Question Number 59279 by Mr X pcx last updated on 07/May/19
$${find}\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{x}}{{sinx}}{dx} \\ $$
Commented by tanmay last updated on 07/May/19
![f(x)=(x/(sinx)) f(0) undefined but when lim_(x→0) (x/(sinx))=1 f((π/2))=(π/2) ∫_0 ^(π/2) (x/(sinx))dx≈(1/2)[lim_(x→0) (x/(sinx))+f((π/2))]×(π/2) ≈(1/2)[1+(π/2)]×(π/2) ≈(1/2)[(π/2)+(π^2 /4)] pls check](https://www.tinkutara.com/question/Q59294.png)
$${f}\left({x}\right)=\frac{{x}}{{sinx}}\: \\ $$$${f}\left(\mathrm{0}\right)\:{undefined} \\ $$$${but}\:{when}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}}{{sinx}}=\mathrm{1} \\ $$$${f}\left(\frac{\pi}{\mathrm{2}}\right)=\frac{\pi}{\mathrm{2}} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{x}}{{sinx}}{dx}\approx\frac{\mathrm{1}}{\mathrm{2}}\left[\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}}{{sinx}}+{f}\left(\frac{\pi}{\mathrm{2}}\right)\right]×\frac{\pi}{\mathrm{2}} \\ $$$$\approx\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{1}+\frac{\pi}{\mathrm{2}}\right]×\frac{\pi}{\mathrm{2}} \\ $$$$\approx\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\pi}{\mathrm{2}}+\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\right] \\ $$$${pls}\:{check} \\ $$
Commented by maxmathsup by imad last updated on 10/May/19
![we have proved that x−(x^3 /6) ≤sinx ≤x for x ∈]0,(π/2)] ⇒ (1/x) ≤ (1/(sinx)) ≤ (1/(x−(x^3 /6))) ⇒1 ≤ (x/(sinx)) ≤ (1/(1−(x^2 /6))) ⇒(π/2) ≤ ∫_0 ^(π/2) (x/(sinx))dx ≤ ∫_0 ^(π/2) (dx/(1−(x^2 /6))) ∫_0 ^(π/2) (dx/(1−(x^2 /6))) =_(x=(√6)t) ∫_0 ^(π/(2(√6))) (((√6)dt)/(1−t^2 )) =((√6)/2) ∫_0 ^(π/(2(√6))) {(1/(1−t)) +(1/(1+t))}dt =((√6)/2)[ln∣((1+t)/(1−t))∣]_0 ^(π/(2(√6))) =((√6)/2){ln(((1+(π/(2(√6))))/(1−(π/(2(√6))))))}=((√6)/2) ln(((2(√6)+π)/(2(√6)−π))) ⇒ (π/2) ≤ ∫_0 ^(π/2) (x/(sinx)) dx ≤ ((√6)/2)ln(((2(√6) +π)/(2(√6)−π))) .if we put α_0 =((√6)/2)ln(((2(√6)+π)/(2(√6)−π))) we can take (((π/2)+α_0 )/2) =(π/4) +(α_0 /2) as approximate value for this integral .](https://www.tinkutara.com/question/Q59462.png)
$$\left.\:\left.{we}\:{have}\:{proved}\:{that}\:\:{x}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}}\:\leqslant{sinx}\:\leqslant{x}\:\:{for}\:{x}\:\in\right]\mathrm{0},\frac{\pi}{\mathrm{2}}\right]\:\Rightarrow \\ $$$$\frac{\mathrm{1}}{{x}}\:\leqslant\:\frac{\mathrm{1}}{{sinx}}\:\leqslant\:\frac{\mathrm{1}}{{x}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}}}\:\Rightarrow\mathrm{1}\:\leqslant\:\frac{{x}}{{sinx}}\:\leqslant\:\frac{\mathrm{1}}{\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{6}}}\:\Rightarrow\frac{\pi}{\mathrm{2}}\:\leqslant\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{x}}{{sinx}}{dx}\:\leqslant\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{dx}}{\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{6}}} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{dx}}{\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{6}}}\:=_{{x}=\sqrt{\mathrm{6}}{t}} \:\:\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}\sqrt{\mathrm{6}}}} \:\:\:\:\frac{\sqrt{\mathrm{6}}{dt}}{\mathrm{1}−{t}^{\mathrm{2}} }\:=\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}\sqrt{\mathrm{6}}}} \:\:\left\{\frac{\mathrm{1}}{\mathrm{1}−{t}}\:+\frac{\mathrm{1}}{\mathrm{1}+{t}}\right\}{dt} \\ $$$$=\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}\left[{ln}\mid\frac{\mathrm{1}+{t}}{\mathrm{1}−{t}}\mid\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}\sqrt{\mathrm{6}}}} \:\:=\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}\left\{{ln}\left(\frac{\mathrm{1}+\frac{\pi}{\mathrm{2}\sqrt{\mathrm{6}}}}{\mathrm{1}−\frac{\pi}{\mathrm{2}\sqrt{\mathrm{6}}}}\right)\right\}=\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}\:{ln}\left(\frac{\mathrm{2}\sqrt{\mathrm{6}}+\pi}{\mathrm{2}\sqrt{\mathrm{6}}−\pi}\right)\:\Rightarrow \\ $$$$\frac{\pi}{\mathrm{2}}\:\leqslant\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{x}}{{sinx}}\:{dx}\:\leqslant\:\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}{ln}\left(\frac{\mathrm{2}\sqrt{\mathrm{6}}\:+\pi}{\mathrm{2}\sqrt{\mathrm{6}}−\pi}\right)\:.{if}\:{we}\:{put}\:\alpha_{\mathrm{0}} =\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}{ln}\left(\frac{\mathrm{2}\sqrt{\mathrm{6}}+\pi}{\mathrm{2}\sqrt{\mathrm{6}}−\pi}\right)\:{we}\:{can} \\ $$$${take}\:\frac{\frac{\pi}{\mathrm{2}}+\alpha_{\mathrm{0}} }{\mathrm{2}}\:=\frac{\pi}{\mathrm{4}}\:+\frac{\alpha_{\mathrm{0}} }{\mathrm{2}}\:{as}\:{approximate}\:{value}\:{for}\:{this}\:{integral}\:. \\ $$
Commented by maxmathsup by imad last updated on 10/May/19
$${sir}\:{Tanmay}\:{what}\:{s}\:{formulae}\:{you}\:{have}\:{used}\:{in}\:{your}\:{answer}. \\ $$
Commented by tanmay last updated on 10/May/19
![i have calculated the area ... ∫_a ^b f(x)dx≈(1/2)[f(a)+f(b)]×(b−a)](https://www.tinkutara.com/question/Q59465.png)
$${i}\:{have}\:{calculated}\:{the}\:{area}\:… \\ $$$$\int_{{a}} ^{{b}} {f}\left({x}\right){dx}\approx\frac{\mathrm{1}}{\mathrm{2}}\left[{f}\left({a}\right)+{f}\left({b}\right)\right]×\left({b}−{a}\right) \\ $$
Commented by maxmathsup by imad last updated on 10/May/19
$${thanks}\:{sir}. \\ $$