How do you integrate $int x/sqrt(144-x^2)dx$ using trigonometric substitution?

Question

2535 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-12-14T08:49:23

Sub $x = 12sinu$ .

Restrict $-frac{pi}{2}<\u<=frac{pi}{2}$ .

Note that $cosu>=0$ .

$frac{dx}{du} = 12cosu$

$int frac{x}{sqrt(144-x^2)} dx = int frac{12sinu}{sqrt(144-(12sinu)^2)} * frac{dx}{du} du$

$= int frac{12sinu}{sqrt{144(1-sin^2u)}} * (12cosu) du$

$= int frac{12sinucosu}{sqrt{cos^2u}} du$

$= 12 int sinu du$

$= -12 cosu + C$ , where $C$ is the constant of integration.

$= -sqrt{144-(12sinu)^2} + C$

$= -sqrt{144-x^2} + C$

How do you integrate int x/sqrt(144-x^2)dxint x/sqrt(144-x^2)dx using trigonometric substitution?