Question Number 59185 by maxmathsup by imad last updated on 09/May/19
![calculate ∫_0 ^1 (x/(sinx))dx let f(x) =sinx ⇒f(x) =f(0) +xf^′ (0) +(x^2 /2)f^((2)) (0)+(x^3 /(3!))f^((3)) (0) +o(x^4 ) but f(0) =0 f^′ (x) =cosx ⇒f^′ (0)=1 f^((2)) (x) =−sinx ⇒f^((2)) (0)=0 f^((3)) (0) =−cos(x) ⇒f^((3)) (0) =−1 ⇒sinx =x−(x^3 /6) +o(x^4 ) ⇒ x−(x^3 /3) ≤sinx ≤x for x ∈]0,1] (1/x) ≤(1/(sinx)) ≤(1/(x−(x^3 /6))) ⇒1≤(x/(sinx)) ≤(1/(1−(x^2 /6))) ⇒ 1 ≤ ∫_0 ^1 (x/(sinx)) dx ≤ ∫_0 ^1 (dx/(1−(x^2 /6))) ∫_0 ^1 (dx/(1−(x^2 /6))) =_(x =(√6)t) ∫_0 ^(1/( (√6))) (((√6)dt)/(1−t^2 )) =((√6)/2) ∫_0 ^(1/( (√6))) ((1/(1−t)) +(1/(1+t)))dt =((√6)/2)[ln∣((1+t)/(1−t))∣]_0 ^(1/( (√6))) =((√6)/2) { ln(((1+(1/( (√6))))/(1−(1/( (√6))))))} =((√6)/2) ln((((√6)+1)/( (√6)−1))) ⇒ 1≤ ∫_0 ^1 (x/(sinx)) dx ≤ ((√6)/2)ln((((√6)+1)/( (√6)−1))) .](https://www.tinkutara.com/question/Q59185.png)
$${calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{x}}{{sinx}}{dx} \\ $$$${let}\:{f}\left({x}\right)\:={sinx}\:\Rightarrow{f}\left({x}\right)\:={f}\left(\mathrm{0}\right)\:+{xf}^{'} \left(\mathrm{0}\right)\:+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}{f}^{\left(\mathrm{2}\right)} \left(\mathrm{0}\right)+\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}{f}^{\left(\mathrm{3}\right)} \left(\mathrm{0}\right)\:+{o}\left({x}^{\mathrm{4}} \right) \\ $$$${but}\:{f}\left(\mathrm{0}\right)\:=\mathrm{0}\:\:\:{f}^{'} \left({x}\right)\:={cosx}\:\Rightarrow{f}^{'} \left(\mathrm{0}\right)=\mathrm{1}\:\:\:\:{f}^{\left(\mathrm{2}\right)} \left({x}\right)\:=−{sinx}\:\Rightarrow{f}^{\left(\mathrm{2}\right)} \left(\mathrm{0}\right)=\mathrm{0} \\ $$$${f}^{\left(\mathrm{3}\right)} \left(\mathrm{0}\right)\:=−{cos}\left({x}\right)\:\Rightarrow{f}^{\left(\mathrm{3}\right)} \left(\mathrm{0}\right)\:=−\mathrm{1}\:\Rightarrow{sinx}\:={x}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}}\:+{o}\left({x}^{\mathrm{4}} \right)\:\Rightarrow \\ $$$$\left.{x}\left.−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}\:\leqslant{sinx}\:\leqslant{x}\:\:\:\:{for}\:{x}\:\in\right]\mathrm{0},\mathrm{1}\right]\:\:\:\frac{\mathrm{1}}{{x}}\:\leqslant\frac{\mathrm{1}}{{sinx}}\:\leqslant\frac{\mathrm{1}}{{x}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}}}\:\Rightarrow\mathrm{1}\leqslant\frac{{x}}{{sinx}}\:\leqslant\frac{\mathrm{1}}{\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{6}}}\:\Rightarrow \\ $$$$\mathrm{1}\:\leqslant\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{x}}{{sinx}}\:{dx}\:\leqslant\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dx}}{\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{6}}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dx}}{\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{6}}}\:=_{{x}\:=\sqrt{\mathrm{6}}{t}} \:\:\:\:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\:\sqrt{\mathrm{6}}}} \:\:\:\:\frac{\sqrt{\mathrm{6}}{dt}}{\mathrm{1}−{t}^{\mathrm{2}} }\:=\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}\:\:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\:\sqrt{\mathrm{6}}}} \:\:\left(\frac{\mathrm{1}}{\mathrm{1}−{t}}\:+\frac{\mathrm{1}}{\mathrm{1}+{t}}\right){dt} \\ $$$$=\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}\left[{ln}\mid\frac{\mathrm{1}+{t}}{\mathrm{1}−{t}}\mid\right]_{\mathrm{0}} ^{\frac{\mathrm{1}}{\:\sqrt{\mathrm{6}}}} \:\:=\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}\:\left\{\:{ln}\left(\frac{\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{6}}}}{\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{6}}}}\right)\right\}\:=\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}\:{ln}\left(\frac{\sqrt{\mathrm{6}}+\mathrm{1}}{\:\sqrt{\mathrm{6}}−\mathrm{1}}\right)\:\Rightarrow \\ $$$$\mathrm{1}\leqslant\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{x}}{{sinx}}\:{dx}\:\leqslant\:\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}{ln}\left(\frac{\sqrt{\mathrm{6}}+\mathrm{1}}{\:\sqrt{\mathrm{6}}−\mathrm{1}}\right)\:. \\ $$$$ \\ $$
Commented by maxmathsup by imad last updated on 09/May/19
![its only a try chang.tan((x/2))=t give I =∫_0 ^1 (x/(sinx))dx =∫_0 ^(tan((1/2))) ((2arctan(t))/((2t)/(1+t^2 ))) ((2dt)/(1+t^2 )) = 2∫_0 ^(tan((1/2))) ((arctan(t))/t) dt let introduce the parametric function f(α) =∫_0 ^(tan((1/2))) ((arctan(tα))/t) dt we have f^′ (α) =∫_0 ^(tan((1/2))) (t/(t(1+α^2 t^2 ))) dt =∫_0 ^(tan((1/2))) (dt/(1+α^2 t^2 )) =_(αt =u) ∫_0 ^(αtan((1/2))) (du/(α(1+u^2 ))) =(1/α) [arctan(u)]_0 ^(αtan((1/2))) = (1/α) arctan(αtan((1/2))) ⇒ f(α) =∫ (1/α) arctan(α tan((1/2)))dα +c ...be continued...](https://www.tinkutara.com/question/Q59396.png)
$${its}\:{only}\:{a}\:{try}\:\:\:{chang}.{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}\:\:{give}\:\: \\ $$$${I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{x}}{{sinx}}{dx}\:=\int_{\mathrm{0}} ^{{tan}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} \:\:\:\frac{\mathrm{2}{arctan}\left({t}\right)}{\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }}\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\:=\:\mathrm{2}\int_{\mathrm{0}} ^{{tan}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} \:\:\:\frac{{arctan}\left({t}\right)}{{t}}\:{dt} \\ $$$${let}\:{introduce}\:{the}\:{parametric}\:{function}\:{f}\left(\alpha\right)\:=\int_{\mathrm{0}} ^{{tan}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} \:\:\frac{{arctan}\left({t}\alpha\right)}{{t}}\:{dt} \\ $$$${we}\:{have}\:{f}^{'} \left(\alpha\right)\:=\int_{\mathrm{0}} ^{{tan}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} \:\:\frac{{t}}{{t}\left(\mathrm{1}+\alpha^{\mathrm{2}} {t}^{\mathrm{2}} \right)}\:{dt}\:=\int_{\mathrm{0}} ^{{tan}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} \:\:\frac{{dt}}{\mathrm{1}+\alpha^{\mathrm{2}} {t}^{\mathrm{2}} } \\ $$$$=_{\alpha{t}\:={u}} \:\:\:\:\int_{\mathrm{0}} ^{\alpha{tan}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} \:\:\:\frac{{du}}{\alpha\left(\mathrm{1}+{u}^{\mathrm{2}} \right)}\:=\frac{\mathrm{1}}{\alpha}\:\left[{arctan}\left({u}\right)\right]_{\mathrm{0}} ^{\alpha{tan}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} \:=\:\frac{\mathrm{1}}{\alpha}\:{arctan}\left(\alpha{tan}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\right)\:\Rightarrow \\ $$$${f}\left(\alpha\right)\:=\int\:\:\frac{\mathrm{1}}{\alpha}\:{arctan}\left(\alpha\:{tan}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\right){d}\alpha\:+{c}\:…{be}\:{continued}… \\ $$