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

Answered by MJS_new last updated on 15/Oct/21

$$0⩽x⩽\frac{1}{2}\Rightarrow$$$$\int \frac{dx}{\sqrt[4]{{(1+x)}^2{(1-x)}^6}}=$$$$=\int \frac{dx}{(1-x)\sqrt{1-x^2}}=$$$$[t=\frac{1+\sqrt{1-x^2}}{x}\to dx=-\frac{x^2\sqrt{1-x^2}}{1+\sqrt{1-x^2}}dt]$$$$=-2\int \frac{dt}{{(t-1)}^2}=\frac{2}{t-1}=...$$$$=\frac{\sqrt{1-x^2}}{1-x}-1+C$$$$\Rightarrow$$$$(-1+\sqrt{3})\underset{0}{\stackrel{1/2}{\int }}\frac{dx}{\sqrt[4]{{(1+x)}^2{(1-x)}^6}}=4-2\sqrt{3}$$

Answered by FongXD last updated on 15/Oct/21

$$={\int }_0^{\frac{1}{2}}\left(\frac{\sqrt{3}-1}{\sqrt[4]{{(1+\mathrm{x})}^2{(1-\mathrm{x})}^2{(1-\mathrm{x})}^4}}\right)\mathrm{dx}$$$$={\int }_0^{\frac{1}{2}}\left[\frac{\sqrt{3}-1}{(1-\mathrm{x})\sqrt[4]{{(1-{\mathrm{x}}^2)}^2}}\right]\mathrm{dx}$$$$\mathrm{let}\mathrm{x}=\sin \theta ,\Rightarrow \mathrm{dx}=\cos \theta \mathrm{d}\theta$$$$={\int }_0^{\frac{\pi }{6}}\frac{\sqrt{3}-1}{(1-\sin \theta )\sqrt[4]{{(1-{\sin }^2\theta )}^2}}\times \cos \theta \mathrm{d}\theta$$$$={\int }_0^{\frac{\pi }{6}}\frac{(\sqrt{3}-1)\mathrm{d}\theta }{1-\sin \theta }={\int }_0^{\frac{\pi }{6}}\frac{(\sqrt{3}-1)\mathrm{d}\theta }{{(\cos \frac{\theta }{2}-\sin \frac{\theta }{2})}^2}$$$$=\frac{\sqrt{3}-1}{2}{\int }_0^{\frac{\pi }{6}}\frac{\mathrm{d}\theta }{{\cos }^2(\frac{\theta }{2}+\frac{\pi }{4})}=\frac{\sqrt{3}-1}{2}{\left[2\tan (\frac{\theta }{2}+\frac{\pi }{4})\right]}_0^{\frac{\pi }{6}}$$$$=(\sqrt{3}-1)(\tan \frac{\pi }{3}-\tan \frac{\pi }{4})={(\sqrt{3}-1)}^2=4-2\sqrt{3}$$$$\mathrm{so}.\overline{)\begin{array}{c}{\int }_0^{\frac{1}{2}}\left(\frac{\sqrt{3}-1}{\sqrt[4]{{(1+\mathrm{x})}^2{(1-\mathrm{x})}^4}}\right)\mathrm{dx}=4-2\sqrt{3}\end{array}}$$

Commented by cortano last updated on 15/Oct/21

$$\mathrm{yes}$$

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