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

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

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Answer 1 · 2015-06-23T21:02:16

This can be done from trig substitution. Notice how

$\sqrt{a^{2} - x^{2}} \propto \sqrt{a^{2} - a^{2} {sin}^{2} θ} \propto \sqrt{4 - x^{2}}$ $sqrt(a^2 - x^2) prop sqrt(a^2 - a^2sin^2theta) prop sqrt(4 - x^2)$
where $a = 2$ $a = 2$

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

$\Rightarrow \int 2 cos θ \cdot 2 cos θ d θ$ $=> int 2costheta*2costhetad theta$

$= 4 \int {cos}^{2} θ d θ$ $= 4int cos^2thetad theta$

Now you can use the identity:
${cos}^{2} θ = \frac{1 + cos (2 θ)}{2}$ $cos^2theta = (1+cos(2theta))/2$

Thus:
$= 2 \int d θ + 2 \int cos (2 θ) d θ$ $= 2int d theta + 2int cos(2theta)d theta$

$= 2 \int d θ + 2 \cdot \frac{1}{2} \int 2 cos (2 θ) d θ$ $= 2int d theta + 2*1/2int 2cos(2theta)d theta$

$= 2 θ + sin (2 θ) + C$ $= 2theta + sin(2theta) + C$

Since $x = 2 sin θ$ $x = 2sintheta$ , $θ = arcsin (\frac{x}{2})$ $theta = arcsin(x/2)$ .
Since $sin (2 θ) = 2 sin θ cos θ$ $sin(2theta) = 2sinthetacostheta$ :

$sin (2 θ) = \frac{x \sqrt{4 - x^{2}}}{2}$ $sin(2theta) = (xsqrt(4-x^2))/2$

$\Rightarrow 2 arcsin (\frac{x}{2}) + \frac{x \sqrt{4 - x^{2}}}{2} + C$ $=> color(blue)(2arcsin(x/2) + (xsqrt(4-x^2))/2 + C)$

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

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