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Question Number 57667 by maxmathsup by imad last updated on 09/Apr/19
calculate U_n =∫_(π/n) ^((2π)/n) (dx/(2 +sinx)) 1) calculate U_n and lim_(n→+∞) nU_n 2) find nature of Σ U_n
$${calculate}\:{U}_{{n}} =\int_{\frac{\pi}{{n}}} ^{\frac{\mathrm{2}\pi}{{n}}} \:\:\:\:\:\frac{{dx}}{\mathrm{2}\:+{sinx}} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{U}_{{n}} \:\:\:\:\:\:\:{and}\:{lim}_{{n}\rightarrow+\infty} \:\:{nU}_{{n}} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{nature}\:{of}\:\Sigma\:{U}_{{n}} \\ $$
Commented by Abdo msup. last updated on 11/Apr/19
1) changement tan((x/2)) =t give U_n = ∫_(tan((π/(2n)))) ^(tan((π/n))) ((2dt)/((1+t^2 )(2+((2t)/(1+t^2 ))))) =∫_(tan((π/(2n)))) ^(tan((π/n))) ((2dt)/(2+2t^2 +2t)) =∫_(tan((π/(2n)))) ^(tan((π/n))) (dt/(t^2 +t +1)) =∫_(tan((π/(2n)))) ^(tan((π/n))) (dt/((t +(1/2))^2 +(3/4))) =_(t+(1/2)=((√3)/2)u) (4/3) ∫_((2tan((π/(2n)))+1)/( (√3))) ^((2tan((π/n))+1)/( (√3))) (1/(1+u^2 )) ((√3)/2) du =(2/( (√3))) [arctan(u)]_((2tan((π/(2n)))+1)/( (√3))) ^((2tan((π/n)) +1)/( (√3))) ⇒ U_n =(2/( (√3))){ arctan(((2tan((π/n))+1)/( (√3))))−arctan(((2tan((π/(2n)))+1)/( (√3))))}
$$\left.\mathrm{1}\right)\:{changement}\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)\:={t}\:{give}\: \\ $$$${U}_{{n}} =\:\int_{{tan}\left(\frac{\pi}{\mathrm{2}{n}}\right)} ^{{tan}\left(\frac{\pi}{{n}}\right)} \:\:\:\:\:\:\frac{\mathrm{2}{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left(\mathrm{2}+\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\right)} \\ $$$$=\int_{{tan}\left(\frac{\pi}{\mathrm{2}{n}}\right)} ^{{tan}\left(\frac{\pi}{{n}}\right)} \:\:\frac{\mathrm{2}{dt}}{\mathrm{2}+\mathrm{2}{t}^{\mathrm{2}} \:+\mathrm{2}{t}}\:=\int_{{tan}\left(\frac{\pi}{\mathrm{2}{n}}\right)} ^{{tan}\left(\frac{\pi}{{n}}\right)} \:\:\:\frac{{dt}}{{t}^{\mathrm{2}} \:+{t}\:+\mathrm{1}} \\ $$$$=\int_{{tan}\left(\frac{\pi}{\mathrm{2}{n}}\right)} ^{{tan}\left(\frac{\pi}{{n}}\right)} \:\:\:\frac{{dt}}{\left({t}\:+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{3}}{\mathrm{4}}}\:=_{{t}+\frac{\mathrm{1}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{u}} \frac{\mathrm{4}}{\mathrm{3}}\:\:\:\int_{\frac{\mathrm{2}{tan}\left(\frac{\pi}{\mathrm{2}{n}}\right)+\mathrm{1}}{\:\sqrt{\mathrm{3}}}} ^{\frac{\mathrm{2}{tan}\left(\frac{\pi}{{n}}\right)+\mathrm{1}}{\:\sqrt{\mathrm{3}}}} \:\:\frac{\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{2}} }\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{du} \\ $$$$=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:\left[{arctan}\left({u}\right)\right]_{\frac{\mathrm{2}{tan}\left(\frac{\pi}{\mathrm{2}{n}}\right)+\mathrm{1}}{\:\sqrt{\mathrm{3}}}} ^{\frac{\mathrm{2}{tan}\left(\frac{\pi}{{n}}\right)\:+\mathrm{1}}{\:\sqrt{\mathrm{3}}}} \:\:\:\:\:\Rightarrow \\ $$$${U}_{{n}} =\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\left\{\:{arctan}\left(\frac{\mathrm{2}{tan}\left(\frac{\pi}{{n}}\right)+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)−{arctan}\left(\frac{\mathrm{2}{tan}\left(\frac{\pi}{\mathrm{2}{n}}\right)+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\right\} \\ $$$$ \\ $$$$ \\ $$
Commented by Abdo msup. last updated on 11/Apr/19
let use arctan((x/y))=(π/2) −arctan((y/x)) for xy>0 ⇒arctan(((2tan((π/n))+1)/( (√3))))=(π/2) −arctan(((√3)/(2tan((π/n))+1))) 2tan((π/n))+1 ∼((2π)/n)+1 ⇒((√3)/(2tan((π/n))+1)) ∼ ((√3)/(1+((2π)/n))) ∼(√3){1−((2π)/n)} ⇒ arctan(((2tan((π/n))+1)/( (√3)))) ∼(π/2) −(√3){1−((2π)/n)} also arctan(((2tan((π/(2n)))+1)/( (√3)))) ∼(π/2) −(√3){1−(π/n)} ⇒ U_n ∼ (2/( (√3))){(π/2) −(√3) +((2π(√3))/n) −(π/2) +(√3)−((π(√3))/n)} U_n ∼(2/( (√3))).((π(√3))/n) ⇒ lim_(n→+∞) nU_n = 2π .
