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

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?

• Substitute $\theta=\sin^{-1}(x)$ and make use of $\sin(2\theta)=2\sin\theta\cos\theta$ – David P Jan 29 '20 at 3:07
• Make a triangle to relate theta with x and sine. – coffeemath Jan 29 '20 at 3:08
• @DavidPeterson thanks very much homie, works like a charm i got it – user745912 Jan 29 '20 at 3:16

\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}
Hint : $$\sin( 2\theta) = 2 \sin( \theta) \cos( \theta)$$ and $$\begin{eqnarray*} \cos( \sin^{-1}( X)) = \sqrt{ 1-(\sin( \sin^{-1}(X))^2} =\sqrt{1-X^2} . \end{eqnarray*}$$
Alternatively write the numerator as $$9-(9-x^2)$$