How do you integrate #int 3/(xsqrt(x^2-9))# by trigonometric substitution?

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Answer 1 · 2016-09-11T00:07:11

#"arcsec"(x/3)+C#

We have:

#int3/(xsqrt(x^2-9))dx#

We will use the substitution #x=3sectheta#. Thus #dx=3secthetatanthetad theta#.

#=int3/(3secthetasqrt(9sec^2theta-9))(3secthetatanthetad theta)#

#=int1/(secthetasqrt(sec^2theta-1))(secthetatanthetad theta)#

Note that #tan^2theta+1=sec^2theta# so #sqrt(sec^2theta-1)=tantheta#:

#=int1/(secthetatantheta)(secthetatanthetad theta)#

#=intd theta#

#=theta+C#

From #x=3sectheta# we see that #theta="arcsec"(x/3)#:

#="arcsec"(x/3)+C#