Question Number 29972 by abdo imad last updated on 14/Feb/18
$${let}\:{give}\:\mid{x}\mid<\mathrm{1}\:{find}\:\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\:\frac{{d}\theta}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} {cos}^{\mathrm{2}} \theta}}\:. \\ $$
Commented by abdo imad last updated on 18/Feb/18
$${we}\:{have}\:\left(\mathrm{1}+{u}^{} \right)^{\alpha} \:\:=\mathrm{1}+\alpha{u}\:+\frac{\alpha\left(\alpha−\mathrm{1}\right)}{\mathrm{2}!}{u}^{\mathrm{2}} \:+… \\ $$$$=\mathrm{1}+\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\alpha\left(\alpha−\mathrm{1}\right)…\left(\alpha−{n}+\mathrm{1}\right)}{{n}!}{u}^{{n}} \:\:\:{with}\mid{u}\mid<\mathrm{1}\:{we}\:{have} \\ $$$$\mid{xcos}\theta\mid<\mathrm{1}\:\Rightarrow\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} {cos}^{\mathrm{2}} \theta}}=\left(\mathrm{1}−\left({xcos}\theta\right)^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$=\mathrm{1}+\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\left(−\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}\right)…\left(−\frac{\mathrm{1}}{\mathrm{2}}−{n}+\mathrm{1}\right)}{{n}!}\:{x}^{\mathrm{2}{n}} \:{cos}^{\mathrm{2}{n}} \theta\:{let} \\ $$$${put}\:{A}_{{n}} =\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\left(−\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}\right)…\left(−\frac{\mathrm{1}}{\mathrm{2}}−{n}+\mathrm{1}\right)\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\frac{{d}\theta}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} {cos}^{\mathrm{2}} \theta}}=\frac{\pi}{\mathrm{2}}\:+\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{A}_{{n}} }{{n}!}{x}^{\mathrm{2}{n}} \:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{cos}^{\mathrm{2}{n}} \theta\:{d}\theta\:{but} \\ $$$${w}_{{n}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{cos}^{\mathrm{2}{n}} \theta{d}\theta\:{is}\:{known}\:\left({wallis}\:{integral}\right)\:{so} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{d}\theta}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} \:{cos}^{\mathrm{2}} \theta}}\:=\:\frac{\pi}{\mathrm{2}}\:+\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{A}_{{n}} \:{B}_{{n}} }{{n}!}\:{x}^{\mathrm{2}{n}} \:. \\ $$