Question Number 57653 by MJS last updated on 09/Apr/19
$$\mathrm{is}\:\mathrm{it}\:\mathrm{possible}\:\mathrm{to}\:\mathrm{find}\:\mathrm{the}\:\mathrm{exact}\:\mathrm{value}\:\mathrm{of}\:{I}? \\ $$$${I}=\underset{\mathrm{0}} {\overset{\pi} {\int}}\mathrm{sin}\:\left(\mathrm{sin}\:{x}\right)\:{dx} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 09/Apr/19
Commented by tanmay.chaudhury50@gmail.com last updated on 09/Apr/19
Commented by maxmathsup by imad last updated on 10/Apr/19
![its only a try I =∫_0 ^(π/2) sin(sinx)dx +∫_(π/2) ^π sin(sinx)dx chang.sinx = t give ∫_0 ^(π/2) sin(sinx)dx =∫_0 ^1 sin(t)(dt/( (√(1−t^2 )))) =∫_0 ^1 ((sint)/( (√(1−t^2 )))) dt let ϕ(t) =sint we have ϕ(t) =ϕ(0)+tϕ^′ (0) +(t^2 /2)ϕ^((2)) (0)+(t^3 /(3!))ϕ^((3)) (0) +... ϕ^′ (t) =cost ⇒ϕ^′ (0) =1 ϕ^((2)) (t) =−sint ⇒ϕ^((2)) (0) =0 ϕ^((3)) (t) =−cost ⇒ϕ^((3)) (0) =−1⇒ϕ(t) =t−(t^3 /6) +0(t^3 ) ⇒ t−(t^3 /6) ≤sint ≤t for t ∈[0,(π/2)] ⇒ ((t−(t^3 /6))/( (√(1−t^2 )))) ≤((sint)/( (√(1−t^2 )))) ≤(t/( (√(1−t^2 )))) ⇒ ∫_0 ^1 (t/( (√(1−t^2 )))) −(1/6) ∫_0 ^1 (t^3 /( (√(1−t^2 ))))dt ≤ ∫_0 ^1 ((sint)/( (√(1−t^2 )))) dt ≤ ∫_0 ^1 ((tdt)/( (√(1−t^2 )))) ∫_0 ^1 (t/( (√(1−t^2 ))))dt =[−(√(1−t^2 ))]_0 ^1 =1 ∫_0 ^1 (t^3 /( (√(1−t^2 )))) dt =_(t =sinθ) ∫_0 ^(π/2) ((sin^3 θ)/(cosθ)) cosθ dθ =∫_0 ^(π/2) sin^3 θ dθ =∫_0 ^(π/2) sinθ(1−cos^2 θ)dθ =∫_0 ^(π/(2 )) sinθ dθ −∫_0 ^(π/2) sinθ cos^2 θ dθ =[−cosθ]_0 ^(π/2) +[(1/3)cos^3 θ]_0 ^(π/2) =1−(1/3) =(2/3) ⇒1−(1/6) (2/3) ≤ ∫_0 ^1 ((sint)/( (√(1−t^2 )))) dt ≤1 ⇒(8/9) ≤ ∫_0 ^1 ((sint)/( (√(1−t^2 )))) ≤1 also ∫_(π/2) ^π sin(sinxdx =_(x =(π/2)+t) ∫_0 ^(π/2) sin(cost)dt =_(t =(π/2)−u) −∫_0 ^(π/2) sin(sinu)(−du) =∫_0 ^(π/2) sin(sinu)du ⇒ I =2 ∫_0 ^(π/2) sin(sinu)du =2 ∫_0 ^1 ((sinx)/( (√(1−x^2 )))) dx ⇒ ((16)/9) ≤ ∫_0 ^π sin(sinx)dx ≤ 2 we can take v_0 =((2+((16)/9))/2) =1+((16)/(18)) =1+(8/9) =((17)/9) ∼ 1,8 as approximate value . desmos give ∫ ∼ 1,78 and 1,78 ∼1,8 so this answer is a better value....](https://www.tinkutara.com/question/Q57679.png)
$${its}\:{only}\:{a}\:{try}\:\:\:{I}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{sin}\left({sinx}\right){dx}\:+\int_{\frac{\pi}{\mathrm{2}}} ^{\pi} \:{sin}\left({sinx}\right){dx}\:\: \\ $$$${chang}.{sinx}\:=\:{t}\:{give}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{sin}\left({sinx}\right){dx}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:{sin}\left({t}\right)\frac{{dt}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{sint}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}\:{dt}\:\:\:{let}\:\:\:\varphi\left({t}\right)\:={sint}\:\:{we}\:{have}\:\varphi\left({t}\right)\:=\varphi\left(\mathrm{0}\right)+{t}\varphi^{'} \left(\mathrm{0}\right)\:+\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\varphi^{\left(\mathrm{2}\right)} \left(\mathrm{0}\right)+\frac{{t}^{\mathrm{3}} }{\mathrm{3}!}\varphi^{\left(\mathrm{3}\right)} \left(\mathrm{0}\right)\:+… \\ $$$$\varphi^{'} \left({t}\right)\:={cost}\:\Rightarrow\varphi^{'} \left(\mathrm{0}\right)\:=\mathrm{1} \\ $$$$\varphi^{\left(\mathrm{2}\right)} \left({t}\right)\:=−{sint}\:\Rightarrow\varphi^{\left(\mathrm{2}\right)} \left(\mathrm{0}\right)\:=\mathrm{0} \\ $$$$\varphi^{\left(\mathrm{3}\right)} \left({t}\right)\:=−{cost}\:\Rightarrow\varphi^{\left(\mathrm{3}\right)} \left(\mathrm{0}\right)\:=−\mathrm{1}\Rightarrow\varphi\left({t}\right)\:={t}−\frac{{t}^{\mathrm{3}} }{\mathrm{6}}\:+\mathrm{0}\left({t}^{\mathrm{3}} \right)\:\Rightarrow \\ $$$${t}−\frac{{t}^{\mathrm{3}} }{\mathrm{6}}\:\leqslant{sint}\:\leqslant{t}\:\:\:\:\:\:{for}\:{t}\:\in\left[\mathrm{0},\frac{\pi}{\mathrm{2}}\right]\:\Rightarrow\:\frac{{t}−\frac{{t}^{\mathrm{3}} }{\mathrm{6}}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}\:\leqslant\frac{{sint}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}\:\leqslant\frac{{t}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{t}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}\:−\frac{\mathrm{1}}{\mathrm{6}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{t}^{\mathrm{3}} }{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{dt}\:\leqslant\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{sint}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}\:{dt}\:\leqslant\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{tdt}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{t}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{dt}\:=\left[−\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\right]_{\mathrm{0}} ^{\mathrm{1}} \:=\mathrm{1} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{t}^{\mathrm{3}} }{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}\:{dt}\:=_{{t}\:={sin}\theta} \:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{sin}^{\mathrm{3}} \theta}{{cos}\theta}\:{cos}\theta\:{d}\theta\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{sin}^{\mathrm{3}} \theta\:{d}\theta \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{sin}\theta\left(\mathrm{1}−{cos}^{\mathrm{2}} \theta\right){d}\theta\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}\:}} {sin}\theta\:{d}\theta\:−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{sin}\theta\:{cos}^{\mathrm{2}} \theta\:{d}\theta\:=\left[−{cos}\theta\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} +\left[\frac{\mathrm{1}}{\mathrm{3}}{cos}^{\mathrm{3}} \theta\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}\:=\frac{\mathrm{2}}{\mathrm{3}}\:\Rightarrow\mathrm{1}−\frac{\mathrm{1}}{\mathrm{6}}\:\frac{\mathrm{2}}{\mathrm{3}}\:\leqslant\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{sint}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}\:{dt}\:\leqslant\mathrm{1}\:\Rightarrow\frac{\mathrm{8}}{\mathrm{9}}\:\leqslant\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{sint}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}\:\leqslant\mathrm{1}\:\:{also} \\ $$$$\int_{\frac{\pi}{\mathrm{2}}} ^{\pi} \:\:{sin}\left({sinxdx}\:=_{{x}\:=\frac{\pi}{\mathrm{2}}+{t}} \:\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{sin}\left({cost}\right){dt}\:=_{{t}\:=\frac{\pi}{\mathrm{2}}−{u}} \:\:−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{sin}\left({sinu}\right)\left(−{du}\right)\right. \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:{sin}\left({sinu}\right){du}\:\:\Rightarrow\:{I}\:=\mathrm{2}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{sin}\left({sinu}\right){du}\:=\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{sinx}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:{dx}\:\Rightarrow \\ $$$$\frac{\mathrm{16}}{\mathrm{9}}\:\leqslant\:\int_{\mathrm{0}} ^{\pi} \:\:{sin}\left({sinx}\right){dx}\:\leqslant\:\mathrm{2}\:\:\:\:\:{we}\:{can}\:{take}\:{v}_{\mathrm{0}} =\frac{\mathrm{2}+\frac{\mathrm{16}}{\mathrm{9}}}{\mathrm{2}}\:=\mathrm{1}+\frac{\mathrm{16}}{\mathrm{18}}\:=\mathrm{1}+\frac{\mathrm{8}}{\mathrm{9}}\:=\frac{\mathrm{17}}{\mathrm{9}}\:\sim\:\mathrm{1},\mathrm{8} \\ $$$${as}\:{approximate}\:{value}\:. \\ $$$${desmos}\:{give}\:\:\int\:\sim\:\:\mathrm{1},\mathrm{78}\:\:{and}\:\:\mathrm{1},\mathrm{78}\:\sim\mathrm{1},\mathrm{8}\:\:{so}\:{this}\:{answer}\:{is}\:{a}\:{better}\:{value}…. \\ $$
Commented by MJS last updated on 10/Apr/19
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{for}\:\mathrm{trying} \\ $$
Commented by Abdo msup. last updated on 10/Apr/19
$${you}\:{are}\:{always}\:{welcome}\:{sir}. \\ $$