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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) 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

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 .