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Question Number 146772 by ZiYangLee last updated on 15/Jul/21

$$\mathrm{If}z=\cos \theta +i\sin \theta ,\mathrm{prove}\mathrm{that}$$$${\cos }^6\theta =\frac{1}{32}(\cos 6\theta +6\cos 4\theta +15\cos 2\theta +10).$$$$\mathrm{Hence}\mathrm{or}\mathrm{otherwise},\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}$$$${\int }_0^a\sqrt{{(a^2-x^2)}^5}dx.$$

Answered by mathmax by abdo last updated on 15/Jul/21

$$\mathrm{\Psi }={\int }_0^{\mathrm{a}}{({\mathrm{a}}^2-{\mathrm{x}}^2)}^{\frac{5}{2}}\mathrm{dx}\mathrm{changement}\mathrm{x}=\mathrm{asint}\mathrm{give}$$$$\mathrm{\Psi }={\int }_0^{\frac{\pi }{2}}{\mathrm{a}}^5\left({\cos }^5\theta \right)\mathrm{acos}\theta \mathrm{d}\theta ={\mathrm{a}}^6{\int }_0^{\frac{\pi }{2}}{\cos }^6\theta \mathrm{d}\theta$$$$=\frac{{\mathrm{a}}^6}{32}\{{\int }_0^{\frac{\pi }{2}}\cos \left(6\theta \right)\mathrm{d}\theta +6{\int }_0^{\frac{\pi }{2}}\cos \left(4\theta \right)\mathrm{d}\theta +15{\int }_0^{\frac{\pi }{2}}\cos \left(2\theta \right)\mathrm{d}\theta +10{\int }_0^{\frac{\pi }{2}}\left[\mathrm{d}\theta \right\}$$$$=\frac{{\mathrm{a}}^6}{32}\{{\left[\frac{1}{6}\sin \left(6\theta \right)\right]}_0^{\frac{\pi }{2}}+6{\left[\frac{1}{4}\sin \left(4\theta \right)\right]}_0^{\frac{\pi }{2}}+15{\left[\frac{1}{2}\sin \left(2\theta \right)\right]}_0^{\frac{\pi }{2}}+5\pi \}$$$$=\frac{{\mathrm{a}}^6}{32}\{0+0+0+5\pi \}=\frac{5\pi }{32}{\mathrm{a}}^6$$$${\cos }^6\theta ={\left(\frac{{\mathrm{e}}^{\mathrm{i}\theta }+{\mathrm{e}}^{-\mathrm{i}\theta }}{2}\right)}^6=\frac{1}{2^6}\sum _{\mathrm{k}=0}^6{\mathrm{C}}_6^{\mathrm{k}}{\left({\mathrm{e}}^{\mathrm{i}\theta }\right)}^{\mathrm{k}}{\left({\mathrm{e}}^{-\mathrm{i}\theta }\right)}^{6-\mathrm{k}}$$$$=\frac{1}{2^6}\sum _{\mathrm{k}=0}^6{\mathrm{C}}_6^{\mathrm{k}}{\mathrm{e}}^{\mathrm{ik}\theta +\mathrm{ik}\theta -6\mathrm{i}\theta }=\frac{1}{2^6}\sum _{\mathrm{k}=0}^6{\mathrm{C}}_6^{\mathrm{k}}{\mathrm{e}}^{\mathrm{i}(2\mathrm{k}-6)\theta }$$$$=.....$$

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