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# let-A-n-0-pi-2-cos-n-xdx-1-calculate-A-0-A-2-and-A-3-2-calculate-A-n-interms-of-n-3-find-0-pi-2-cos-8-xdx-

Question Number 65770 by mathmax by abdo last updated on 03/Aug/19
$${let}\:{A}_{{n}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{cos}^{{n}} {xdx} \\$$$$\left.\mathrm{1}\right)\:{calculate}\:{A}_{\mathrm{0}} ,{A}_{\mathrm{2}} \:{and}\:{A}_{\mathrm{3}} \\$$$$\left.\mathrm{2}\right){calculate}\:{A}_{{n}} {interms}\:{of}\:{n} \\$$$$\left.\mathrm{3}\right)\:{find}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{cos}^{\mathrm{8}} {xdx}\: \\$$
Commented by ~ À ® @ 237 ~ last updated on 04/Aug/19
$$\:{A}_{\mathrm{0}} =\left[{x}\right]_{\mathrm{0}_{} } ^{\frac{\pi}{\mathrm{2}}} =\frac{\pi}{\mathrm{2}} \\$$$$\mathrm{2}{A}_{{n}} \:=\:\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left({sinx}\right)^{\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\mathrm{1}} \left({cosx}\right)^{\mathrm{2}\left(\frac{{n}+\mathrm{1}}{\mathrm{2}}\right)−\mathrm{1}} {dx}={B}\left(\frac{\mathrm{1}}{\mathrm{2}}.\:\:\frac{{n}+\mathrm{1}}{\mathrm{2}}\right) \\$$$${Then}\:{A}_{{n}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}.\:\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{{n}+\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left(\frac{{n}+\mathrm{2}}{\mathrm{2}}\right)} \\$$$${when}\:{using}\:{the}\:{Lagrange}\:{formula}\:\underset{{k}=\mathrm{1}} {\overset{{p}−\mathrm{1}} {\prod}}\Gamma\left({z}+\frac{{k}}{{p}}\right)=\left(\mathrm{2}\pi\right)^{\frac{{p}−\mathrm{1}}{\mathrm{2}}} {p}^{\frac{\mathrm{1}}{\mathrm{2}}−{pz}} \Gamma\left({pz}\right) \\$$$$\Gamma\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)=\sqrt{\pi\:}\:\frac{\Gamma\left(\mathrm{2}{n}\right)}{\mathrm{2}^{\mathrm{2}{n}−\mathrm{1}} }\:\:\forall\:{n}>\mathrm{1} \\$$$${so}\:{we}\:{can}\:{find}\: \\$$$${A}_{\mathrm{2}{n}} =\:\frac{\sqrt{\pi}}{\mathrm{2}}.\frac{\frac{\sqrt{\pi}\:\Gamma\left(\mathrm{2}{n}\right)}{\mathrm{2}^{\mathrm{2}{n}−\mathrm{1}} }}{\Gamma\left({n}+\mathrm{1}\right)}=\:\frac{\left(\mathrm{2}{n}−\mathrm{1}\right)!}{\mathrm{2}^{\mathrm{2}{n}} \:{n}!}\:\pi \\$$$${A}_{\mathrm{2}{n}+\mathrm{1}} =\:\frac{\sqrt{\pi}}{\mathrm{2}}.\:\frac{\Gamma\left({n}+\mathrm{1}\right)}{\:\sqrt{\pi}\:.\frac{\Gamma\left(\mathrm{2}{n}+\mathrm{2}\right)}{\mathrm{2}^{\mathrm{2}{n}+\mathrm{1}} }}=\frac{\mathrm{2}^{\mathrm{2}{n}} .{n}!}{\left(\mathrm{2}{n}+\mathrm{1}\right)!} \\$$$${So}\:\:\:{A}_{\mathrm{1}} =\mathrm{1}\:\:\:\:\:\:{A}_{\mathrm{3}} =\:\frac{\mathrm{2}^{\mathrm{2}} .\mathrm{1}!}{\mathrm{3}!}\:\:\:\:\:\:{A}_{\mathrm{8}} =\frac{\mathrm{7}!\pi}{\mathrm{2}^{\mathrm{8}} \mathrm{4}!}\:\: \\$$
Commented by mathmax by abdo last updated on 04/Aug/19
$${thank}\:{you}\:{sir}. \\$$