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Question Number 187095 by normans last updated on 13/Feb/23

$$$$$${\int }_0^1\frac{1}{\sqrt{1-{\mathit{x}}^2}}+\frac{1}{\sqrt{1-{\mathit{x}}^2}}\mathit{d}\mathit{x}$$$$$$

Commented by Frix last updated on 13/Feb/23

$$\mathrm{This}\mathrm{is}\mathrm{too}\mathrm{hard}\mathrm{since}\mathrm{we}\mathrm{have}\mathrm{not}\mathrm{yet}$$$$\mathrm{learned}\mathrm{how}\mathrm{to}\mathrm{add}\frac{1}{x}+\frac{1}{x}.\mathrm{Is}\mathrm{it}\frac{1+1}{x+x}?\mathrm{Or}$$$$\mathrm{maybe}\frac{1}{x}+\frac{1}{x}=\frac{1}{x}-\frac{x}{1}?\mathrm{We}\mathrm{need}\mathrm{a}\mathrm{genius}$$$$\mathrm{to}\mathrm{help}\mathrm{us}\mathrm{out}...$$

Answered by MJS_new last updated on 13/Feb/23

$$\mathrm{the}\mathrm{answer}\mathrm{is}$$$$-\mathrm{i}\ln (-1)$$

Answered by CElcedricjunior last updated on 14/Feb/23

$${\int }_0^1(\frac{1}{\sqrt{1-{\mathit{x}}^2}}+\frac{1}{\sqrt{1-{\mathit{x}}^2}})\mathit{d}\mathit{x}=2{\int }_0^1\frac{\mathit{d}\mathit{x}}{\sqrt{1-{\mathit{x}}^2}}★Moivre$$$$=2{\left[\mathit{a}\mathit{r}\mathit{c}\mathit{s}\mathit{i}\mathit{n}\left(\mathit{x}\right)\right]}_0^1◼\mathit{c}\mathit{e}\mathit{l}\mathit{c}\mathit{e}\mathit{d}\mathit{r}\mathit{i}\mathit{c}\mathit{j}\mathit{u}\mathit{n}\mathit{i}\mathit{o}\mathit{r}$$$$=2\left[\frac{\mathit{\pi }}{2}\right]$$$$=>{\int }_0^1(\frac{1}{\sqrt{1-{\mathit{x}}^2}}+\frac{1}{\sqrt{1-{\mathit{x}}^2}})\mathit{d}\mathit{x}=\mathit{\pi }$$

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