# How do you integrate (x^5)(sqrt(4 - x^2)) dx?

Jun 11, 2015

This is of the form:

$\sqrt{{a}^{2} - {x}^{2}}$

which looks like:

$\sqrt{1 - 1 {\sin}^{2} x} = \cos x$

So, let's do the following substitution. Let:

$x = a \sin \theta$
$\mathrm{dx} = a \cos \theta d \theta$

where $a = \sqrt{4} = 2$

thus:

${x}^{5} = 32 {\sin}^{5} \theta$
$\sqrt{4 - {x}^{2}} = \sqrt{4 - 4 \sin \theta} = 2 \cos \theta$
$\mathrm{dx} = 2 \cos \theta d \theta$

$\int {x}^{5} \sqrt{4 - {x}^{2}} \mathrm{dx} = \int 32 {\sin}^{5} \theta \cdot 4 {\cos}^{2} \theta d \theta$

$= 128 \int {\sin}^{5} \theta {\cos}^{2} \theta d \theta$

$= 128 \int {\left({\sin}^{2} \theta\right)}^{2} \sin \theta {\cos}^{2} \theta d \theta$

$= 128 \int {\left(1 - {\cos}^{2} \theta\right)}^{2} {\cos}^{2} \theta \sin \theta d \theta$

Now, let:
$w = \cos \theta$
$\mathrm{dw} = - \sin \theta d \theta$

We can then get:

$= - 128 \int {\left(1 - {w}^{2}\right)}^{2} {w}^{2} \mathrm{dw}$

$= - 128 \int \left(1 - 2 {w}^{2} + {w}^{4}\right) {w}^{2} \mathrm{dw}$

$= - 128 \int {w}^{2} - 2 {w}^{4} + {w}^{6} \mathrm{dw}$

$= - 128 \left[{w}^{3} / 3 - \frac{2}{5} {w}^{5} + {w}^{7} / 7\right]$

Build a triangle; $\frac{x}{2} = \sin \theta$, so $\frac{\sqrt{4 - {x}^{2}}}{2} = \cos \theta$

Thus, we can re-substitute back in the previous values.

$= - \frac{128}{3} {\cos}^{3} \theta + \frac{256}{5} {\cos}^{5} \theta - \frac{128}{7} {\cos}^{7} \theta$

$= - {\cancel{128}}^{16} / 3 {\left(4 - {x}^{2}\right)}^{\frac{3}{2}} / \cancel{8} + {\cancel{256}}^{8} / 5 {\left(4 - {x}^{2}\right)}^{\frac{5}{2}} / \cancel{32} - \frac{\cancel{128}}{7} {\left(4 - {x}^{2}\right)}^{\frac{7}{2}} / \cancel{128}$

$= - \frac{16}{3} {\left(4 - {x}^{2}\right)}^{\frac{3}{2}} + \frac{8}{5} {\left(4 - {x}^{2}\right)}^{\frac{5}{2}} - \frac{1}{7} {\left(4 - {x}^{2}\right)}^{\frac{7}{2}} + C$

Jun 11, 2015

Use a $u$ substitution to get: $- \frac{16}{3} {\left(4 - {x}^{2}\right)}^{\frac{3}{2}} + \frac{8}{5} {\left(4 - {x}^{2}\right)}^{\frac{5}{2}} - \frac{1}{7} {\left(4 - {x}^{2}\right)}^{\frac{7}{2}} + C$

#### Explanation:

Although this integral can be evaluated using a trigonometric substitution, it can also be done with a $u$ substitution.

$\int {x}^{5} \sqrt{4 - {x}^{2}} \mathrm{dx}$

Let $u = 4 - {x}^{2}$, so $\mathrm{du} = - 2 x \mathrm{dx}$ and we can rewrite the integral:

$\int {x}^{5} \sqrt{4 - {x}^{2}} \mathrm{dx} = - \frac{1}{2} \int {x}^{4} \sqrt{4 - {x}^{2}} \left(- 2 x\right) \mathrm{dx}$

With our choice of $u = 4 - {x}^{2}$, we also get ${x}^{2} = 4 - u$

so ${x}^{4} = {\left({x}^{2}\right)}^{2} = {\left(4 - u\right)}^{2} = 16 - 8 u + {u}^{2}$

Substituting in the integral yields:

$- \frac{1}{2} \int \left(16 - 8 u + {u}^{2}\right) {u}^{\frac{1}{2}} \mathrm{du}$

Distribute the ${u}^{\frac{1}{2}}$ and integrate term by term.

$- \frac{1}{2} \int \left(16 {u}^{\frac{1}{2}} - 8 {u}^{\frac{3}{2}} + {u}^{\frac{5}{2}}\right) \mathrm{du}$

$= - \frac{1}{2} \left[16 \left(\frac{2}{3} {u}^{\frac{3}{2}}\right) - 8 \left(\frac{2}{5} {u}^{\frac{5}{2}}\right) + \left(\frac{2}{7} {u}^{\frac{7}{2}}\right)\right] + C$

Now simplify and rewrite as you see fit.

$= - \frac{16}{3} {\left(4 - {x}^{2}\right)}^{\frac{3}{2}} + \frac{8}{5} {\left(4 - {x}^{2}\right)}^{\frac{5}{2}} - \frac{1}{7} {\left(4 - {x}^{2}\right)}^{\frac{7}{2}} + C$