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Question Number 29972 by abdo imad last updated on 14/Feb/18

let give ∣x∣<1 find ∫_0 ^(π/2) (dθ/(√(1−x^2 cos^2 θ))) .

$${let}\:{give}\:\mid{x}\mid<\mathrm{1}\:{find}\:\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\:\frac{{d}\theta}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} {cos}^{\mathrm{2}} \theta}}\:. \\ $$

Commented byabdo imad last updated on 18/Feb/18

we have (1+u^ )^α =1+αu +((α(α−1))/(2!))u^2 +... =1+Σ_(n=1) ^∞ ((α(α−1)...(α−n+1))/(n!))u^n with∣u∣<1 we have ∣xcosθ∣<1 ⇒ (1/(√(1−x^2 cos^2 θ)))=(1−(xcosθ)^2 )^(−(1/2)) =1+Σ_(n=1) ^∞ (((−(1/2))(−(1/2)−1)...(−(1/2)−n+1))/(n!)) x^(2n) cos^(2n) θ let put A_n =(−(1/2))(−(1/2)−1)...(−(1/2)−n+1) ⇒ ∫_0 ^(π/2) (dθ/(√(1−x^2 cos^2 θ)))=(π/2) +Σ_(n=1) ^∞ (A_n /(n!))x^(2n) ∫_0 ^(π/2) cos^(2n) θ dθ but w_n =∫_0 ^(π/2) cos^(2n) θdθ is known (wallis integral) so ∫_0 ^(π/2) (dθ/(√(1−x^2 cos^2 θ))) = (π/2) +Σ_(n=1) ^∞ ((A_n B_n )/(n!)) x^(2n) .

$${we}\:{have}\:\left(\mathrm{1}+{u}^{} \right)^{\alpha} \:\:=\mathrm{1}+\alpha{u}\:+\frac{\alpha\left(\alpha−\mathrm{1}\right)}{\mathrm{2}!}{u}^{\mathrm{2}} \:+... \\ $$ $$=\mathrm{1}+\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\alpha\left(\alpha−\mathrm{1}\right)...\left(\alpha−{n}+\mathrm{1}\right)}{{n}!}{u}^{{n}} \:\:\:{with}\mid{u}\mid<\mathrm{1}\:{we}\:{have} \\ $$ $$\mid{xcos}\theta\mid<\mathrm{1}\:\Rightarrow\:\frac{\mathrm{1}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} {cos}^{\mathrm{2}} \theta}}=\left(\mathrm{1}−\left({xcos}\theta\right)^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \\ $$ $$=\mathrm{1}+\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\left(−\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}\right)...\left(−\frac{\mathrm{1}}{\mathrm{2}}−{n}+\mathrm{1}\right)}{{n}!}\:{x}^{\mathrm{2}{n}} \:{cos}^{\mathrm{2}{n}} \theta\:{let} \\ $$ $${put}\:{A}_{{n}} =\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\left(−\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}\right)...\left(−\frac{\mathrm{1}}{\mathrm{2}}−{n}+\mathrm{1}\right)\:\Rightarrow \\ $$ $$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\frac{{d}\theta}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} {cos}^{\mathrm{2}} \theta}}=\frac{\pi}{\mathrm{2}}\:+\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{A}_{{n}} }{{n}!}{x}^{\mathrm{2}{n}} \:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{cos}^{\mathrm{2}{n}} \theta\:{d}\theta\:{but} \\ $$ $${w}_{{n}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{cos}^{\mathrm{2}{n}} \theta{d}\theta\:{is}\:{known}\:\left({wallis}\:{integral}\right)\:{so} \\ $$ $$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{d}\theta}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} \:{cos}^{\mathrm{2}} \theta}}\:=\:\frac{\pi}{\mathrm{2}}\:+\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{A}_{{n}} \:{B}_{{n}} }{{n}!}\:{x}^{\mathrm{2}{n}} \:. \\ $$

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