Question Number 49956 by maxmathsup by imad last updated on 12/Dec/18
$${find}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{cos}\left({n}\:{arcosx}\right){dx}\:\:{with}\:{n}\:{integr}\:{natural}. \\ $$
Commented by Abdo msup. last updated on 13/Dec/18
![we have cos(n arcosx) +i sin(narcosx) =(cos(arcosx )+isin(arcosx))^n =(x +i (√(1−x^2 )))^n and ∫_0 ^1 cos(narcosx)dx =Re( ∫_0 ^1 (x+i(√(1−x^2 )))^n dx) but (x+i(√(1−x^2 )))^n =Σ_(k=0) ^n C_n ^k (i(√(1−x^2 )))^k x^(n−k) =Σ_(k=0) ^n i^k C_n ^k (1−x^2 )^(k/2) x^(n−k) =Σ_(p=0) ^([(n/2)]) (−1)^p C_n ^(2p) (1−x^2 )^p x^(n−2p) +Σ_(p=0) ^([((n−1)/2)]) i(−1)^p C_n ^(2p+1) (1−x^2 )^((2p+1)/2) x^(n−2p−1) ⇒ ∫_0 ^1 cos(narcosx)dx =(∫_0 ^1 Σ_(p=0) ^([(n/2)]) (−1)^p C_n ^(2p) (1−x^2 )^p x^(n−2p) dx) =Σ_(p=0) ^([(n/2)]) (−1)^p C_n ^(2p) ∫_0 ^1 (1−x^2 )^p x^(n−2p) dx changement x =sin t give ∫_0 ^1 (1−x^2 )^p x^(n−2p) dx =∫_0 ^(π/2) cos^(2p) t sin^(n−2p) t cost dt =∫_0 ^(π/2) cos^(2p+1) t sin^(n−2p) t dt ...be continued...](https://www.tinkutara.com/question/Q50098.png)
$${we}\:{have}\:{cos}\left({n}\:{arcosx}\right)\:+{i}\:{sin}\left({narcosx}\right) \\ $$$$=\left({cos}\left({arcosx}\:\right)+{isin}\left({arcosx}\right)\right)^{{n}} \\ $$$$=\left({x}\:+{i}\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)^{{n}} \:{and}\:\int_{\mathrm{0}} ^{\mathrm{1}} {cos}\left({narcosx}\right){dx} \\ $$$$={Re}\left(\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\left({x}+{i}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)^{{n}} \:{dx}\right)\:{but} \\ $$$$\left({x}+{i}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)^{{n}} \:=\sum_{{k}=\mathrm{0}} ^{{n}} \:\:{C}_{{n}} ^{{k}} \:\left({i}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)^{{k}} \:{x}^{{n}−{k}} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}} \:\:{i}^{{k}} \:{C}_{{n}} ^{{k}} \:\:\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\frac{{k}}{\mathrm{2}}} \:{x}^{{n}−{k}} \\ $$$$=\sum_{{p}=\mathrm{0}} ^{\left[\frac{{n}}{\mathrm{2}}\right]} \:\:\left(−\mathrm{1}\right)^{{p}} \:{C}_{{n}} ^{\mathrm{2}{p}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{{p}} \:\:{x}^{{n}−\mathrm{2}{p}} \\ $$$$+\sum_{{p}=\mathrm{0}} ^{\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]} \:\:{i}\left(−\mathrm{1}\right)^{{p}} \:{C}_{{n}} ^{\mathrm{2}{p}+\mathrm{1}} \:\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\frac{\mathrm{2}{p}+\mathrm{1}}{\mathrm{2}}} \:\:{x}^{{n}−\mathrm{2}{p}−\mathrm{1}} \:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:{cos}\left({narcosx}\right){dx}\:=\left(\int_{\mathrm{0}} ^{\mathrm{1}} \:\sum_{{p}=\mathrm{0}} ^{\left[\frac{{n}}{\mathrm{2}}\right]} \left(−\mathrm{1}\right)^{{p}} \:{C}_{{n}} ^{\mathrm{2}{p}} \:\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{{p}} \:{x}^{{n}−\mathrm{2}{p}} {dx}\right) \\ $$$$=\sum_{{p}=\mathrm{0}} ^{\left[\frac{{n}}{\mathrm{2}}\right]} \:\:\left(−\mathrm{1}\right)^{{p}} \:{C}_{{n}} ^{\mathrm{2}{p}} \:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{{p}} \:{x}^{{n}−\mathrm{2}{p}} \:{dx}\:\:{changement} \\ $$$${x}\:={sin}\:{t}\:{give}\: \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{{p}} \:{x}^{{n}−\mathrm{2}{p}} \:{dx}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{cos}^{\mathrm{2}{p}} {t}\:{sin}^{{n}−\mathrm{2}{p}} {t}\:\:{cost}\:{dt}\: \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:{cos}^{\mathrm{2}{p}+\mathrm{1}} {t}\:{sin}^{{n}−\mathrm{2}{p}} {t}\:{dt}\:\:…{be}\:{continued}… \\ $$
Answered by Smail last updated on 13/Dec/18
![t=arcos(x)⇒dt=((−dx)/( (√(1−x^2 ))))=((−dx)/( (√(1−cos^2 t)))) dx=−sin(t)dt A=−∫_(π/2) ^0 sin(t)cos(nt)dt =∫_0 ^(π/2) sin(t)cos(nt)dt by parts u=cos(nt)⇒u′=−nsin(nt) v′=sin(t)⇒v=−cos(t) A=−[cos(t)cos(nt)]_0 ^(π/2) −n∫_0 ^(π/2) sin(nt)cos(t)dt by parts u=sin(nt)⇒u′=ncos(nt) v′=cos(t)⇒v=sin(t) A=1−n[sin(t)sin(nt)]_0 ^(π/2) +n^2 ∫_0 ^(π/2) sin(t)cos(nt)dt A=1−nsin(((nπ)/2))+n^2 A A(1−n^2 )=1−nsin(((nπ)/2)) A=((nsin(((nπ)/2))−1)/(n^2 −1))](https://www.tinkutara.com/question/Q50069.png)
$${t}={arcos}\left({x}\right)\Rightarrow{dt}=\frac{−{dx}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}=\frac{−{dx}}{\:\sqrt{\mathrm{1}−{cos}^{\mathrm{2}} {t}}} \\ $$$${dx}=−{sin}\left({t}\right){dt} \\ $$$${A}=−\int_{\pi/\mathrm{2}} ^{\mathrm{0}} {sin}\left({t}\right){cos}\left({nt}\right){dt} \\ $$$$=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} {sin}\left({t}\right){cos}\left({nt}\right){dt} \\ $$$${by}\:{parts}\: \\ $$$${u}={cos}\left({nt}\right)\Rightarrow{u}'=−{nsin}\left({nt}\right) \\ $$$${v}'={sin}\left({t}\right)\Rightarrow{v}=−{cos}\left({t}\right) \\ $$$${A}=−\left[{cos}\left({t}\right){cos}\left({nt}\right)\right]_{\mathrm{0}} ^{\pi/\mathrm{2}} −{n}\int_{\mathrm{0}} ^{\pi/\mathrm{2}} {sin}\left({nt}\right){cos}\left({t}\right){dt} \\ $$$${by}\:{parts} \\ $$$${u}={sin}\left({nt}\right)\Rightarrow{u}'={ncos}\left({nt}\right) \\ $$$${v}'={cos}\left({t}\right)\Rightarrow{v}={sin}\left({t}\right) \\ $$$${A}=\mathrm{1}−{n}\left[{sin}\left({t}\right){sin}\left({nt}\right)\right]_{\mathrm{0}} ^{\pi/\mathrm{2}} +{n}^{\mathrm{2}} \int_{\mathrm{0}} ^{\pi/\mathrm{2}} {sin}\left({t}\right){cos}\left({nt}\right){dt} \\ $$$${A}=\mathrm{1}−{nsin}\left(\frac{{n}\pi}{\mathrm{2}}\right)+{n}^{\mathrm{2}} {A} \\ $$$${A}\left(\mathrm{1}−{n}^{\mathrm{2}} \right)=\mathrm{1}−{nsin}\left(\frac{{n}\pi}{\mathrm{2}}\right) \\ $$$${A}=\frac{{nsin}\left(\frac{{n}\pi}{\mathrm{2}}\right)−\mathrm{1}}{{n}^{\mathrm{2}} −\mathrm{1}} \\ $$
Commented by Abdo msup. last updated on 13/Dec/18
$${thanks}\:{sir}\:{Smail}.. \\ $$
Commented by Smail last updated on 14/Dec/18
$${you}\:{are}\:{quite}\:{welcome} \\ $$