Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 57653 by MJS last updated on 09/Apr/19

$$\mathrm{is}\mathrm{it}\mathrm{possible}\mathrm{to}\mathrm{find}\mathrm{the}\mathrm{exact}\mathrm{value}\mathrm{of}I?$$$$I=\underset{0}{\stackrel{\pi }{\int }}\sin \left(\sin x\right)dx$$

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

$$itsonlyatryI={\int }_0^{\frac{\pi }{2}}sin\left(sinx\right)dx+{\int }_{\frac{\pi }{2}}^{\pi }sin\left(sinx\right)dx$$$$chang.sinx=tgive{\int }_0^{\frac{\pi }{2}}sin\left(sinx\right)dx={\int }_0^{1}sin\left(t\right)\frac{dt}{\sqrt{1-t^2}}$$$$={\int }_0^1\frac{sint}{\sqrt{1-t^2}}dtlet\phi \left(t\right)=sintwehave\phi \left(t\right)=\phi \left(0\right)+t{\phi }^{{}^{\prime }}\left(0\right)+\frac{t^2}{2}{\phi }^{\left(2\right)}\left(0\right)+\frac{t^3}{3!}{\phi }^{\left(3\right)}\left(0\right)+...$$$${\phi }^{{}^{\prime }}\left(t\right)=cost\Rightarrow {\phi }^{{}^{\prime }}\left(0\right)=1$$$${\phi }^{\left(2\right)}\left(t\right)=-sint\Rightarrow {\phi }^{\left(2\right)}\left(0\right)=0$$$${\phi }^{\left(3\right)}\left(t\right)=-cost\Rightarrow {\phi }^{\left(3\right)}\left(0\right)=-1\Rightarrow \phi \left(t\right)=t-\frac{t^3}{6}+0\left(t^3\right)\Rightarrow$$$$t-\frac{t^3}{6}⩽sint⩽tfort\in [0,\frac{\pi }{2}]\Rightarrow \frac{t-\frac{t^3}{6}}{\sqrt{1-t^2}}⩽\frac{sint}{\sqrt{1-t^2}}⩽\frac{t}{\sqrt{1-t^2}}\Rightarrow$$$${\int }_0^1\frac{t}{\sqrt{1-t^2}}-\frac{1}{6}{\int }_0^1\frac{t^3}{\sqrt{1-t^2}}dt⩽{\int }_0^1\frac{sint}{\sqrt{1-t^2}}dt⩽{\int }_0^1\frac{tdt}{\sqrt{1-t^2}}$$$${\int }_0^1\frac{t}{\sqrt{1-t^2}}dt={[-\sqrt{1-t^2}]}_0^1=1$$$${\int }_0^1\frac{t^3}{\sqrt{1-t^2}}dt{=}_{t=sin\theta }{\int }_0^{\frac{\pi }{2}}\frac{{sin}^3\theta }{cos\theta }cos\theta d\theta ={\int }_0^{\frac{\pi }{2}}{sin}^3\theta d\theta$$$$={\int }_0^{\frac{\pi }{2}}sin\theta (1-{cos}^2\theta )d\theta ={\int }_0^{\frac{\pi }{2}}sin\theta d\theta -{\int }_0^{\frac{\pi }{2}}sin\theta {cos}^2\theta d\theta ={[-cos\theta ]}_0^{\frac{\pi }{2}}+{\left[\frac{1}{3}{cos}^3\theta \right]}_0^{\frac{\pi }{2}}$$$$=1-\frac{1}{3}=\frac{2}{3}\Rightarrow 1-\frac{1}{6}\frac{2}{3}⩽{\int }_0^1\frac{sint}{\sqrt{1-t^2}}dt⩽1\Rightarrow \frac{8}{9}⩽{\int }_0^1\frac{sint}{\sqrt{1-t^2}}⩽1also$$$${\int }_{\frac{\pi }{2}}^{\pi }sin(sinxdx{=}_{x=\frac{\pi }{2}+t}{\int }_0^{\frac{\pi }{2}}sin\left(cost\right)dt{=}_{t=\frac{\pi }{2}-u}-{\int }_0^{\frac{\pi }{2}}sin\left(sinu\right)(-du)$$$$={\int }_0^{\frac{\pi }{2}}sin\left(sinu\right)du\Rightarrow I=2{\int }_0^{\frac{\pi }{2}}sin\left(sinu\right)du=2{\int }_0^1\frac{sinx}{\sqrt{1-x^2}}dx\Rightarrow$$$$\frac{16}{9}⩽{\int }_0^{\pi }sin\left(sinx\right)dx⩽2wecantake{v}_0=\frac{2+\frac{16}{9}}{2}=1+\frac{16}{18}=1+\frac{8}{9}=\frac{17}{9}\sim 1,8$$$$asapproximatevalue.$$$$desmosgive\int \sim 1,78and1,78\sim 1,8sothisanswerisabettervalue....$$

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

$$youarealwayswelcomesir.$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com