How to integrate the following ??

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How to integrate the following ??

$\int \frac{d x}{{(1 - x^{2})}^{\frac{3}{2}}}$ $intdx/(1 - x^2)^(3/2)$

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Answer 1 · 2017-10-01T14:50:34

The answer is $= \frac{x}{\sqrt{1 - x^{2}}} + C$ $=x/sqrt(1-x^2)+C$

Explanation:

We need $\int {sec}^{2} x d x = tan x$ $intsec^2x dx=tanx$

We perform this integral by substitution

Let $x = sin θ$ $x=sintheta$ , $\Rightarrow$ $=>$ , $d x = cos θ d θ$ $dx=costheta d theta$

$1 - x^{2} = 1 - {sin}^{2} θ = {cos}^{2} θ$ $1-x^2=1-sin^2theta=cos^2theta$

$tan θ = \frac{sin θ}{cos θ} = \frac{x}{\sqrt{1 - x^{2}}}$ $tan theta=sintheta/cos theta=x/sqrt(1-x^2)$

Therefore,

$\int \frac{d x}{{(1 - x^{2})}^{\frac{3}{2}}} = \int \frac{cos θ d θ}{{({cos}^{2})}^{\frac{3}{2}}}$ $int(dx)/(1-x^2)^(3/2)=int(costhetad theta)/(cos^2)^(3/2)$

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

$= \int {sec}^{2} θ d θ$ $=intsec^2theta d theta$

$= tan θ$ $=tantheta$

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

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How to integrate the following ??

$\int \frac{d x}{{(1 - x^{2})}^{\frac{3}{2}}}$ $intdx/(1 - x^2)^(3/2)$

1 Answer

Explanation:

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How to integrate the following ??

intdx/(1 - x^2)^(3/2)∫dx(1−x2)32intdx/(1 - x^2)^(3/2)

1 Answer

Explanation:

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$\int \frac{d x}{{(1 - x^{2})}^{\frac{3}{2}}}$ $intdx/(1 - x^2)^(3/2)$