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 $C_4=C_3+ln7$ .

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

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