Evaluate the integral $\int \frac{x^2}{\sqrt{9-x^2}}dx$

Question

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?

Answer 1

\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}

Answer 2

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*}

Answer 3

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

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