How do you integrate #int 1/sqrt(x^2-6x+4) # using trigonometric substitution?

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Answer 1 · 2018-03-30T04:25:09

Here,

#I=int1/sqrt(x^2-6x+4)dx#

#I=int1/sqrt(x^2-6x+9-5)dx#

#I=int1/sqrt((x-3)^2-(sqrt5^2))dx#

Taking ,#x-3=sqrt5secu=>x=3+sqrt5secu#

#:.dx=sqrt5secutanudu#

So,

#I=int(sqrt5secutanu)/sqrt((sqrt5secu)^2-(sqrt5)^2)du#

#I=int(sqrt5secutanu)/(sqrt5sqrt(sec^2u-1))du#

#=int((secutanu)/tanu)du#

#=intsecudu#

#=ln|secu+tanu|+c#

#=ln|secu+sqrt(sec^2u-1)|+c#

#=ln|(x-3)/sqrt5+sqrt(((x-3)/sqrt5)^2-1)|+c#

#=ln|(x-3)/sqrt5+sqrt((x-3)^2-5)/sqrt5|+c#

#=ln|((x-3)+sqrt(x^2-6x+9-5))/sqrt5|+c#

#=ln|(x-3)+sqrt(x^2-6x+4)|-lnsqrt5+c#

#=ln|(x-3)+sqrt(x^2-6x+4)|+C,where,C=c-lnsqrt5#