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

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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 #