Integration Questions

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

$$\\$$$$\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}\:−\:\boldsymbol{{x}}^{\mathrm{2}} }}\:\:+\:\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}\:−\:\boldsymbol{{x}}^{\mathrm{2}} }}\:\:\boldsymbol{{dx}}\:\:\:\:\:\: \\$$$$\\$$

Commented by Frix last updated on 13/Feb/23

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

Answered by MJS_new last updated on 13/Feb/23

$$\mathrm{the}\:\mathrm{answer}\:\mathrm{is} \\$$$$−\mathrm{i}\:\mathrm{ln}\:\left(−\mathrm{1}\right) \\$$

Answered by CElcedricjunior last updated on 14/Feb/23

$$\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\boldsymbol{{x}}^{\mathrm{2}} }}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\boldsymbol{{x}}^{\mathrm{2}} }}\right)\boldsymbol{{dx}}=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\boldsymbol{{dx}}}{\:\sqrt{\mathrm{1}−\boldsymbol{{x}}^{\mathrm{2}} }}\:\bigstar{Moivre} \\$$$$=\mathrm{2}\left[\boldsymbol{{arcsin}}\left(\boldsymbol{{x}}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \blacksquare\boldsymbol{{cel}}\:\:\boldsymbol{{cedric}}\:\boldsymbol{{junior}} \\$$$$=\mathrm{2}\left[\frac{\boldsymbol{\pi}}{\mathrm{2}}\right] \\$$$$=>\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\boldsymbol{{x}}^{\mathrm{2}} }}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\boldsymbol{{x}}^{\mathrm{2}} }}\right)\boldsymbol{{dx}}=\boldsymbol{\pi} \\$$