Question Number 55571 by maxmathsup by imad last updated on 26/Feb/19
$${let}\:{u}_{{n}} =\:\int_{\frac{\pi}{{n}+\mathrm{1}}} ^{\frac{\pi}{{n}}} \sqrt{{tan}\left({x}\right)}{dx}\:\:{with}\:{n}\geqslant\mathrm{3} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{U}_{{n}} \:{interms}\:{of}\:{n}\:\:\:{and}\:{calculate}\:{lim}_{{n}\rightarrow+\infty\:\:} \:{U}_{{n}} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{nature}\:{of}\:{the}\:{serie}\:\sum_{{n}\geqslant\mathrm{3}} \:{U}_{{n}} \\ $$
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](https://www.tinkutara.com/question/Q55758.png)
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{proved}\:{that}\: \\ $$$$\int\sqrt{{tanx}}{dx}\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left\{{ln}\left(\sqrt{\frac{{tanx}−\sqrt{\mathrm{2}{tanx}}+\mathrm{1}}{{tanx}\:+\sqrt{\mathrm{2}{tanx}}+\mathrm{1}}}\:+{arctan}\left(\sqrt{\mathrm{2}{tanx}}+\mathrm{1}\right)+{arctan}\left(\sqrt{\mathrm{2}{tanx}}−\mathrm{1}\right)\right\}\:+{c}\right. \\ $$$$\left.\Rightarrow{u}_{{n}} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left[{ln}\sqrt{\frac{{tannx}−\sqrt{\mathrm{2}{tanx}}+\mathrm{1}}{{tanx}\:+\sqrt{\mathrm{2}{tanx}}+\mathrm{1}}}\right)\:+{arctan}\left(\sqrt{\mathrm{2}{tanx}}+\mathrm{1}\right)+{arctan}\left(\sqrt{\mathrm{2}{tanx}}−\mathrm{1}\right)\right]_{\frac{\pi}{{n}+\mathrm{1}}} ^{\frac{\pi}{{n}}} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:{A}_{{n}} −\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:{B}_{{n}} \:\:{with} \\ $$$${A}_{{n}} ={ln}\left(\sqrt{\frac{{tan}\left(\frac{\pi}{{n}}\right)−\sqrt{\mathrm{2}{tan}\left(\frac{\pi}{{n}}\right)}+\mathrm{1}}{{tan}\left(\frac{\pi}{{n}}\right)+\sqrt{\mathrm{2}{tan}\left(\frac{\pi}{{n}}\right)}+\mathrm{1}}}\right)+{arctan}\left(\sqrt{\mathrm{2}{tan}\left(\frac{\pi}{{n}}\right)}+\mathrm{1}\right) \\ $$$$\left.+{arctan}\left(\sqrt{\mathrm{2}{tan}\left(\frac{\pi}{{n}}\right)}−\mathrm{1}\right)\right\}\:{and} \\ $$$${B}_{{n}} ={ln}\left(\sqrt{\frac{{tan}\left(\frac{\pi}{{n}+\mathrm{1}}\right)−\sqrt{\mathrm{2}{tan}\left(\frac{\pi}{{n}+\mathrm{1}}\right)}+\mathrm{1}}{{tan}\left(\frac{\pi}{{n}+\mathrm{1}}\right)+\sqrt{\mathrm{2}{tan}\left(\frac{\pi}{{n}+\mathrm{1}}\right)}+\mathrm{1}}}\right)+{arctan}\left(\sqrt{\mathrm{2}{tan}\left(\frac{\pi}{{n}+\mathrm{1}}\right)}+\mathrm{1}\right) \\ $$$$+{arctan}\left(\sqrt{\mathrm{2}{tan}\left(\frac{\pi}{{n}+\mathrm{1}}\right)}−\mathrm{1}\right\} \\ $$$${we}\:{have}\:{lim}_{{n}\rightarrow+\infty} \:{A}_{{n}} ={ln}\left(\frac{\mathrm{1}}{\mathrm{1}}\right)+{arctan}\left(\mathrm{1}\right)+{arctan}\left(−\mathrm{1}\right)\:=\mathrm{0} \\ $$$${lim}_{{n}\rightarrow+\infty} {B}_{{n}} ={ln}\left(\frac{\mathrm{1}}{\mathrm{1}}\right)+{arctan}\left(\mathrm{1}\right)+{arctan}\left(−\mathrm{1}\right)=\mathrm{0}\:\Rightarrow{lim}_{{n}\rightarrow+\infty} {u}_{{n}} =\mathrm{0} \\ $$