Question Number 92134 by mathmax by abdo last updated on 05/May/20
$${find}\:\int_{−\mathrm{1}} ^{\mathrm{1}} \:\frac{{e}^{{x}} }{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx}\: \\ $$
Commented by mathmax by abdo last updated on 05/May/20
$${we}\:{use}\:{the}\:{approximation}\:\:\int_{−\mathrm{1}} ^{\mathrm{1}} \:\frac{{f}\left({x}\right)}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx}\:\sim\frac{\pi}{{n}}\sum_{{k}=\mathrm{1}} ^{{n}} \:{f}\left({cos}\left(\frac{\left(\mathrm{2}{k}−\mathrm{1}\right)\pi}{\mathrm{2}{n}}\right)\right) \\ $$$$\Rightarrow\int_{−\mathrm{1}} ^{\mathrm{1}} \:\frac{{e}^{{x}} }{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx}\:\sim\frac{\pi}{{n}}\sum_{{k}=\mathrm{1}} ^{{n}} \:{e}^{{cos}\left(\frac{\left(\mathrm{2}{k}−\mathrm{1}\right)\pi}{\mathrm{2}{n}}\right)} \\ $$$${let}\:{take}\:{n}=\mathrm{3}\:\Rightarrow\int_{−\mathrm{1}} ^{\mathrm{1}} \:\frac{{e}^{{x}} }{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx}\:\sim\frac{\pi}{\mathrm{3}}\left\{\:{e}^{{cos}\left(\frac{\pi}{\mathrm{6}}\right)} \:+{e}^{{cos}\left(\frac{\mathrm{3}\pi}{\mathrm{6}}\right)} \:+{e}^{{cos}\left(\frac{\mathrm{5}\pi}{\mathrm{6}}\right)} \right\} \\ $$$$=\frac{\pi}{\mathrm{3}}\left\{\:{e}^{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} \:+\:\mathrm{1}\:+\:{e}^{−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} \right\}\:=\frac{\pi}{\mathrm{3}}\left\{\mathrm{2}{ch}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)+\mathrm{1}\right\}\:\Rightarrow \\ $$$$\int_{−\mathrm{1}} ^{\mathrm{1}} \:\frac{{e}^{{x}} }{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx}\:\sim\:\frac{\pi}{\mathrm{3}}\:+\frac{\mathrm{2}\pi}{\mathrm{3}}{ch}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right) \\ $$