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

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