How to calculate $\int \frac{x^3}{\sqrt{9-x^2}}$

Question

According to my textbook this is : $\frac{\sqrt{9-x^2}^3}{3} -9 \sqrt{9-x^2}$. However my solution is completely different. Here's my work:

$ x = 3\sin \theta, dx = 3 \cos\theta d\theta$

$$\int \frac{27\cos^3 \theta}{\sqrt{9-9\sin^2\theta}} = \int \frac{27\cos^3\theta}{3\cos^2\theta} = 9\int\cos^2 \theta$$

$$9\left(\int \frac{1}{2} d\theta+ \int \cos 2\theta\right) = 9\left(\frac{1}{2}\theta + \frac{\sin 2\theta}{2}\right)$$

Solving for $\theta$ yields:

$$\frac{9}{2}\arcsin\left(\frac{x}{3}\right) + \frac{9}{4}\sin\left(2\arcsin\left(\frac{x}{3}\right)\right)$$

What's wrong with my solution?

Answer 1

You have two mistakes here.

$x = 3 \sin \theta$ I like that

$\int\frac {x^3}{\sqrt {9-x^2}} \ dx = \int \frac {27\sin^3 \theta}{\sqrt {9-9sin^2\theta}} (3\cos \theta) \ d\theta$

Your first mistake is at this step. Do you see where you went wrong?

next: $\sqrt {9-9\sin^2\theta} = 3\cos \theta$ not $3\cos^2 \theta$

$\int 27\sin^3 \theta \ d\theta$

Can you get home from here? If you are stuck try: $\sin^2 \theta = 1-\cos^2\theta$

Update:

"I still can't see how $27(-\cos(\arcsin(\frac{x}{3})) + \frac{\cos^3(\arcsin(\frac{x}{3}))}{3})$ is equivalent to the solution given in the textbook."

$\sin (\arcsin \frac {x}{3}) = \frac {x}{3}\\ \sin^2 (\arcsin \frac {x}{3}) = \frac {x^2}{9}\\ 1-\sin^2 (\arcsin \frac {x}{3}) = 1-\frac {x^2}{9}\\ \cos^2 (\arcsin \frac {x}{3}) = 1-\frac {x^2}{9}\\ \cos (\arcsin \frac {x}{3}) = \sqrt {1-\frac {x^2}{9}}\\ \cos (\arcsin \frac {x}{3}) = \frac {\sqrt {9-x^2}}{3}\\$

This can also be done geometrically, and I suggest you prove it this way (an exercise left to the reader).

$27(-\cos(\arcsin(\frac{x}{3}) + \frac{\cos^3(\arcsin(\frac{x}{3}))}{3})\\ 27(-\sqrt {1 - \frac{x^2}{9}} + \frac{\left(\sqrt {1-\frac {x^2}{9}}\right)^3}{3})\\ -9\sqrt {9 - x^2} + \frac{\left(\sqrt {9- x^2}\right)^3}{3} $

Answer 2

After doing your substitution, the numerator should become $81\sin^3\theta\cos\theta.$

Anyway, my suggestion is that you do $x^2=y$ and $2x\,\mathrm dx=\mathrm dy$.

Answer 3

With your substitution we have: $$ \int \frac{x^3}{\sqrt{9-x^2}}dx=\int \frac{81\sin^3 \theta \cos \theta}{\sqrt{9-9\sin^2\theta}}d \theta $$

that cannot be simplified as in OP.

Answer 4

substituting $$u=x^2$$ then we have $$du=2xdx$$ and our integral will be $$\frac{1}{2}\int\frac{u}{\sqrt{9-u}}du$$ no let $$s=9-u$$ and with $$ds=-du$$our integral is $$-\frac{1}{2}\int\left(\frac{9}{\sqrt{s}}-\sqrt{s}\right)ds$$

Answer 5

Let $u = 9-x^2$, $dx=\frac{du}{-2x}$ $$\int \frac{x^3}{\sqrt{9-x^2}}dx=-1/2\int \frac{9-u}{\sqrt{u}}du=-1/2[9(2)u^{1/2}-(2/3)u^{3/2}]\\=-9(9-x^2)^{1/2}+1/3(9-x^2)^{1/3}$$

Answer 6

$$I=\int \frac{x^3}{\sqrt{9-x^2}}dx$$ Substitute $u=9-x^2$ then $du=-2xdx$ $$I=\int \frac{x^3}{\sqrt{9-x^2}}dx=-\frac12\int \frac {9-u}{\sqrt u}du$$

Answer 7

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

Answer 8

Plenty of good answers, but as the expression goes 'There's more than one way to skin a Cat' so I will provide another.

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

Employ integration by parts:

$$\int u(x)v'(x) = u(x)v(x) - \int v(x)u'(x)$$

\begin{align} u(x) &= x^2 & v'(x) &= \frac{x}{\sqrt{9 - x^2}} \\ u'(x) &= 2x & v(x) &= -\sqrt{9 - x^2} \end{align}

Thus

\begin{align} \int \frac{x^3}{\sqrt{9-x^2}}\:dx &= x^2 \cdot -\sqrt{9 - x^2} - \int -\sqrt{9 - x^2} \cdot 2x \:dx \\ &= -x^2\sqrt{9 - x^2} + 2 \int x\sqrt{9 - x^2}\:dx \\ & = -x^2\sqrt{9 - x^2} + 2 \cdot -\frac{1}{3}\left(9 - x^2\right)^{\frac{3}{2}} + C \\ &= -x^2\sqrt{9 - x^2}- \frac{2}{3}\left(9 - x^2\right)^{\frac{3}{2}} + C \end{align}

Where $C$ is a constant of integration

Answer 9

Why not one more method?

\begin{align} \int \frac{x^3}{\sqrt{9 - x^2}}\:dx &= \int x \cdot \frac{x^2}{\sqrt{9 - x^2}}\:dx = - \int x \cdot \frac{-x^2}{\sqrt{9 - x^2}}\:dx \\ &=- \int x \cdot \frac{-x^2 - 9 + 9}{\sqrt{9 - x^2}}\:dx = -\int x \cdot \left[ \frac{9 - x^2}{\sqrt{9 -x^2}} - \frac{9}{\sqrt{9 - x^2}} \right]\:dx\\ &= -\int x\sqrt{9 = x^2}\:dx + \int \frac{9x}{\sqrt{9 - x^2}}\:dx \\ &= \frac{1}{3}\left(9 - x^2\right)^{\frac{3}{2}} - 9\sqrt{9 - x^2} + C \end{align}

Where $C$ is a constant of integration.

Answer 10

Substitute $u^2=9-x^2$ and integrate to obtain: $$-9u + \frac{u^3}{3} +C.$$ (Observe that $2u\ du=-2x\ dx.$)