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How to evaluate the integral $x^2/\sqrt{9-x^2}$

So I use trigonometric substitution where $x = 3\sin(\theta)$

And I got down to:

$$ \frac 9 2 (\theta - \sin(2\theta)/2) + C $$

What do I do from here?

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    $\begingroup$ Substitute $\theta=\sin^{-1}(x)$ and make use of $\sin(2\theta)=2\sin\theta\cos\theta$ $\endgroup$ – David P Jan 29 '20 at 3:07
  • $\begingroup$ Make a triangle to relate theta with x and sine. $\endgroup$ – coffeemath Jan 29 '20 at 3:08
  • $\begingroup$ @DavidPeterson thanks very much homie, works like a charm i got it $\endgroup$ – user745912 Jan 29 '20 at 3:16
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\begin{align} x & = 3\sin\theta \\[8pt] \frac x 3 & = \sin\theta \\[8pt] \arcsin \frac x 3 & = \theta \\[8pt] \sin(2\theta) & = 2\sin\theta\cos\theta \\[8pt] & = 2\cdot \frac x 3 \cdot \cos\left(\arcsin\frac x 3 \right) \\[8pt] & = \frac{2x} 3 \cdot \frac{\sqrt{3^2-x^2}} 3. \end{align}

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  • $\begingroup$ thx homie, i got exactly what u got $\endgroup$ – user745912 Jan 29 '20 at 3:22
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Hint : $ \sin( 2\theta) = 2 \sin( \theta) \cos( \theta)$ and \begin{eqnarray*} \cos( \sin^{-1}( X)) = \sqrt{ 1-(\sin( \sin^{-1}(X))^2} =\sqrt{1-X^2} . \end{eqnarray*}

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Alternatively write the numerator as $$9-(9-x^2)$$

Use http://www.sosmath.com/tables/integral/integ13/integ13.html

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