How do you find the integral of $\sqrt{9 - x^{2}} d x$ $sqrt(9-x^2)dx$ ?

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How do you find the integral of $\sqrt{9 - x^{2}} d x$ $sqrt(9-x^2)dx$ ?

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Answer 1 · 2015-08-17T09:58:28

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

Explanation:

${sin}^{2} A + {cos}^{2} A = 1$ $sin^2 A + cos^2 A = 1$
$cos 2 A = 2 {cos}^{2} A - 1$ $cos 2A = 2cos^2 A - 1$
$sin 2 A = 2 sin A cos A$ $sin 2A = 2sin Acos A$

Use substitution.
$x = 3 sin t$ $x = 3 sin t$

$\int \sqrt{9 - x^{2}} d x = \int 3 \sqrt{1 - {sin}^{2} t} \times 3 cos t d t = 9 \int {cos}^{2} t d t$ $int sqrt(9-x^2)dx = int 3sqrt(1-sin^2 t) xx 3 cos t dt = 9 int cos^2 t dt$

$9 \int {cos}^{2} t d x = \frac{9}{2} \int (1 + cos 2 t) d t = \frac{9}{2} (t + \frac{1}{2} sin 2 t) + C$ $9int cos^2 tdx = 9/2 int (1+cos 2t)dt = 9/2 (t + 1/2sin 2t) + C$

In terms of $x$ $x$ :
$\int \sqrt{9 - x^{2}} d x = \frac{9}{2} ({sin}^{- 1} (\frac{x}{3}) + \frac{x}{3} \sqrt{1 - {(\frac{x}{3})}^{2}}) + C$ $int sqrt(9-x^2)dx = 9/2(sin^(-1) (x/3) + x/3sqrt(1-(x/3)^2)) + C$

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How do you find the integral of $\sqrt{9 - x^{2}} d x$ $sqrt(9-x^2)dx$ ?

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How do you find the integral of $\sqrt{9 - x^{2}} d x$ $sqrt(9-x^2)dx$ ?