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

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

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

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

$=intsqrt3 cos theta cos theta d theta$

$=sqrt 3intcos^2 theta d theta$

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

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

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

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

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