Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

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 .

$$\left.\mathrm{2}\right)\:{we}\:{have}\:{U}_{{n}} \sim\frac{\mathrm{2}\pi}{{n}}\:\:\:{and}\:{the}\:{serie}\:\Sigma\:\frac{\mathrm{2}\pi}{{n}}\:{diverges}\:{so} \\ $$$$\Sigma{U}_{{n}} \:{is}\:{divergent}\:. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com