How do you integrate #int 1/sqrt(x^2-49)dx# using trigonometric substitution?

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Answer 1 · 2016-01-04T06:37:17

For #x>7#,
#int 1/sqrt{x^2 - 49} dx = ln(sqrt{x^2 - 49} + x) + C#.

For #x<-7#,
#int 1/sqrt{x^2 - 49} dx = -ln(sqrt{x^2 - 49} - x) + C#.

Where #C# is the constant of integration.

Use the identity #sec^2u - 1 -= tan^2u#.

Substitute #x = 7secu#.

For #x>7#, let #0<\u<\pi/2#.

#frac{dx}{du} = 7 secu tanu#

#int 1/sqrt{x^2 - 49} dx = int 1/sqrt{x^2 - 49} frac{dx}{du} du #

#= int 1/sqrt{(7secu)^2 - 49} (7 secu tanu) du#

#= int frac{7 secu tanu}{7sqrt{sec^2u - 1}} du#

#= int secu frac{tanu}{sqrt{tan^2u}} du#

Since #0<\u<\pi/2#, #sqrt{tan^2u} -= tanu#.

#int secu frac{tanu}{sqrt{tan^2u}} du = int secu du#

#= ln|secu+tanu| + C_1#, where #C_1# is an integration constant.

#= ln(secu+tanu) + C_1#

#= ln(7secu+7tanu) + C_2#, where #C_2=C_1-ln7#.

#= ln(x + sqrt{x^2 - 49}) + C_2#

For #x<-7#, let #pi/2<\u<\pi#, then #sqrt{tan^2u} -= -tanu#.

#int secu frac{tanu}{sqrt{tan^2u}} du = -int secu du#

#= -ln|secu+tanu| + C_3#, where #C_3# is an integration constant.

#= -ln(-secu-tanu) + C_3#

#= -ln(-7secu-7tanu) + C_4#, where #C_4=C_3+ln7#.

#= -ln(-x+sqrt{x^2-49}) + C_4#