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

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Answer 1 · 2018-03-27T20:05:55

#1/3 ln|(3x-1) + sqrt(9x^2-6x)| + C#

#int dx/sqrt(9x^2-6x)#

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

Let : #3x - 1 = sec(t) -> dx = 1/3(sec(t) tan(t))dt#

Subtitute :

#int dx/sqrt((3x-1)^2-1)# = #1/3int (sec(t)tan(t) dt)/sqrt(sec^2(t)-1)#

= #1/3int (sec(t)tan(t) dt)/(tan(t))#

= #1/3int sec(t)dt#

= #1/3ln|sec(t) + tan(t)| + C#

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