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

Question

Evelyn Agard · Accepted Answer

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#

Emma Aldridge · Answer

undefined To integrate $ \int \frac{1}{\sqrt{x^2 - 49}} \, dx $ using trigonometric substitution, let $ x = 7\sec(\theta) $. Then, $ dx = 7\sec(\theta)\tan(\theta) \, d\theta $. Substitute these expressions into the integral and simplify, then integrate the resulting expression in terms of $ \theta $. Finally, substitute back $ x $ in terms of $ \theta $ to obtain the final answer.