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Question Number 206962 by Ghisom last updated on 01/May/24

$$\underset{0}{\stackrel{1}{\int }}\frac{\sqrt{1-x}}{\sqrt{1-\sqrt{1-x}}+\sqrt{1+\sqrt{1-x}}}dx=?$$

Answered by lepuissantcedricjunior last updated on 02/May/24

$${\int }_0^1\frac{\sqrt{1-\mathit{x}}}{\sqrt{1-\sqrt{1-\mathit{x}}}+\sqrt{1+\sqrt{1-\mathit{x}}}}\mathit{d}\mathit{x}=\mathit{k}$$$$\mathit{k}={\int }_0^1\frac{\sqrt{1-\mathit{x}}}{{\left(\sqrt{1-\sqrt{1-\mathit{x}}}\right)}^2-{\left(\sqrt{1+\sqrt{1-\mathit{x}}}\right)}^2}\times \frac{\sqrt{1-\sqrt{1-\mathit{x}}}-\sqrt{1+\sqrt{1-\mathit{x}}}}{1}\mathit{d}\mathit{x}$$$$\mathit{k}={\int }_0^1\frac{\sqrt{1-\mathit{x}}(\sqrt{1-\sqrt{1-\mathit{x}}}-\sqrt{1+\sqrt{1-x}})}{-2\sqrt{1-\sqrt{\mathit{x}}}}\mathit{d}\mathit{x}$$$$\mathit{k}=-\frac{1}{2}{\int }_0^1\left(\sqrt{1-\sqrt{1-\mathit{x}}}\right)\mathit{d}\mathit{x}+\frac{1}{2}{\int }_0^1\left(\sqrt{1+\sqrt{1-\mathit{x}}}\right)\mathit{d}\mathit{x}$$$$\mathit{p}\mathit{o}\mathit{s}\mathit{o}\mathit{n}\mathit{s}\{\begin{array}{l}1-\sqrt{1-\mathit{x}}={\mathit{u}}^2\\ 1+\sqrt{1-\mathit{x}}={\mathit{v}}^2\end{array}=>\{\begin{array}{l}\mathit{x}=1-{(1-{\mathit{u}}^2)}^2\\ \mathit{x}=1-{({\mathit{v}}^2-1)}^2\end{array}=>\{\begin{array}{l}\mathit{d}\mathit{x}=2\mathit{u}(1-{\mathit{u}}^2)\mathit{d}\mathit{u}\\ \mathit{d}\mathit{x}=-2\mathit{v}({\mathit{v}}^2-1)\mathit{d}\mathit{v}\end{array}$$$$\mathit{q}\mathit{d}:\{\begin{array}{l}\{\begin{array}{l}\mathit{x}\to 0\\ \mathit{x}\to 1\end{array}\\ \{\begin{array}{l}\mathit{x}\to 1\\ \mathit{x}\to 0\end{array}\end{array}=>\{\begin{array}{l}\{\begin{array}{l}\mathit{u}\to 0\\ \mathit{u}\to 1\end{array}\\ \{\begin{array}{l}\mathit{v}\to 1\\ \mathit{v}\to 2\end{array}\end{array}$$$$\mathit{k}=-{\int }_0^1{\mathit{u}}^2(1-{\mathit{u}}^2)\mathit{d}\mathit{u}-{\int }_1^2{\mathit{v}}^2({\mathit{v}}^2-1)\mathit{d}\mathit{v}$$$$={[\frac{1}{3}{\mathit{u}}^3-\frac{1}{5}{\mathit{u}}^5]}_0^1-{[\frac{1}{5}{\mathit{v}}^5-\frac{1}{3}{\mathit{v}}^3]}_1^2$$$$=[\frac{1}{3}-\frac{1}{5}-\frac{32}{5}+\frac{8}{3}+\frac{1}{5}-\frac{1}{3}]$$$$=\frac{8}{3}-\frac{32}{5}=\frac{40-96}{15}=-\frac{56}{15}$$

Commented by Frix last updated on 02/May/24

$$\mathrm{Error}:$$$$\mathit{q}\mathit{d}:\{\begin{array}{l}\{\begin{array}{l}\mathit{x}\to 0\\ \mathit{x}\to 1\end{array}\\ \{\begin{array}{l}\mathit{x}\to 1\\ \mathit{x}\to 0\end{array}\end{array}=>\{\begin{array}{l}\{\begin{array}{l}\mathit{u}\to 0\\ \mathit{u}\to 1\end{array}\\ \{\begin{array}{l}\mathit{v}\to 1\\ \mathit{v}\to \sqrt{2}\end{array}\end{array}$$

Answered by Frix last updated on 02/May/24

$$\underset{0}{\stackrel{1}{\int }}\frac{\sqrt{1-x}}{\sqrt{1-\sqrt{1-x}}+\sqrt{1+\sqrt{1-x}}}\times \frac{\sqrt{1+\sqrt{1-x}}-\sqrt{1-\sqrt{1-x}}}{\sqrt{1+\sqrt{1-x}}-\sqrt{1-\sqrt{1-x}}}dx=$$$$=\underset{0}{\stackrel{1}{\int }}\frac{\sqrt{1+\sqrt{1-x}}-\sqrt{1-\sqrt{1-x}}}{2}dx=$$$$=\frac{1}{2}\underset{0}{\stackrel{1}{\int }}\sqrt{1+\sqrt{1-x}}dx-\frac{1}{2}\underset{0}{\stackrel{1}{\int }}\sqrt{1-\sqrt{1-x}}dx$$$$$$$$\frac{1}{2}\underset{0}{\stackrel{1}{\int }}\sqrt{1+\sqrt{1-x}}dx=$$$$t=\sqrt{1+\sqrt{1-x}}\iff x=2t^2-t^4\wedge t⩾1$$$$\Rightarrow dx=4(t-t^3)dt$$$$=2\underset{\sqrt{2}}{\stackrel{1}{\int }}{t}^2-t^{4}dt=2{[\frac{t^3}{3}-\frac{t^5}{5}]}_{\sqrt{2}}^1=\frac{4(1+\sqrt{2})}{15}$$$$$$$$\frac{1}{2}\underset{0}{\stackrel{1}{\int }}\sqrt{1-\sqrt{1-x}}dx=$$$$t=\sqrt{1-\sqrt{1-x}}\iff x=2t^2-t^4\wedge 0⩽t⩽1$$$$\Rightarrow dx=4(t-t^3)dt$$$$=2\underset{0}{\stackrel{1}{\int }}{t}^2-t^{4}dt=2{[\frac{t^3}{3}-\frac{t^5}{5}]}_0^1=\frac{4}{15}$$$$$$$$\Rightarrow$$$$\underset{0}{\stackrel{1}{\int }}\frac{\sqrt{1-x}}{\sqrt{1-\sqrt{1-x}}+\sqrt{1+\sqrt{1-x}}}dx=\frac{4\sqrt{2}}{15}$$

Commented by Ghisom last updated on 02/May/24

$$\mathrm{thank}\mathrm{you}$$

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