$${let}\:{use}\:{arctan}\left(\frac{{x}}{{y}}\right)=\frac{\pi}{\mathrm{2}}\:−{arctan}\left(\frac{{y}}{{x}}\right)\:\:{for}\:{xy}>\mathrm{0} \\ $$$$\Rightarrow{arctan}\left(\frac{\mathrm{2}{tan}\left(\frac{\pi}{{n}}\right)+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)=\frac{\pi}{\mathrm{2}}\:−{arctan}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}{tan}\left(\frac{\pi}{{n}}\right)+\mathrm{1}}\right) \\ $$$$\mathrm{2}{tan}\left(\frac{\pi}{{n}}\right)+\mathrm{1}\:\sim\frac{\mathrm{2}\pi}{{n}}+\mathrm{1}\:\Rightarrow\frac{\sqrt{\mathrm{3}}}{\mathrm{2}{tan}\left(\frac{\pi}{{n}}\right)+\mathrm{1}}\:\sim\:\frac{\sqrt{\mathrm{3}}}{\mathrm{1}+\frac{\mathrm{2}\pi}{{n}}} \\ $$$$\sim\sqrt{\mathrm{3}}\left\{\mathrm{1}−\frac{\mathrm{2}\pi}{{n}}\right\}\:\Rightarrow \\ $$$${arctan}\left(\frac{\mathrm{2}{tan}\left(\frac{\pi}{{n}}\right)+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\:\sim\frac{\pi}{\mathrm{2}}\:−\sqrt{\mathrm{3}}\left\{\mathrm{1}−\frac{\mathrm{2}\pi}{{n}}\right\}\:{also} \\ $$$${arctan}\left(\frac{\mathrm{2}{tan}\left(\frac{\pi}{\mathrm{2}{n}}\right)+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\:\sim\frac{\pi}{\mathrm{2}}\:−\sqrt{\mathrm{3}}\left\{\mathrm{1}−\frac{\pi}{{n}}\right\}\:\Rightarrow \\ $$$${U}_{{n}} \:\sim\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\left\{\frac{\pi}{\mathrm{2}}\:−\sqrt{\mathrm{3}}\:+\frac{\mathrm{2}\pi\sqrt{\mathrm{3}}}{{n}}\:−\frac{\pi}{\mathrm{2}}\:+\sqrt{\mathrm{3}}−\frac{\pi\sqrt{\mathrm{3}}}{{n}}\right\} \\ $$$${U}_{{n}} \sim\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}.\frac{\pi\sqrt{\mathrm{3}}}{{n}}\:\Rightarrow\:{lim}_{{n}\rightarrow+\infty} \:{nU}_{{n}} =\:\mathrm{2}\pi\:. \\ $$
Commented by Abdo msup. last updated on 11/Apr/19
2) we have U_n ∼((2π)/n) and the serie Σ ((2π)/n) diverges so ΣU_n is divergent .
$$\left.\mathrm{2}\right)\:{we}\:{have}\:{U}_{{n}} \sim\frac{\mathrm{2}\pi}{{n}}\:\:\:{and}\:{the}\:{serie}\:\Sigma\:\frac{\mathrm{2}\pi}{{n}}\:{diverges}\:{so} \\ $$$$\Sigma{U}_{{n}} \:{is}\:{divergent}\:. \\ $$

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