Integrating $\int\frac{x^3}{\sqrt{9-x^2}}dx$ via trig substitution

Question

What I have done so far: Substituting $$x=3\sin(t)\Rightarrow dx=3\cos(t)dt$$ converting our integral to $$I=\int\frac{x^3}{\sqrt{9-x^2}}dx=\int \frac{27\sin^3(t) dt}{3\sqrt{\cos^2(t)}}3\cos(t)dt\\ \Rightarrow \frac{I}{27}=\int \sin^3(t)dt=-\frac{\sin^2(t)\cos(t)}{3}+\frac{2}{3}\int\sin(t)dt\\ \Rightarrow I=18\cos(t)-9\sin^2(t)\cos(t)+c$$ Which is where I'm stuck. I can substitute back in for the $\sin(t)$ terms but don't know hot to deal with the $\cos(t)$ terms and back out information about $x$.

edited: because I miswrote the problem, I am sorry.

Answer 1

A nice trick to do in this situation is to write

$$ \sin^3t = (1-\cos^2t)\sin t. $$

You can now do a secondary $u$-substitution where you can take $u = \cos t$ and $du = -\sin tdt$.

Edit

Where you left off you had

$$ \frac{I}{27} \;\; =\;\; \int \sin^3t dt \;\; =\;\; \int (1-\cos^2t)\sin t dt $$

If we now take the $u$-substitution I suggested above we find

$$ \frac{I}{27} \;\; =\;\;- \int (1 - u^2) du \;\; =\;\; \frac{1}{3}u^3 - u + C $$

for some constant $C$. We can do the first re-substitution

$$ \frac{I}{27} \;\; =\;\; \frac{1}{3} \cos^3t - \cos t+C $$

and now you can do the other substitution by noting that $\cos t = \sqrt{1 - \sin^2t} = \sqrt{1 - \frac{x^2}{9}} = \frac{1}{3} \sqrt{9 - x^2}$.

Additional Edit

If we plug back in the above expressions we get

\begin{eqnarray*} \frac{I}{27} & = & \cos t\left (\frac{1}{3}\cos^2t - 1\right ) + C \\ & = & \frac{1}{3} \sqrt{9-x^2}\left (\frac{1}{27}(9-x^2) - 1\right ) + C\\ I & = & \frac{1}{3} \sqrt{9-x^2} (9- x^2 - 27) + C \\ & = & -\frac{1}{3} \sqrt{9-x^2}(x^2 + 18) + C. \end{eqnarray*}

The answer you saw on Wolfram Alpha was the result of taking through the calculation as far as it could go.

Answer 2

Hint:

Think of Pythagoras' identity (which you used to remove the square root).

Your substitution was not done rigourously: it has to correspond to a bijective substitution. You should have set $t=\arcsin\dfrac x3$, i.e. $-\frac\pi2\le t\le\frac\pi2$.

Answer 3

Substitution $x=3\sin{y} \rightarrow dx=3\cos{y}$ $dy$

Then

$27 \int \frac{\sin^3{y}}{3\cos{y}}.3\cos{y}$ $dy$

$=\frac{27}{4}\int 4\sin^3{y}$ $dy$

Use Identity

$4\sin^3{y}=3\sin{y}-\sin{3y}$

$\frac{27}{4}\int 4\sin^3{y}$ $dy$ $=\frac{27}{4}\int 3\sin{y}-\sin{3y}$ $dy$

$=\frac{27}{4}(-3\cos{y}-\frac{\cos{3y}}{3})$

From substitution,

$\cos{y}=\frac{3}{\sqrt{9-x^2}}$

$\cos{3y}=4\cos^3{y}-3\cos{y}$

$\therefore \frac{27}{4}(-3\cos{y}-\frac{\cos{3y}}{3})=\frac{27}{4}[-\frac{9}{\sqrt{9-x^2}}-\frac{1}{3}(\frac{108}{(9-x^2)^{\frac{3}{2}}}-\frac{9}{\sqrt{9-x^2}})]$

You should be able to carry forward.Good luck!