Question-156743

Question Number 156743 by cortano last updated on 15/Oct/21

Answered by MJS_new last updated on 15/Oct/21

0 ⩽ x ⩽ \frac{1}{2} \Rightarrow

\int \frac{d x}{\sqrt[4]{{(1 + x)}^{2} {(1 - x)}^{6}}} =

= \int \frac{d x}{(1 - x) \sqrt{1 - x^{2}}} =

[t = \frac{1 + \sqrt{1 - x^{2}}}{x} \to d x = - \frac{x^{2} \sqrt{1 - x^{2}}}{1 + \sqrt{1 - x^{2}}} d t]

= - 2 \int \frac{d t}{{(t - 1)}^{2}} = \frac{2}{t - 1} = \dots

= \frac{\sqrt{1 - x^{2}}}{1 - x} - 1 + C

\Rightarrow

(- 1 + \sqrt{3}) \underset{0}{\int^{1 / 2}} \frac{d x}{\sqrt[4]{{(1 + x)}^{2} {(1 - x)}^{6}}} = 4 - 2 \sqrt{3}

Answered by FongXD last updated on 15/Oct/21

= \int_{0}^{\frac{1}{2}} (\frac{\sqrt{3} - 1}{\sqrt[4]{{(1 + x)}^{2} {(1 - x)}^{2} {(1 - x)}^{4}}}) dx

= \int_{0}^{\frac{1}{2}} [\frac{\sqrt{3} - 1}{(1 - x) \sqrt[4]{{(1 - x^{2})}^{2}}}] dx

let x = \sin θ, \Rightarrow dx = \cos θ d θ

= \int_{0}^{\frac{π}{6}} \frac{\sqrt{3} - 1}{(1 - \sin θ) \sqrt[4]{{(1 - \sin^{2} θ)}^{2}}} \times \cos θ d θ

= \int_{0}^{\frac{π}{6}} \frac{(\sqrt{3} - 1) d θ}{1 - \sin θ} = \int_{0}^{\frac{π}{6}} \frac{(\sqrt{3} - 1) d θ}{{(\cos \frac{θ}{2} - \sin \frac{θ}{2})}^{2}}

= \frac{\sqrt{3} - 1}{2} \int_{0}^{\frac{π}{6}} \frac{d θ}{\cos^{2} (\frac{θ}{2} + \frac{π}{4})} = \frac{\sqrt{3} - 1}{2} {[2 \tan (\frac{θ}{2} + \frac{π}{4})]}_{0}^{\frac{π}{6}}

= (\sqrt{3} - 1) (\tan \frac{π}{3} - \tan \frac{π}{4}) = {(\sqrt{3} - 1)}^{2} = 4 - 2 \sqrt{3}

so . \begin{matrix} \int_{0}^{\frac{1}{2}} (\frac{\sqrt{3} - 1}{\sqrt[4]{{(1 + x)}^{2} {(1 - x)}^{4}}}) dx = 4 - 2 \sqrt{3} \end{matrix}

Commented by cortano last updated on 15/Oct/21

yes

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