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

Answered by Olaf_Thorendsen last updated on 11/Jul/21

$$\mathrm{\Omega }={\int }_0^{\frac{1}{2}}\frac{1+\sqrt{3}}{\sqrt[4]{{(x+1)}^2{(1-x)}^6}}dx$$$$\mathrm{Let}x=\cos 2u$$$$\mathrm{\Omega }={\int }_{\frac{\pi }{4}}^{\frac{\pi }{6}}\frac{1+\sqrt{3}}{\sqrt[4]{{(\cos 2u+1)}^2{(1-\cos 2u)}^6}}(-2\sin 2udu)$$$$\mathrm{\Omega }=2{\int }_{\frac{\pi }{6}}^{\frac{\pi }{4}}\frac{1+\sqrt{3}}{\sqrt[4]{{\left({2\cos }^{2}u\right)}^2{\left({2\sin }^{2}u\right)}^6}}\sin 2udu$$$$\mathrm{\Omega }={\int }_{\frac{\pi }{6}}^{\frac{\pi }{4}}\frac{1+\sqrt{3}}{\cos u{\sin }^{3}u}\sin u\cos udu$$$$\mathrm{\Omega }={\int }_{\frac{\pi }{6}}^{\frac{\pi }{4}}\frac{1+\sqrt{3}}{{\sin }^{2}u}du$$$$\mathrm{\Omega }=(1+\sqrt{3}){[-\cot u]}_{\frac{\pi }{6}}^{\frac{\pi }{4}}$$$$\mathrm{\Omega }=(1+\sqrt{3})(\sqrt{3}-1)$$$$\mathrm{\Omega }=2$$

Answered by gsk2684 last updated on 11/Jul/21

$$\underset{0}{\stackrel{\frac{1}{2}}{\int }}\frac{1+\sqrt{3}}{{(x+1)}^{\frac{1}{2}}{(1-x)}^{\frac{3}{2}}}dx$$$$\underset{0}{\stackrel{\frac{1}{2}}{\int }}\frac{1+\sqrt{3}}{{(x+1)}^2{\left(\frac{1-x}{1+x}\right)}^{\frac{3}{2}}}dx,put\frac{1-x}{1+x}=t\Rightarrow \frac{-2}{{(1+x)}^2}dx=dt$$$$\underset{1}{\stackrel{\frac{1}{3}}{\int }}\frac{1+\sqrt{3}}{t^{\frac{3}{2}}}\frac{dt}{-2}=\frac{-1}{2}(1+\sqrt{3}){\left[\frac{t^{\frac{-1}{2}}}{\frac{-1}{2}}\right]}_1^{\frac{1}{3}}$$$$=(1+\sqrt{3})(\frac{1}{\sqrt{\frac{1}{3}}}-1)=2$$

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