Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 115943 by A8;15: last updated on 29/Sep/20

Commented by A8;15: last updated on 29/Sep/20

thanks sir

Commented by mathdave last updated on 29/Sep/20

Commented by mathdave last updated on 29/Sep/20

Commented by Tawa11 last updated on 06/Sep/21

$$\mathrm{great}\mathrm{sir}$$

Answered by TANMAY PANACEA last updated on 29/Sep/20

$$f\left(x\right)=\frac{1}{{(\sqrt{5}-cosx)}^3}$$$$f\left(0\right)=\frac{1}{{(\sqrt{5}-1)}^3}$$$$f\left(\pi \right)=\frac{1}{{(\sqrt{5}+1)}^3}$$$$\frac{1}{{(\sqrt{5}-1)}^3}>f\left(x\right)>\frac{1}{{(\sqrt{5}+1)}^3}$$$${\int }_0^{\pi }\frac{dx}{{(\sqrt{5}-1)}^3}>{\int }_0^{\pi }\frac{dx}{{(\sqrt{5}-cosx)}^3}>{\int }_0^{\pi }\frac{dx}{{(\sqrt{5}+1)}^3}$$$$\frac{\pi }{{(\sqrt{5}-1)}^3}>I>\frac{\pi }{{(\sqrt{5}+1)}^3}$$

Commented by A8;15: last updated on 29/Sep/20

thanks sir

Answered by maths mind last updated on 29/Sep/20

$${\int }_0^{\pi }\frac{dx}{{(\sqrt{5}-cos\left(x\right))}^3}={\int }_0^{\pi }\frac{dx}{{(\sqrt{5}+cos\left(x\right))}^3}$$$$\Rightarrow {\int }_0^{\pi }\frac{dx}{{(\sqrt{5}-cos\left(x\right))}^3}=\frac{1}{2}{\int }_0^{2\pi }\frac{dx}{{(\sqrt{5}+cos\left(x\right))}^3}$$$$cos\left(x\right)=\frac{e^{ix}+e^{-ix}}{2}$$$$=\frac{1}{2}{\int }_0^{2\pi }\frac{8e^{3ix}}{{(\sqrt{5}{e}^{ix}+e^{2ix}+1)}^3}dx$$$$z=e^{ix}\Rightarrow dz={ie}^{ix}dx$$$$=\frac{4}{i}{\int }_C\frac{z^2}{{(z^2+\sqrt{5}z+1)}^3},Cunitecircl$$$$\text{Missing left or extra right}$$$$=only\frac{-\sqrt{5}+1}{2}\in C$$$$=8\pi Res\left(\frac{z^2}{{(z-z_1)}^3{(z-z_2)}^3}{\mid }_{z_2}\right)$$$$$$

