Question-146106

Question Number 146106 by smallEinstein last updated on 10/Jul/21

Answered by Olaf_Thorendsen last updated on 11/Jul/21

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

Let x = \cos 2 u

Ω = \int_{\frac{π}{4}}^{\frac{π}{6}} \frac{1 + \sqrt{3}}{\sqrt[4]{{(\cos 2 u + 1)}^{2} {(1 - \cos 2 u)}^{6}}} (- 2 \sin 2 u d u)

Ω = 2 \int_{\frac{π}{6}}^{\frac{π}{4}} \frac{1 + \sqrt{3}}{\sqrt[4]{{({2 \cos}^{2} u)}^{2} {({2 \sin}^{2} u)}^{6}}} \sin 2 u d u

Ω = \int_{\frac{π}{6}}^{\frac{π}{4}} \frac{1 + \sqrt{3}}{\cos u \sin^{3} u} \sin u \cos u d u

Ω = \int_{\frac{π}{6}}^{\frac{π}{4}} \frac{1 + \sqrt{3}}{\sin^{2} u} d u

Ω = (1 + \sqrt{3}) {[- \cot u]}_{\frac{π}{6}}^{\frac{π}{4}}

Ω = (1 + \sqrt{3}) (\sqrt{3} - 1)

Ω = 2

Answered by gsk2684 last updated on 11/Jul/21

\underset{0}{\int^{\frac{1}{2}}} \frac{1 + \sqrt{3}}{{(x + 1)}^{\frac{1}{2}} {(1 - x)}^{\frac{3}{2}}} d x

\underset{0}{\int^{\frac{1}{2}}} \frac{1 + \sqrt{3}}{{(x + 1)}^{2} {(\frac{1 - x}{1 + x})}^{\frac{3}{2}}} d x, p u t \frac{1 - x}{1 + x} = t \Rightarrow \frac{- 2}{{(1 + x)}^{2}} d x = d t

\underset{1}{\int^{\frac{1}{3}}} \frac{1 + \sqrt{3}}{t^{\frac{3}{2}}} \frac{d t}{- 2} = \frac{- 1}{2} (1 + \sqrt{3}) {[\frac{t^{\frac{- 1}{2}}}{\frac{- 1}{2}}]}_{1}^{\frac{1}{3}}

= (1 + \sqrt{3}) (\frac{1}{\sqrt{\frac{1}{3}}} - 1) = 2

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