# Trig substitution for $\sqrt{9-x^2}$

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.

• 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 $\sin\theta>0$. – bjorn93 Mar 19 '20 at 7:33

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.

• The assignment had this simplification of the answer but with the $\frac{1}{3}$ carried through to the second term, $-\left (6 + \frac{x^2}{3} \right ) ...$ – Conner M. Mar 19 '20 at 15:18

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...

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 $$u=x^2$$):

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

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}

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.

• I didn't downvote your answer, so let's agree to disagree. – Toby Mak Mar 19 '20 at 7:44

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.

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