How do you evaluate #int# #dx/(x^2sqrt(x^2 - 9))# with x = 3sec(#theta#)?

Question

Evaluate the integral using the indicated trigonometric substitution. (Use C for the constant of integration.)

#int# #dx/(x^2sqrt(x^2 - 9))#, x = 3sec(#theta#)

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Answer 1 · 2018-07-05T11:29:58

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

Let #x = 3sectheta#

#=>dx = 3sectheta tantheta d theta#

So
#I=int(3sectheta tantheta d theta)/(9sec^2thetasqrt(9sec^2theta- 9))#

#=int(3sectheta tantheta d theta)/(27sec^2thetasqrt(sec^2theta- 1))#

#=1/9int(sectheta tantheta d theta)/(sec^2theta tan theta)#

#=1/9intcosthetad theta#

#=1/9sintheta +C#

#=1/9tantheta/sectheta +C#

#=1/9(sqrt(sec^2theta-1))/sectheta +C#

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

#=1/9(sqrt(x^2-9))/x +C#