How do you find $\int \frac{x^{2}}{\sqrt{81 - x^{2}}} d x$ $int (x^2) / (sqrt(81 - x^2)) dx$ using trigonometric substitution?

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How do you find $\int \frac{x^{2}}{\sqrt{81 - x^{2}}} d x$ $int (x^2) / (sqrt(81 - x^2)) dx$ using trigonometric substitution?

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Answer 1 · 2015-12-06T20:01:26

$= \frac{81}{2} {sin}^{- 1} (\frac{x}{9}) - \frac{x \sqrt{81 - x^{2}}}{2} + C$ $=81/2sin^(-1)(x/9)-(xsqrt(81-x^2))/2+C$

Explanation:

Let $x = 9 sin θ$ $x=9sintheta$
Then $d x = 9 cos θ d θ$ $dx=9costheta d theta$ and $sin θ = \frac{x}{9}$ $sintheta=x/9$

Then the original integral becomes

$9 \int \frac{81 {sin}^{2} θ}{\sqrt{81 - 81 {sin}^{2} θ}} cos θ d θ$ $9int(81sin^2theta)/(sqrt(81-81sin^2theta))costheta d theta$

Now using the trig identities ${sin}^{2} θ + {cos}^{2} θ = 1$ $sin^2theta+cos^2theta = 1$ and
$sin 2 θ = 2 sin θ cos θ$ $sin2theta=2sinthetacostheta$ , together with factorizations and laws of surds, we may compute the integral as

$9 \times 81 \int \frac{{sin}^{2} θ}{9 \sqrt{1 - {sin}^{2} θ}} cos θ d θ$ $9xx81int(sin^2theta)/(9sqrt(1-sin^2theta))costheta d theta$

$= 81 \int \frac{{sin}^{2} θ}{cos θ} cos θ d θ$ $=81int(sin^2theta)/costheta costheta d theta$

$= 81 [\frac{θ}{2} - \frac{sin 2 θ}{4}] + C$ $=81[theta/2-(sin2theta)/4]+C$

$= \frac{81}{2} {sin}^{- 1} (\frac{x}{9}) - \frac{81}{2} \cdot \frac{x}{9} \cdot \frac{\sqrt{81 - x^{2}}}{9} + C$ $=81/2sin^(-1)(x/9)-81/2*x/9*sqrt(81-x^2)/9+C$

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How do you find $\int \frac{x^{2}}{\sqrt{81 - x^{2}}} d x$ $int (x^2) / (sqrt(81 - x^2)) dx$ using trigonometric substitution?

1 Answer

Explanation:

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How do you find∫x2√81−x2dxint (x^2) / (sqrt(81 - x^2)) dx using trigonometric substitution?

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Explanation:

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How do you find $\int \frac{x^{2}}{\sqrt{81 - x^{2}}} d x$ $int (x^2) / (sqrt(81 - x^2)) dx$ using trigonometric substitution?