

#1/sqrt(1/(cos^2(u))-1) = 1/|tan(u)|#

#1/sqrt(3)int1/cos(u)du#

#1/sqrt(3)intcos(u)/cos^2(u)du#

let's #t = sin(u)#

#dt = cos(u)du#

#= 1/sqrt(3)int1/(1-t^2)dt#

do partial fraction to get

#1/sqrt(3)(int1/(2(t+1))dt-int1/(2(t-1))dt#)

which is

#1/(2sqrt(3))[ln(|t+1|)-ln(|t-1|)]#

Substitute back for #t = sin(u)#

#1/(2sqrt(3))[ln(|sin(u)+1|)-ln(|sin(u)-1|)]#

Substitute back for #u = arccos(sqrt(17)/(sqrt(3)x-2sqrt(3))) #

#1/(2sqrt(3))[ln(|sin(arccos(sqrt(17)/(sqrt(3)x-2sqrt(3))))+1|)-ln(|sin(arccos(sqrt(17)/(sqrt(3)x-2sqrt(3))))-1|)]#

assuming that
#sin(arccos(sqrt(17)/(sqrt(3)x-2sqrt(3)))) = sqrt(1-17/(sqrt(3) x-2 sqrt(3))^2)#

You finally have

#1/(2sqrt(3))[ln(|sqrt(1-17/(sqrt(3) x-2 sqrt(3))^2)+1|)-ln(|sqrt(1-17/(sqrt(3) x-2 sqrt(3))^2)-1|)]#