let u_n = ∫_(π/(n+1)) ^(π/n) (√(tan(x)))dx with n≥3 1) calculate U_n interms of n and calculate lim_(n→+∞ ) U_n 2) find nature of the serie Σ_(n≥3) U


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Question Number 55571 by maxmathsup by imad last updated on 26/Feb/19

$l e t u_{n} = \int_{\frac{π}{n + 1}}^{\frac{π}{n}} \sqrt{t a n (x)} d x w i t h n ⩾ 3$ $1) c a l c u l a t e U_{n} i n t e r m s o f n a n d c a l c u l a t e {l i m}_{n \to + \infty} U_{n}$ $2) f i n d n a t u r e o f t h e s e r i e \sum_{n ⩾ 3} U_{n}$
Commented by maxmathsup by imad last updated on 03/Mar/19
$1) we have proved that ∫(√(tanx))dx =(1/(√2)){ln((√((tanx−(√(2tanx))+1)/(tanx +(√(2tanx))+1))) +arctan((√(2tanx))+1)+arctan((√(2tanx))−1)} +c ⇒u_n =(1/(√2))[ln(√((tannx−(√(2tanx))+1)/(tanx +(√(2tanx))+1)))) +arctan((√(2tanx))+1)+arctan((√(2tanx))−1)]_(π/(n+1)) ^(π/n) =(1/(√2)) A_n −(1/(√2)) B_n with A_n =ln((√((tan((π/n))−(√(2tan((π/n))))+1)/(tan((π/n))+(√(2tan((π/n))))+1))))+arctan((√(2tan((π/n))))+1) +arctan((√(2tan((π/n))))−1)} and B_n =ln((√((tan((π/(n+1)))−(√(2tan((π/(n+1)))))+1)/(tan((π/(n+1)))+(√(2tan((π/(n+1)))))+1))))+arctan((√(2tan((π/(n+1)))))+1) +arctan((√(2tan((π/(n+1)))))−1} we have lim_(n→+∞) A_n =ln((1/1))+arctan(1)+arctan(−1) =0 lim_(n→+∞) B_n =ln((1/1))+arctan(1)+arctan(−1)=0 ⇒lim_(n→+∞) u_n =0$
$1) w e h a v e p r o v e d t h a t$ $\int \sqrt{t a n x} d x = \frac{1}{\sqrt{2}} {l n (\sqrt{\frac{t a n x - \sqrt{2 t a n x} + 1}{t a n x + \sqrt{2 t a n x} + 1}} + a r c t a n (\sqrt{2 t a n x} + 1) + a r c t a n (\sqrt{2 t a n x} - 1)} + c$ ${\Rightarrow u_{n} = \frac{1}{\sqrt{2}} [l n \sqrt{\frac{t a n n x - \sqrt{2 t a n x} + 1}{t a n x + \sqrt{2 t a n x} + 1}}) + a r c t a n (\sqrt{2 t a n x} + 1) + a r c t a n (\sqrt{2 t a n x} - 1)]}_{\frac{π}{n + 1}}^{\frac{π}{n}}$ $= \frac{1}{\sqrt{2}} A_{n} - \frac{1}{\sqrt{2}} B_{n} w i t h$ $A_{n} = l n (\sqrt{\frac{t a n (\frac{π}{n}) - \sqrt{2 t a n (\frac{π}{n})} + 1}{t a n (\frac{π}{n}) + \sqrt{2 t a n (\frac{π}{n})} + 1}}) + a r c t a n (\sqrt{2 t a n (\frac{π}{n})} + 1)$ $+ a r c t a n (\sqrt{2 t a n (\frac{π}{n})} - 1)} a n d$ $B_{n} = l n (\sqrt{\frac{t a n (\frac{π}{n + 1}) - \sqrt{2 t a n (\frac{π}{n + 1})} + 1}{t a n (\frac{π}{n + 1}) + \sqrt{2 t a n (\frac{π}{n + 1})} + 1}}) + a r c t a n (\sqrt{2 t a n (\frac{π}{n + 1})} + 1)$ $+ a r c t a n (\sqrt{2 t a n (\frac{π}{n + 1})} - 1}$ $w e h a v e {l i m}_{n \to + \infty} A_{n} = l n (\frac{1}{1}) + a r c t a n (1) + a r c t a n (- 1) = 0$ ${l i m}_{n \to + \infty} B_{n} = l n (\frac{1}{1}) + a r c t a n (1) + a r c t a n (- 1) = 0 \Rightarrow {l i m}_{n \to + \infty} u_{n} = 0$
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