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Question Number 98424 by mathmax by abdo last updated on 13/Jun/20

calculste A_n =∫_(−(1/2)) ^(1/2) x^n (√((1−x)/(1+x)))dx find nature of the serie Σ A_n

$$\mathrm{calculste}\:\mathrm{A}_{\mathrm{n}} =\int_{−\frac{\mathrm{1}}{\mathrm{2}}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\mathrm{x}^{\mathrm{n}} \sqrt{\frac{\mathrm{1}−\mathrm{x}}{\mathrm{1}+\mathrm{x}}}\mathrm{dx} \\ $$$$\mathrm{find}\:\mathrm{nature}\:\mathrm{of}\:\mathrm{the}\:\mathrm{serie}\:\Sigma\:\mathrm{A}_{\mathrm{n}} \\ $$

Answered by maths mind last updated on 14/Jun/20

x=cos(2a) ⇒∫_((5π)/6) ^(π/6) cos^n (2a)(√(tg^2 (a))).−2sin(2a)da..A =∫−4cos^n (2a)tg(a)sin(a)cos(a)da =∫−4cos^n (2a)sin^2 (a) =2∫cos^n (2a)(cos(2a)−1)da =2∫_0 ^(π/6) {cos^(n+1) (2a)−cos^n (2a)}da =2∫cos^(n+1) (2a)−2∫cos^n (2a)da =∫cos^(n+1) (s)ds−∫cos^n (s)ds..I B(x;a,b)=∫_0 ^x t^(a−1) (1−t)^(b−1) dt ...incompletd Betta function ∫cos^n (t)dt, u=cos^2 (t) −2∫u^((n/2)+1) (1−u)^(1/2) du=−2β(u;(n/2)+2,(3/2))+c .−2β(cos^2 (t);(n/2)+2,(3/2))...I ∫_((5π)/6) ^(π/2) cos^n (t)dt..withe t=π−c ∫_(π/2) ^(π/6) (−1)^n cos^n (t) A=−2∫_((5π )/6) ^(π/2) cos^(n+1) (t)+2∫_((5π)/6) ^(π/2) cos^n (t)dt+2∫_(π/2) ^(π/6) cos^(n+1) (t)dt−2∫_(π/2) ^(π/6) cos^n (t)dt expresse withe..I ∣An∣≤∫∣x∣^n (√3)=(√3)(2^(n+1) /(n+1)) ⇒ΣA_n Cv absilutly ⇒ΣA_n Cv

$${x}={cos}\left(\mathrm{2}{a}\right) \\ $$$$\Rightarrow\int_{\frac{\mathrm{5}\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{6}}} {cos}^{{n}} \left(\mathrm{2}{a}\right)\sqrt{{tg}^{\mathrm{2}} \left({a}\right)}.−\mathrm{2}{sin}\left(\mathrm{2}{a}\right){da}..{A} \\ $$$$=\int−\mathrm{4}{cos}^{{n}} \left(\mathrm{2}{a}\right){tg}\left({a}\right){sin}\left({a}\right){cos}\left({a}\right){da} \\ $$$$=\int−\mathrm{4}{cos}^{{n}} \left(\mathrm{2}{a}\right){sin}^{\mathrm{2}} \left({a}\right) \\ $$$$=\mathrm{2}\int{cos}^{{n}} \left(\mathrm{2}{a}\right)\left({cos}\left(\mathrm{2}{a}\right)−\mathrm{1}\right){da} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \left\{{cos}^{{n}+\mathrm{1}} \left(\mathrm{2}{a}\right)−{cos}^{{n}} \left(\mathrm{2}{a}\right)\right\}{da} \\ $$$$=\mathrm{2}\int{cos}^{{n}+\mathrm{1}} \left(\mathrm{2}{a}\right)−\mathrm{2}\int{cos}^{{n}} \left(\mathrm{2}{a}\right){da} \\ $$$$=\int{cos}^{{n}+\mathrm{1}} \left({s}\right){ds}−\int{cos}^{{n}} \left({s}\right){ds}..{I} \\ $$$${B}\left({x};{a},{b}\right)=\int_{\mathrm{0}} ^{{x}} {t}^{{a}−\mathrm{1}} \left(\mathrm{1}−{t}\right)^{{b}−\mathrm{1}} {dt}\:...{incompletd}\:{Betta}\:{function} \\ $$$$\int{cos}^{{n}} \left({t}\right){dt},\:\:\:{u}={cos}^{\mathrm{2}} \left({t}\right) \\ $$$$−\mathrm{2}\int{u}^{\frac{{n}}{\mathrm{2}}+\mathrm{1}} \left(\mathrm{1}−{u}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} {du}=−\mathrm{2}\beta\left({u};\frac{{n}}{\mathrm{2}}+\mathrm{2},\frac{\mathrm{3}}{\mathrm{2}}\right)+{c} \\ $$$$.−\mathrm{2}\beta\left({cos}^{\mathrm{2}} \left({t}\right);\frac{{n}}{\mathrm{2}}+\mathrm{2},\frac{\mathrm{3}}{\mathrm{2}}\right)...{I} \\ $$$$\int_{\frac{\mathrm{5}\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{2}}} {cos}^{{n}} \left({t}\right){dt}..{withe} \\ $$$${t}=\pi−{c} \\ $$$$\int_{\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{6}}} \left(−\mathrm{1}\right)^{{n}} {cos}^{{n}} \left({t}\right) \\ $$$${A}=−\mathrm{2}\int_{\frac{\mathrm{5}\pi\:}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{2}}} {cos}^{{n}+\mathrm{1}} \left({t}\right)+\mathrm{2}\int_{\frac{\mathrm{5}\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{2}}} {cos}^{{n}} \left({t}\right){dt}+\mathrm{2}\int_{\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{6}}} {cos}^{{n}+\mathrm{1}} \left({t}\right){dt}−\mathrm{2}\int_{\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{6}}} {cos}^{{n}} \left({t}\right){dt}\:\: \\ $$$${expresse}\:{withe}..{I} \\ $$$$\mid{An}\mid\leqslant\int\mid{x}\mid^{{n}} \sqrt{\mathrm{3}}=\sqrt{\mathrm{3}}\frac{\mathrm{2}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}} \\ $$$$\Rightarrow\Sigma{A}_{{n}} \:\:{Cv}\:{absilutly}\:\Rightarrow\Sigma{A}_{{n}} \:\:{Cv} \\ $$$$ \\ $$

Commented by abdomathmax last updated on 15/Jun/20

thank you sir.

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Commented by maths mind last updated on 15/Jun/20

withe pleasur

$${withe}\:{pleasur} \\ $$

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