How do you evaluate $\int \frac{x}{\sqrt{1 - x^{2}}}$ $int x/sqrt(1-x^2)$ from $[- \frac{1}{2}, 0]$ $[-1/2, 0]$ ?

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How do you evaluate $\int \frac{x}{\sqrt{1 - x^{2}}}$ $int x/sqrt(1-x^2)$ from $[- \frac{1}{2}, 0]$ $[-1/2, 0]$ ?

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Answer 1 · 2016-12-13T14:01:42

The answer is $= \frac{\sqrt{3} - 2}{2} = - 0.134$ $=(sqrt3-2)/2=-0.134$

Explanation:

We do the integration by substitution

$\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C (n \neq - 1)$ $intx^ndx=x^(n+1)/(n+1)+C (n!=-1)$

Let $u = 1 - x^{2}$ $u=1-x^2$ , $\Rightarrow$ $=>$ , $d u = - 2 x d x$ $du=-2xdx$

$\int \frac{x d x}{\sqrt{1 - x^{2}}} = - \frac{1}{2} \int \frac{d u}{\sqrt{u}}$ $int(xdx)/sqrt(1-x^2)=-1/2int(du)/sqrtu$

$= - \frac{1}{2} \int u^{- \frac{1}{2}} d u = - \frac{1}{2} \frac{u^{\frac{1}{2}}}{\frac{1}{2}}$ $=-1/2intu^(-1/2)du=-1/2u^(1/2)/(1/2)$

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

So,

$\int_{- \frac{1}{2}}^{0} \frac{x d x}{\sqrt{1 - x^{2}}} = {[- \sqrt{1 - x^{2}}]}_{- \frac{1}{2}}^{0}$ $int_(-1/2)^0(xdx)/sqrt(1-x^2)=[-sqrt(1-x^2)]_(-1/2)^0$

$= - 1 + \sqrt{1 - \frac{1}{4}} = - 1 + (\frac{\sqrt{3}}{2})$ $=-1+sqrt(1-1/4)=-1+(sqrt3/2)$

$= \frac{\sqrt{3} - 2}{2} = - 0.134$ $=(sqrt3-2)/2=-0.134$

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How do you evaluate $\int \frac{x}{\sqrt{1 - x^{2}}}$ $int x/sqrt(1-x^2)$ from $[- \frac{1}{2}, 0]$ $[-1/2, 0]$ ?

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Explanation:

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