How do you integrate #int dx/(9x^2-25)^(3/2)# using trig substitutions?

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Answer 1 · 2016-11-24T10:49:18

The answer is #=-x/(25*sqrt(9x^2-25))+C#

We use #sec^2u=1+tan^2u#

Let's use the substitution

#x=(5secu)/3# #=>#, #dx=(5secutanudu)/3#

#intdx/(9x^2-25)^(3/2)=int((5secutanudu)/3)/(9*25/9sec^2u-25)^(3/2)#

#=int(5secutanudu)/(3*125(tan^3u))#

#=1/75int(secudu)/tan^2u#

#=1/75int(cos^2udu)/(cosu sin^2u)#

#=1/75int(cosudu)/(sin^2u)#

Let #v=sinu#,#=>##dv=cosudu#

Therefore,

#1/75int(cosudu)/(sin^2u)=1/75int(dv)/v^2#

#=1/75intv^(-2)dv=1/75v^(-1)/-1=-1/(75v)#

#=-1/(75sinu)#

#x=(5secu)/3# #=>#, #cosu=5/(3x)#

#sin^2u=1-cos^2u=1-25/(9x^2)=(9x^2-25)/(9x^2)#

#1/sinu=(3x)/(sqrt(9x^2-25))#

Therefore,

#intdx/(9x^2-25)^(3/2)=(-1/75)*(3x)/(sqrt(9x^2-25))+C#

#=-x/(25*sqrt(9x^2-25))+C#