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


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Question Number 57667 by maxmathsup by imad last updated on 09/Apr/19

$c a l c u l a t e U_{n} = \int_{\frac{π}{n}}^{\frac{2 π}{n}} \frac{d x}{2 + s i n x}$ $1) c a l c u l a t e U_{n} a n d {l i m}_{n \to + \infty} {n U}_{n}$ $2) f i n d n a t u r e o f Σ U_{n}$
Commented by Abdo msup. last updated on 11/Apr/19
$1) changement tan((x/2)) =t give U_n = ∫_(tan((π/(2n)))) ^(tan((π/n))) ((2dt)/((1+t^2 )(2+((2t)/(1+t^2 ))))) =∫_(tan((π/(2n)))) ^(tan((π/n))) ((2dt)/(2+2t^2 +2t)) =∫_(tan((π/(2n)))) ^(tan((π/n))) (dt/(t^2 +t +1)) =∫_(tan((π/(2n)))) ^(tan((π/n))) (dt/((t +(1/2))^2 +(3/4))) =_(t+(1/2)=((√3)/2)u) (4/3) ∫_((2tan((π/(2n)))+1)/(√3)) ^((2tan((π/n))+1)/(√3)) (1/(1+u^2 )) ((√3)/2) du =(2/(√3)) [arctan(u)]_((2tan((π/(2n)))+1)/(√3)) ^((2tan((π/n)) +1)/(√3)) ⇒ U_n =(2/(√3)){ arctan(((2tan((π/n))+1)/(√3)))−arctan(((2tan((π/(2n)))+1)/(√3)))}$
$1) c h a n g e m e n t t a n (\frac{x}{2}) = t g i v e$ $U_{n} = \int_{t a n (\frac{π}{2 n})}^{t a n (\frac{π}{n})} \frac{2 d t}{(1 + t^{2}) (2 + \frac{2 t}{1 + t^{2}})}$ $= \int_{t a n (\frac{π}{2 n})}^{t a n (\frac{π}{n})} \frac{2 d t}{2 + 2 t^{2} + 2 t} = \int_{t a n (\frac{π}{2 n})}^{t a n (\frac{π}{n})} \frac{d t}{t^{2} + t + 1}$ $= \int_{t a n (\frac{π}{2 n})}^{t a n (\frac{π}{n})} \frac{d t}{{(t + \frac{1}{2})}^{2} + \frac{3}{4}} =_{t + \frac{1}{2} = \frac{\sqrt{3}}{2} u} \frac{4}{3} \int_{\frac{2 t a n (\frac{π}{2 n}) + 1}{\sqrt{3}}}^{\frac{2 t a n (\frac{π}{n}) + 1}{\sqrt{3}}} \frac{1}{1 + u^{2}} \frac{\sqrt{3}}{2} d u$ $= \frac{2}{\sqrt{3}} {[a r c t a n (u)]}_{\frac{2 t a n (\frac{π}{2 n}) + 1}{\sqrt{3}}}^{\frac{2 t a n (\frac{π}{n}) + 1}{\sqrt{3}}} \Rightarrow$ $U_{n} = \frac{2}{\sqrt{3}} {a r c t a n (\frac{2 t a n (\frac{π}{n}) + 1}{\sqrt{3}}) - a r c t a n (\frac{2 t a n (\frac{π}{2 n}) + 1}{\sqrt{3}})}$
Commented by Abdo msup. last updated on 11/Apr/19
$let use arctan((x/y))=(π/2) −arctan((y/x)) for xy>0 ⇒arctan(((2tan((π/n))+1)/(√3)))=(π/2) −arctan(((√3)/(2tan((π/n))+1))) 2tan((π/n))+1 ∼((2π)/n)+1 ⇒((√3)/(2tan((π/n))+1)) ∼ ((√3)/(1+((2π)/n))) ∼(√3){1−((2π)/n)} ⇒ arctan(((2tan((π/n))+1)/(√3))) ∼(π/2) −(√3){1−((2π)/n)} also arctan(((2tan((π/(2n)))+1)/(√3))) ∼(π/2) −(√3){1−(π/n)} ⇒ U_n ∼ (2/(√3)){(π/2) −(√3) +((2π(√3))/n) −(π/2) +(√3)−((π(√3))/n)} U_n ∼(2/(√3)).((π(√3))/n) ⇒ lim_(n→+∞) nU_n = 2π .$
$l e t u s e a r c t a n (\frac{x}{y}) = \frac{π}{2} - a r c t a n (\frac{y}{x}) f o r x y > 0$ $\Rightarrow a r c t a n (\frac{2 t a n (\frac{π}{n}) + 1}{\sqrt{3}}) = \frac{π}{2} - a r c t a n (\frac{\sqrt{3}}{2 t a n (\frac{π}{n}) + 1})$ $2 t a n (\frac{π}{n}) + 1 \sim \frac{2 π}{n} + 1 \Rightarrow \frac{\sqrt{3}}{2 t a n (\frac{π}{n}) + 1} \sim \frac{\sqrt{3}}{1 + \frac{2 π}{n}}$ $\sim \sqrt{3} {1 - \frac{2 π}{n}} \Rightarrow$ $a r c t a n (\frac{2 t a n (\frac{π}{n}) + 1}{\sqrt{3}}) \sim \frac{π}{2} - \sqrt{3} {1 - \frac{2 π}{n}} a l s o$ $a r c t a n (\frac{2 t a n (\frac{π}{2 n}) + 1}{\sqrt{3}}) \sim \frac{π}{2} - \sqrt{3} {1 - \frac{π}{n}} \Rightarrow$ $U_{n} \sim \frac{2}{\sqrt{3}} {\frac{π}{2} - \sqrt{3} + \frac{2 π \sqrt{3}}{n} - \frac{π}{2} + \sqrt{3} - \frac{π \sqrt{3}}{n}}$ $U_{n} \sim \frac{2}{\sqrt{3}} . \frac{π \sqrt{3}}{n} \Rightarrow {l i m}_{n \to + \infty} {n U}_{n} = 2 π .$
Commented by Abdo msup. last updated on 11/Apr/19

$2) w e h a v e U_{n} \sim \frac{2 π}{n} a n d t h e s e r i e Σ \frac{2 π}{n} d i v e r g e s s o$ $Σ U_{n} i s d i v e r g e n t .$
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