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

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#)

Answer 1

Isabella Ackerman

#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#

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Answer 2

Luke Alvarez

Substitute ( x = 3\sec(\theta) ) into the integral expression and convert ( dx ) in terms of ( d\theta ). Then, evaluate the integral using trigonometric identities and properties of integrals.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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

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#)

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#)