How do you integrate #int sqrt(3(1-x^2))dx# using trigonometric substitution?

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Answer 1 · 2016-04-05T19:36:03

#int sqrt(3(1-x^2)) dx=sqrt3/4sin2theta+sqrt3/2 theta +C#

#x=sintheta, dx=cos theta d theta#

#intsqrt(3(1-sin^2theta))*cos theta d theta=intsqrt(3(cos^2theta)) cos theta d theta#

#=intsqrt3 cos theta cos theta d theta#

#=sqrt 3intcos^2 theta d theta#

#=sqrt3 int1/2 (cos2 theta+1) d theta #

#=sqrt3/2 int (cos2 theta+1) d theta#

#=sqrt3/2 [1/2 sin2theta+theta]#

#=sqrt3/4sin2theta+sqrt3/2 theta +C#