c3v2/2 J V9 _ x? dx
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Let $x = 3\sin{u}$, so $dx = 3\cos{u} du$. The integral becomes: $$\int \frac{3v^2}{2} J \sqrt{9 - (3\sin{u})^2} (3\cos{u} du)$$ Simplify the expression inside the square root: $$\int \frac{3v^2}{2} J \sqrt{9(1 - \sin^2{u})} (3\cos{u} du)$$ Now, use the Show more…
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