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I have an integral that trig substitution could be used to simplify.

$$ \int\frac{x^3dx}{\sqrt{9-x^2}} $$

The first step is where I'm not certain I have it correct. I know that, say, $\sin \theta = \sqrt{1-cos^2 \theta}$, but is it correct in this case $3\sin \theta = \sqrt{9 - (3\cos \theta)^2}$?

Setting then $x = 3\cos \theta; dx = -3\sin \theta d\theta$

$$-\int \frac{(3\cos\theta)^3}{3\sin\theta}3\sin\theta d\theta$$

$$-27\int\cos^3\theta d\theta$$

$$-27\int(1-\sin ^2\theta)\cos \theta d\theta$$

Substituting again, $u=\sin \theta; du=\cos \theta d\theta$

$$-27\int(1-u^2)du $$

$$-27u + 9u^3 + C$$

$$-27\sin \theta + 9 \sin^3 \theta + C$$

$$-9\sqrt{9-x^2} + 3\sin\theta\cdot 3\sin\theta\cdot \sin \theta + C$$

$$-9\sqrt{9-x^2} + (\sqrt{9-x^2})^2 \cdot \frac{\sqrt{9-x^2}}{3} + C$$

$$-9\sqrt{9-x^2} + \frac{1}{3}(9-x^2)(9-x^2)^{\frac{1}{2}} + C$$

$$-9\sqrt{9-x^2} + \frac{1}{3}(9-x^2)^\frac{3}{2} + C $$

I guess I have more doubts that I've done the algebra correctly than the substitution, but in any case I'm not getting the correct answer. Have I calculated correctly? Is the answer simplified completely?

EDIT

Answer needed to be simplified further:

$$-9\sqrt{9-x^2} + \frac{1}{3}(\sqrt{9-x^2}^2 \sqrt{9-x^2}) + C$$

$$-9\sqrt{9-x^2} + \frac{1}{3}((9-x^2)\sqrt{9-x^2}) + C$$

$$\sqrt{9-x^2} \left (-9 + \frac{1}{3}(9-x^2) \right ) + C$$

$$\sqrt{9-x^2} \left (-6 - \frac{x^2}{3} \right ) + C$$

$$ \bbox[5px,border:2px solid red] { - \left ( 6+ \frac{x^2}{3} \right ) \sqrt{9-x^2} } $$

This is the answer the assignment was looking for.

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    $\begingroup$ Everything looks correct. You can differentiate your answer to check it. You might also want to clarify that we take $\theta\in(0,\pi)$ in $x=3\cos\theta$. This ensures $\sin\theta>0$. $\endgroup$ – bjorn93 Mar 19 '20 at 7:33
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You work is correct, You may further simplify $$I=-9\sqrt{9-x^2}+\frac{1}{3}\left(\sqrt{9-x^2}\right)^3 +C= \sqrt{9-x^2} \left (-9+\frac{9-x^2}{3}\right)+C$$ $$\implies I=-\frac{1}{3}\sqrt{9-x^2}~~(18+x^2)+C,$$ which is the final correct answer.

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  • $\begingroup$ The assignment had this simplification of the answer but with the $\frac{1}{3}$ carried through to the second term, $-\left (6 + \frac{x^2}{3} \right ) ...$ $\endgroup$ – Conner M. Mar 19 '20 at 15:18
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Maybe the following hint, based on Differential binomial, seems to be the same as @Toby's. However, you focused just on a trig substitution. You can take $$(9-x^2)=t^2,$$ and simplify the integral...

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You can solve without the trigonometric substitution, decomposing as

$$\frac{x^3}{\sqrt{9-x^2}}=9\frac{x}{\sqrt{9-x^2}}-x\sqrt{9-x^2}$$ and the antiderivatives are immediate (by $u=x^2$):

$$-9\sqrt{9-x^2}-\frac13(9-x^2)^{3/2}.$$

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alternative is to use $u=\sqrt{9-x^{2}}$

$$ \begin{aligned} \int{\frac{x^{3}}{\sqrt{9-x^{2}}}\ dx}&=\int{(u^{2}-9)\ du}\\ &=\frac{u^{3}}{3}-9u+c\\ &=\frac{\left(\sqrt{9-x^{2}}\right)^{3}}{3}-9\sqrt{9-x^{2}}+c \end{aligned} $$

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What you have done is absolutely correct, except where you forgot to mention that $\theta$ is in $(0, \pi)$, but you can simplify your answer further.

The book's answer might be something like $-\frac{1}{3} \sqrt{9-x^2} (x^2+18)$, which you can get by factoring out a factor of $\sqrt{9-x^2}$:

$$-9\sqrt{9-x^2} + \frac{1}{3}(9-x^2)(9-x^2)^{\frac{1}{2}} + C$$ $$= \sqrt{9-x^2} \left(-9 + \frac{1}{3}(9-x^2) \right)+ C$$

and you can surely continue from here.

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  • $\begingroup$ I didn't downvote your answer, so let's agree to disagree. $\endgroup$ – Toby Mak Mar 19 '20 at 7:44
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Firstly, $$\sin\theta=\sqrt{1-\cos^2\theta}$$ is wrong. Try $\theta=-\frac{\pi}{2}$.

But for $\theta\in(0,\pi)$ we see that $-3<3\cos\theta<3$ and $x=3\cos\theta$ gets any value from $(-3,3)$.

Also, for these values of $\theta$ we obtain a right formula: $$\sin\theta=\sqrt{1-\cos^2\theta}.$$ Secondly, $$\cos^3\theta=\frac{1}{4}(\cos3\theta+3\cos\theta)$$ and we can evaluate the integral shorter.

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  • $\begingroup$ Why did you get that damned downvote?!!! $\endgroup$ – Mikasa Mar 19 '20 at 13:29
  • $\begingroup$ @mrs I don't know. $\endgroup$ – Michael Rozenberg Mar 19 '20 at 14:07
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    $\begingroup$ Lets make it recovered! $\endgroup$ – Mikasa Mar 20 '20 at 8:23

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