Question #34e5a

#I=intx^2/(x^2-9)^(3/2)dx#

Use the substitution #x=3sectheta#. This implies that #dx=3secthetatanthetad theta#. Importantly, #x^2-9=9sec^2theta-9=9(sec^2theta-1)=9tan^2theta#.

#I=int(9sec^2theta)/(9tan^2theta)^(3/2)(3secthetatanthetad theta)#

#I=int(27sec^3thetatantheta)/(27tan^3theta)d theta#

#I=intsec^3theta/tan^2thetad theta#

#I=int1/cos^3thetacos^2theta/sin^2theta#

#I=intcsc^2thetasecthetad theta#

Use #csc^2theta=cot^2theta+1#:

#I=int(cot^2theta+1)secthetad theta#

#I=int(cos^2theta/sin^2theta 1/costheta+sectheta)d theta#

#I=int(cotthetacsctheta+sectheta)d theta#

These have common integrals:

#I=-csctheta+lnabs(sectheta+tantheta)+C#

Our original substitution was #sectheta=x/3#. This means that #costheta=3/x#, so we have a right triangle where the side adjacent to #theta# is #3# and the hypotenuse is #x#. Through the Pythagorean Theorem, the side opposite #theta# is #sqrt(x^2-9)#. Then:

• #tantheta="opp"/"adj"=sqrt(x^2-9)/3#
• #csctheta=1/sintheta="hyp"/"opp"=x/sqrt(x^2-9)#

Hence:

#I=lnabs(x/3+sqrt(x^2+9)/3)-x/sqrt(x^2-9)+C#

Factor #1/3# from the natural logarithm. Using #log(AB)=log(A)+log(B)#, a #ln(1/3)# will be spit out of the logarithm but it can be neglected because as a constant it will be absorbed into #C#, the constant of integration.

#I=lnabs(x+sqrt(x^2-9))-x/sqrt(x^2-9)+C#