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Question Number 159480 by ZiYangLee last updated on 17/Nov/21

Commented by bobhans last updated on 18/Nov/21

$$\{\begin{array}{l}\mathrm{z}=\cos \theta +\mathrm{i}\sin \theta \\ \frac{1}{\mathrm{z}}=\cos \theta -\mathrm{i}\sin \theta \end{array};\{\begin{array}{l}{\mathrm{z}}^6=\cos 6\theta +\mathrm{i}\sin 6\theta \\ \frac{1}{{\mathrm{z}}^6}=\cos 6\theta -\mathrm{i}\sin 6\theta \end{array}$$$${\mathrm{z}}^6+\frac{1}{{\mathrm{z}}^6}=2\cos 6\theta$$$$\Rightarrow {\left({\mathrm{z}}^3\right)}^2+{\left(\frac{1}{{\mathrm{z}}^3}\right)}^2=2\cos 6\theta$$$$\Rightarrow {({\mathrm{z}}^3+\frac{1}{{\mathrm{z}}^3})}^2-2=2\cos 6\theta$$$$\Rightarrow {\left(2\cos 3\theta \right)}^2-2=2\cos 6\theta$$$$\Rightarrow 4{(4\cos {}^3\theta -3\cos \theta )}^2-2=2\cos 6\theta$$$$\Rightarrow 4(16\cos {}^6\theta -24\cos {}^4\theta +9\cos {}^2\theta )=2+2\cos 6\theta$$$$\Rightarrow 16\cos {}^6\theta -24{\left(\frac{1+\cos 2\theta }{2}\right)}^2+9\left(\frac{1+\cos 2\theta }{2}\right)=\frac{1}{2}(1+\cos 6\theta )$$$$\Rightarrow 16\cos {}^6\theta =6(1+2\cos 2\theta +\frac{1+\cos 4\theta }{2})-9\left(\frac{1+\cos 2\theta }{2}\right)+\frac{1}{2}(1+\cos 6\theta )$$$$\Rightarrow 16\cos {}^6\theta =9+12\cos 2\theta +3\cos 4\theta -\frac{9}{2}-\frac{9}{2}\cos 2\theta +\frac{1}{2}+\frac{1}{2}\cos 6\theta$$$$\cos {}^6\theta =\frac{1}{16}(5+\frac{15}{2}\cos 2\theta +3\cos 4\theta +\frac{1}{2}\cos 6\theta )$$$$\cos {}^6\theta =\frac{1}{32}(10+15\cos 2\theta +6\cos 4\theta +\cos 6\theta )$$

Answered by mathmax by abdo last updated on 18/Nov/21

$$\mathrm{f}\left(\mathrm{a}\right)={\int }_0^{\mathrm{a}}{({\mathrm{a}}^2-{\mathrm{x}}^2)}^{\frac{5}{2}}\mathrm{dx}\mathrm{changement}\mathrm{x}=\mathrm{asint}\mathrm{give}$$$$\mathrm{f}\left(\mathrm{a}\right)={\int }_0^{\frac{\pi }{2}}{({\mathrm{a}}^2-{\mathrm{a}}^2{\sin }^2\mathrm{t})}^{\frac{5}{2}}\mathrm{acost}\mathrm{dt}=\mathrm{a}{\int }_0^{\frac{\pi }{2}}{\mathrm{a}}^5{\left({\cos }^2\mathrm{t}\right)}^{\frac{5}{2}}\mathrm{cost}\mathrm{dt}$$$$={\mathrm{a}}^6{\int }_0^{\frac{\pi }{2}}{\cos }^6\mathrm{t}\mathrm{dt}\mathrm{we}\mathrm{have}\mathrm{the}\mathrm{formulae}$$$$2{\int }_0^{\frac{\pi }{2}}{\cos }^{2\mathrm{p}-1}\mathrm{x}{\sin }^{2\mathrm{q}-1}\mathrm{x}\mathrm{dx}=\mathrm{B}(\mathrm{p},\mathrm{q})=\frac{\mathrm{\Gamma }\left(\mathrm{p}\right).\mathrm{\Gamma }\left(\mathrm{q}\right)}{\mathrm{\Gamma }(\mathrm{p}+\mathrm{q})}$$$$2\mathrm{p}-1=6\mathrm{and}2\mathrm{q}-1=0\Rightarrow \mathrm{p}=\frac{7}{2}\mathrm{and}\mathrm{q}=\frac{1}{2}\Rightarrow$$$$\mathrm{f}\left(\mathrm{a}\right)=\frac{{\mathrm{a}}^6}{2}\times 2{\int }_0^{\frac{\pi }{2}}{\cos }^{2\left(\frac{7}{2}\right)-1}\mathrm{t}{\sin }^{2\left(\frac{1}{2}\right)-1}\mathrm{tdt}$$$$=\frac{{\mathrm{a}}^6}{2}\mathrm{B}(\frac{7}{2},\frac{1}{2})=\frac{{\mathrm{a}}^6}{2}\times \frac{\mathrm{\Gamma }\left(\frac{7}{2}\right).\mathrm{\Gamma }\left(\frac{1}{2}\right)}{\mathrm{\Gamma }(\frac{7}{2}+\frac{1}{2})}=\frac{\sqrt{\pi }}{2}{\mathrm{a}}^6.\frac{\mathrm{\Gamma }\left(\frac{7}{2}\right)}{3!}$$$$=\frac{\sqrt{\pi }}{12}{\mathrm{a}}^6\mathrm{\Gamma }\left(\frac{7}{2}\right)\mathrm{we}\mathrm{know}\mathrm{\Gamma }(\mathrm{x}+1)=\mathrm{x}\mathrm{\Gamma }\left(\mathrm{x}\right)\Rightarrow$$$$\mathrm{\Gamma }\left(\frac{7}{2}\right)=\mathrm{\Gamma }(\frac{5}{2}+1)=\frac{5}{2}\mathrm{\Gamma }\left(\frac{5}{2}\right)=\frac{5}{2}\mathrm{\Gamma }(\frac{3}{2}+1)=\frac{5}{2}.\frac{3}{2}\mathrm{\Gamma }\left(\frac{3}{2}\right)$$$$=\frac{15}{4}\mathrm{\Gamma }(\frac{1}{2}+1)=\frac{15}{4}.\frac{1}{2}\mathrm{\Gamma }\left(\frac{1}{2}\right)=\frac{15}{8}\sqrt{\pi }\Rightarrow$$$$\mathrm{f}\left(\mathrm{a}\right)=\frac{\sqrt{\pi }}{12}{\mathrm{a}}^6.\frac{15\sqrt{\pi }}{8}=\frac{15\pi {\mathrm{a}}^6}{12.8}=\frac{3.5\pi {\mathrm{a}}^6}{3.4.8}=\frac{5\pi {\mathrm{a}}^6}{32}$$

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