Commented by A8;15: last updated on 29/Sep/20

thanks

Answered by mathmax by abdo last updated on 29/Sep/20

$$\mathrm{let}\mathrm{f}\left(\mathrm{a}\right)={\int }_0^{\pi }\frac{\mathrm{dx}}{\mathrm{a}-\mathrm{cosx}}\mathrm{with}\mathrm{a}>1\mathrm{we}{\mathrm{havef}}^{{}^{\prime }}\left(\mathrm{a}\right)=-{\int }_0^{\pi }\frac{\mathrm{dx}}{{(\mathrm{a}-\mathrm{cosx})}^2}$$$$\mathrm{and}{\mathrm{f}}^{\left(2\right)}\left(\mathrm{a}\right)={\int }_0^{\pi }\frac{2(\mathrm{a}-\mathrm{cosx})}{{(\mathrm{a}-\mathrm{cosx})}^4}\mathrm{dx}=2{\int }_0^{\pi }\frac{\mathrm{dx}}{{(\mathrm{a}-\mathrm{cosx})}^3}$$$$\Rightarrow {\int }_0^{\pi }\frac{\mathrm{dx}}{{(\sqrt{5}-\mathrm{cosx})}^3}=\frac{1}{2}{\mathrm{f}}^{\left(2\right)}\left(\sqrt{5}\right)\mathrm{let}\mathrm{explicit}\mathrm{f}\left(\mathrm{a}\right)$$$$\mathrm{changement}\mathrm{t}=\tan \left(\frac{\mathrm{x}}{2}\right)\mathrm{give}\mathrm{f}\left(\mathrm{a}\right)={\int }_0^{\mathrm{\infty }}\frac{2\mathrm{dt}}{(1+{\mathrm{t}}^2)(\mathrm{a}-\frac{1-{\mathrm{t}}^2}{1+{\mathrm{t}}^2})}$$$$=2{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dt}}{\mathrm{a}+{\mathrm{at}}^2-1+{\mathrm{t}}^2}=2{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dt}}{(\mathrm{a}+1){\mathrm{t}}^2+\mathrm{a}-1}$$$$=\frac{2}{(\mathrm{a}+1)}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dt}}{{\mathrm{t}}^2+\frac{\mathrm{a}-1}{\mathrm{a}+1}}{=}_{\mathrm{t}=\sqrt{\frac{\mathrm{a}-1}{\mathrm{a}+1}}\mathrm{u}}\frac{2}{\mathrm{a}+1}.\frac{\mathrm{a}+1}{\mathrm{a}-1}{\int }_0^{\mathrm{\infty }}\frac{1}{1+{\mathrm{u}}^2}\frac{\sqrt{\mathrm{a}-1}}{\sqrt{\mathrm{a}+1}}\mathrm{du}$$$$=\frac{2}{\sqrt{{\mathrm{a}}^2-1}}\times \frac{\pi }{2}=\frac{\pi }{\sqrt{{\mathrm{a}}^2-1}}\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)=\pi {\left\{{({\mathrm{a}}^2-1)}^{-\frac{1}{2}}\right\}}^{\left(1\right)}$$$$=\pi (-\frac{1}{2}\left(2\mathrm{a}\right){({\mathrm{a}}^2-1)}^{-\frac{3}{2}})=-\pi \mathrm{a}{({\mathrm{a}}^2-1)}^{-\frac{3}{2}}$$$$\Rightarrow {\mathrm{f}}^{{}^{″}}\left(\mathrm{a}\right)=-\pi \{{({\mathrm{a}}^2-1)}^{-\frac{3}{2}}+\mathrm{a}(-\frac{3}{2})\left(2\mathrm{a}\right){({\mathrm{a}}^2-1)}^{-\frac{5}{2}}\}$$$$=-\pi \{{({\mathrm{a}}^2-1)}^{-\frac{3}{2}}-{3\mathrm{a}}^2{({\mathrm{a}}^2-1)}^{-\frac{5}{2}}\}\Rightarrow$$$${\mathrm{f}}^{{}^{″}}\left(\sqrt{5}\right)=-\pi \{4^{-\frac{3}{2}}-15{\left(4\right)}^{-\frac{5}{2}}\}$$$$=-\frac{\pi }{4^{\frac{3}{2}}}+\frac{15\pi }{4^{\frac{5}{2}}}=-\frac{\pi }{8}+\frac{15\pi }{32}=\frac{-4\pi +15\pi }{32}=\frac{11\pi }{32}\Rightarrow$$$${\int }_0^{\pi }\frac{\mathrm{dx}}{{(\sqrt{5}-\mathrm{cosx})}^3}=\frac{11\pi }{64}$$

Answered by MJS_new last updated on 30/Sep/20

$$\underset{0}{\stackrel{\pi }{\int }}\frac{dx}{{(a-\cos x)}^3}=$$$$[t=\tan \frac{x}{2}\to dx=\frac{2dt}{t^2+1}]$$$$=\frac{2}{{(a+1)}^3}\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{{(t^2+1)}^2}{{(t^2+\frac{a-1}{a+1})}^3}dt=$$$$[b=\frac{a-1}{a+1}]$$$$=\frac{2}{{(a+1)}^3}\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{{(t^2+1)}^2}{{(t^2+b)}^3}dt=$$$$\left[\mathrm{Ostrogradski}\right]$$$$={[-\frac{(b-1)t((5b+3)t^2+b(3b+5))}{4{(a+1)}^3{b}^2{(t^2+b)}^2}]}_0^{\mathrm{\infty }}+$$$$+\frac{3b^2+2b+3}{4{(a+1)}^3{b}^2}\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{dt}{t^2+b}$$$$\mathrm{obviously}\mathrm{the}\mathrm{first}\mathrm{part}\mathrm{is}\mathrm{equal}\mathrm{to}\mathrm{zero}$$$$\frac{3b^2+2b+3}{4{(a+1)}^3{b}^2}\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{dt}{t^2+b}=$$$$={\left[\frac{3b^2+2b+3}{4{(a+1)}^3{b}^{5/2}}\arctan \frac{t}{\sqrt{b}}\right]}_0^{\mathrm{\infty }}=$$$$=\frac{3b^2+2b+3}{8{(a+1)}^3{b}^{5/2}}\pi =$$$$=\frac{2a^2+1}{2{(a^2-1)}^{5/2}}\pi =$$$$=\frac{11}{64}\pi$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com