How do you integrate $\int \sqrt{3 (1 - x^{2})} d x$ $int sqrt(3(1-x^2))dx$ using trigonometric substitution?

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How do you integrate $\int \sqrt{3 (1 - x^{2})} d x$ $int sqrt(3(1-x^2))dx$ using trigonometric substitution?

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Answer 1 · 2016-04-05T19:36:03

$\int \sqrt{3 (1 - x^{2})} d x = \frac{\sqrt{3}}{4} sin 2 θ + \frac{\sqrt{3}}{2} θ + C$ $int sqrt(3(1-x^2)) dx=sqrt3/4sin2theta+sqrt3/2 theta +C$

Explanation:

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

$\int \sqrt{3 (1 - {sin}^{2} θ)} \cdot cos θ d θ = \int \sqrt{3 ({cos}^{2} θ)} cos θ d θ$ $intsqrt(3(1-sin^2theta))*cos theta d theta=intsqrt(3(cos^2theta)) cos theta d theta$

$= \int \sqrt{3} cos θ cos θ d θ$ $=intsqrt3 cos theta cos theta d theta$

$= \sqrt{3} \int {cos}^{2} θ d θ$ $=sqrt 3intcos^2 theta d theta$

$= \sqrt{3} \int \frac{1}{2} (cos 2 θ + 1) d θ$ $=sqrt3 int1/2 (cos2 theta+1) d theta$

$= \frac{\sqrt{3}}{2} \int (cos 2 θ + 1) d θ$ $=sqrt3/2 int (cos2 theta+1) d theta$

$= \frac{\sqrt{3}}{2} [\frac{1}{2} sin 2 θ + θ]$ $=sqrt3/2 [1/2 sin2theta+theta]$

$= \frac{\sqrt{3}}{4} sin 2 θ + \frac{\sqrt{3}}{2} θ + C$ $=sqrt3/4sin2theta+sqrt3/2 theta +C$

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How do you integrate $\int \sqrt{3 (1 - x^{2})} d x$ $int sqrt(3(1-x^2))dx$ using trigonometric substitution?

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Explanation:

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How do you integrate $\int \sqrt{3 (1 - x^{2})} d x$ $int sqrt(3(1-x^2))dx$ using trigonometric substitution?