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Evaluate $\int \sqrt{4-9 x^2} d x$
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We start by rewriting :
$\int \sqrt{4-9 x^2} d x=\int \sqrt{4\left(1-(3 x / 2)^2\right)} d x=\int 2 \sqrt{1-(3 x / 2)^2} d x$
Now let $3 x / 2=\sin u$ so $(3 / 2) d x=\cos u d u$ or $d x=(2 / 3) \cos u d u$. Then
\begin{aligned} \int 2 \sqrt{1-(3 x / 2)^2} d x &=\int 2 \sqrt{1-\sin ^2 u}(2 / 3) \cos u d u=\frac{4}{3} \int \cos ^2 u d u \\ &=\frac{4 u}{6}+\frac{4 \sin 2 u}{12}+C \\ &=\frac{2 \arcsin (3 x / 2)}{3}+\frac{2 \sin u \cos u}{3}+C \\ &=\frac{2 \arcsin (3 x / 2)}{3}+\frac{2 \sin (\arcsin (3 x / 2)) \cos (\arcsin (3 x / 2))}{3}+C \\ &=\frac{2 \arcsin (3 x / 2)}{3}+\frac{2(3 x / 2) \sqrt{1-(3 x / 2)^2}}{3}+C \\ &=\frac{2 \arcsin (3 x / 2)}{3}+\frac{x \sqrt{4-9 x^2}}{2}+C \end{aligned}
by Platinum (141,884 points)

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