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Evaluate the integral of (3-x2)1/2

Compute the following integral.

    \[ \int \sqrt{3-x^2} \, dx. \]


To evaluate the integral we let

    \begin{align*}  x &= \sqrt{3} \sin u & \implies \qquad dx &= \sqrt{3} \cos u \, du \\  x^2 &= 3 \sin^2 u, & \implies \qquad u &= \arcsin \frac{x}{\sqrt{3}}. \end{align*}

Therefore we have

    \begin{align*}  \int \sqrt{3-x^2} \, dx &= \int \sqrt{3-3 \sin^2 u} (\sqrt{3} \cos u \, du) \\[9pt]  &= 3 \int \sqrt{1-\sin^2 u} (\cos u) \, du \\[9pt]  &= 3 \int \cos^2 u \, du \\[9pt]  &= \frac{3}{2} \int (\cos (2u) + 1) \, du \\[9pt]  &= \frac{3}{4} \sin (2u) + \frac{3}{2} u + C \\[9pt]  &= \frac{3}{4} \sin \left(2 \arcsin \frac{x}{\sqrt{3}} \right) + \frac{3}{2} \arcsin \frac{x}{\sqrt{3}} + C \\[9pt]  &= \frac{3}{2} \sin \left( \arcsin \frac{x}{\sqrt{3}} \right) \cos \left( \arcsin \frac{x}{\sqrt{3}} \right) + \frac{3}{2} \arcsin \frac{x}{\sqrt{3}} + C \\[9pt]  &= \frac{3}{2} \frac{x \sqrt{3-x^2}}{3} + \frac{3}{2} \arcsin \frac{x}{\sqrt{3}} + C \\[9pt]  &= \frac{1}{2} x \sqrt{3-x^2} + \frac{3}{2} \arcsin \frac{x}{\sqrt{3}} + C. \end{align*}

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