How do you evaluate the integral $\int \frac{d x}{{(1 - x^{2})}^{\frac{5}{2}}}$ $int dx/(1-x^2)^(5/2)$ ?

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How do you evaluate the integral $\int \frac{d x}{{(1 - x^{2})}^{\frac{5}{2}}}$ $int dx/(1-x^2)^(5/2)$ ?

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Answer 1 · 2017-08-15T16:52:16

$\frac{x (3 - 2 x^{2})}{3 {(1 - x^{2})}^{\frac{3}{2}}} + C$ $(x(3-2x^2))/(3(1-x^2)^(3/2))+C$

Explanation:

Let $x = sin θ$ $x=sintheta$ . This implies that $d x = cos θ d θ$ $dx=costhetad theta$ .

$\int \frac{d x}{{(1 - x^{2})}^{\frac{5}{2}}}$ $intdx/(1-x^2)^(5/2)$

$= \int \frac{cos θ d θ}{{(1 - {sin}^{2} θ)}^{\frac{5}{2}}}$ $=int(costhetad theta)/(1-sin^2theta)^(5/2)$

$= \int \frac{cos θ}{{({cos}^{2} θ)}^{\frac{5}{2}}} d θ$ $=intcostheta/(cos^2theta)^(5/2)d theta$

$= \int {sec}^{4} θ d θ$ $=intsec^4thetad theta$

$= \int {sec}^{2} θ (1 + {tan}^{2} θ) d θ$ $=intsec^2theta(1+tan^2theta)d theta$

Let $u = tan θ$ $u=tantheta$ so $d u = {sec}^{2} θ d θ$ $du=sec^2thetad theta$ .

$= \int (1 + u^{2}) d u$ $=int(1+u^2)du$

$= u + \frac{1}{3} u^{3} + C$ $=u+1/3u^3+C$

$= tan θ + \frac{1}{3} {tan}^{3} θ + C$ $=tantheta+1/3tan^3theta+C$

$= \frac{1}{3} tan θ (3 + {tan}^{2} θ) + C$ $=1/3tantheta(3+tan^2theta)+C$

Where $x = sin θ$ $x=sintheta$ , we have a triangle where the opposite side is $x$ $x$ and the hypotenuse is $1$ $1$ . Thus, the adjacent side is $\sqrt{1 - x^{2}}$ $sqrt(1-x^2)$ and $tan θ = x / \sqrt{1 - x^{2}}$ $tantheta=x//sqrt(1-x^2)$ .

$= \frac{x}{3 \sqrt{1 - x^{2}}} (3 + \frac{x^{2}}{1 - x^{2}})$ $=x/(3sqrt(1-x^2))(3+x^2/(1-x^2))$

$= \frac{1}{3 \sqrt{1 - x^{2}}} (\frac{3 - x^{2}}{1 - x^{2}})$ $=1/(3sqrt(1-x^2))((3-x^2)/(1-x^2))$

$= \frac{x (3 - 2 x^{2})}{3 {(1 - x^{2})}^{\frac{3}{2}}} + C$ $=(x(3-2x^2))/(3(1-x^2)^(3/2))+C$

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How do you evaluate the integral $\int \frac{d x}{{(1 - x^{2})}^{\frac{5}{2}}}$ $int dx/(1-x^2)^(5/2)$ ?

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Explanation:

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