How do you integrate $(x^{5}) (\sqrt{4 - x^{2}}) d x$ $(x^5)(sqrt(4 - x^2)) dx$ ?

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How do you integrate $(x^{5}) (\sqrt{4 - x^{2}}) d x$ $(x^5)(sqrt(4 - x^2)) dx$ ?

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Answer 1 · 2015-06-11T01:36:03

This is of the form:

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

which looks like:

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

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

$x = a sin θ$ $x = asintheta$
$d x = a cos θ d θ$ $dx = acosthetad theta$

where $a = \sqrt{4} = 2$ $a = sqrt4 = 2$

thus:

$x^{5} = 32 {sin}^{5} θ$ $x^5 = 32sin^5theta$
$\sqrt{4 - x^{2}} = \sqrt{4 - 4 sin θ} = 2 cos θ$ $sqrt(4-x^2) = sqrt(4-4sintheta) = 2costheta$
$d x = 2 cos θ d θ$ $dx = 2costhetad theta$

$\int x^{5} \sqrt{4 - x^{2}} d x = \int 32 {sin}^{5} θ \cdot 4 {cos}^{2} θ d θ$ $int x^5sqrt(4-x^2)dx = int 32sin^5theta*4cos^2thetad theta$

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

$= 128 \int {({sin}^{2} θ)}^{2} sin θ {cos}^{2} θ d θ$ $= 128int (sin^2theta)^2sinthetacos^2thetad theta$

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

Now, let:
$w = cos θ$ $w = costheta$
$d w = - sin θ d θ$ $dw = -sinthetad theta$

We can then get:

$= - 128 \int {(1 - w^{2})}^{2} w^{2} d w$ $= -128int (1-w^2)^2w^2dw$

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

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

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

Build a triangle; $\frac{x}{2} = sin θ$ $x/2 = sintheta$ , so $\frac{\sqrt{4 - x^{2}}}{2} = cos θ$ $sqrt(4-x^2)/2 = costheta$

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

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

$= -cancel(128)^(16)/3(4-x^2)^(3/2)/cancel(8) + cancel(256)^8/5(4-x^2)^(5/2)/cancel(32) - cancel(128)/7(4-x^2)^(7/2)/cancel(128)$

$= -16/3(4-x^2)^(3/2) + 8/5(4-x^2)^(5/2) - 1/7(4-x^2)^(7/2) + C$

Answer 2 · 2015-06-11T02:52:32

Use a $u$ substitution to get: $-16/3(4-x^2)^(3/2) +8/5 (4-x^2)^(5/2) - 1/7 (4-x^2)^(7/2) +C$

Explanation:

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

$int x^5 sqrt(4-x^2) dx$

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

$int x^5 sqrt(4-x^2) dx = -1/2 int x^4 sqrt(4-x^2) (-2x) dx$

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

so $x^4 = (x^2)^2 = (4-u)^2 = 16 - 8u +u^2$

Substituting in the integral yields:

$-1/2 int (16 - 8u +u^2) u^(1/2) du$

Distribute the $u^(1/2)$ and integrate term by term.

$-1/2 int (16u^(1/2) - 8u^(3/2) +u^(5/2)) du$

$=-1/2[16 (2/3 u^(3/2)) - 8 (2/5 u^(5/2)) +(2/7u^(7/2))]+C$

Now simplify and rewrite as you see fit.

$= -16/3(4-x^2)^(3/2) +8/5 (4-x^2)^(5/2) - 1/7 (4-x^2)^(7/2) +C$

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