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Question Number 159480 by ZiYangLee last updated on 17/Nov/21
Commented by bobhans last updated on 18/Nov/21
{ ((z=cos θ +i sin θ)),(((1/z)=cos θ−i sin θ)) :} ; { ((z^6 =cos 6θ+i sin 6θ)),(((1/z^6 )=cos 6θ−i sin 6θ)) :} z^6 +(1/z^6 ) = 2cos 6θ ⇒(z^3 )^2 +((1/z^3 ))^2 =2cos 6θ ⇒(z^3 +(1/z^3 ))^2 −2=2cos 6θ ⇒(2cos 3θ)^2 −2=2cos 6θ ⇒4(4cos^3 θ−3cos θ)^2 −2=2cos 6θ ⇒4(16cos^6 θ−24cos^4 θ+9cos^2 θ)=2+2cos 6θ ⇒16cos^6 θ−24(((1+cos 2θ)/2))^2 +9(((1+cos 2θ)/2))=(1/2)(1+cos 6θ) ⇒16cos^6 θ=6(1+2cos 2θ+((1+cos 4θ)/2))−9(((1+cos 2θ)/2))+(1/2)(1+cos 6θ) ⇒16cos^6 θ=9+12cos 2θ+3cos 4θ−(9/2)−(9/2)cos 2θ+(1/2)+(1/2)cos 6θ cos^6 θ=(1/(16))(5+((15)/2)cos 2θ+3cos 4θ+(1/2)cos 6θ) cos^6 θ=(1/(32))(10+15cos 2θ+6cos 4θ+cos 6θ)
$$\:\begin{cases}{\mathrm{z}=\mathrm{cos}\:\theta\:+\mathrm{i}\:\mathrm{sin}\:\theta}\\{\frac{\mathrm{1}}{\mathrm{z}}=\mathrm{cos}\:\theta−\mathrm{i}\:\mathrm{sin}\:\theta}\end{cases}\:;\:\begin{cases}{\mathrm{z}^{\mathrm{6}} =\mathrm{cos}\:\mathrm{6}\theta+\mathrm{i}\:\mathrm{sin}\:\mathrm{6}\theta}\\{\frac{\mathrm{1}}{\mathrm{z}^{\mathrm{6}} }=\mathrm{cos}\:\mathrm{6}\theta−\mathrm{i}\:\mathrm{sin}\:\mathrm{6}\theta}\end{cases} \\ $$$$\:\mathrm{z}^{\mathrm{6}} +\frac{\mathrm{1}}{\mathrm{z}^{\mathrm{6}} }\:=\:\mathrm{2cos}\:\mathrm{6}\theta \\ $$$$\Rightarrow\left(\mathrm{z}^{\mathrm{3}} \right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{z}^{\mathrm{3}} }\right)^{\mathrm{2}} =\mathrm{2cos}\:\mathrm{6}\theta \\ $$$$\Rightarrow\left(\mathrm{z}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{z}^{\mathrm{3}} }\right)^{\mathrm{2}} −\mathrm{2}=\mathrm{2cos}\:\mathrm{6}\theta \\ $$$$\Rightarrow\left(\mathrm{2cos}\:\mathrm{3}\theta\right)^{\mathrm{2}} −\mathrm{2}=\mathrm{2cos}\:\mathrm{6}\theta \\ $$$$\Rightarrow\mathrm{4}\left(\mathrm{4cos}\:^{\mathrm{3}} \theta−\mathrm{3cos}\:\theta\right)^{\mathrm{2}} −\mathrm{2}=\mathrm{2cos}\:\mathrm{6}\theta \\ $$$$\Rightarrow\mathrm{4}\left(\mathrm{16cos}\:^{\mathrm{6}} \theta−\mathrm{24cos}\:^{\mathrm{4}} \theta+\mathrm{9cos}\:^{\mathrm{2}} \theta\right)=\mathrm{2}+\mathrm{2cos}\:\mathrm{6}\theta \\ $$$$\:\Rightarrow\mathrm{16cos}\:^{\mathrm{6}} \theta−\mathrm{24}\left(\frac{\mathrm{1}+\mathrm{cos}\:\mathrm{2}\theta}{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{9}\left(\frac{\mathrm{1}+\mathrm{cos}\:\mathrm{2}\theta}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\mathrm{cos}\:\mathrm{6}\theta\right) \\ $$$$\Rightarrow\mathrm{16cos}\:^{\mathrm{6}} \theta=\mathrm{6}\left(\mathrm{1}+\mathrm{2cos}\:\mathrm{2}\theta+\frac{\mathrm{1}+\mathrm{cos}\:\mathrm{4}\theta}{\mathrm{2}}\right)−\mathrm{9}\left(\frac{\mathrm{1}+\mathrm{cos}\:\mathrm{2}\theta}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\mathrm{cos}\:\mathrm{6}\theta\right) \\ $$$$\Rightarrow\mathrm{16cos}\:^{\mathrm{6}} \theta=\mathrm{9}+\mathrm{12cos}\:\mathrm{2}\theta+\mathrm{3cos}\:\mathrm{4}\theta−\frac{\mathrm{9}}{\mathrm{2}}−\frac{\mathrm{9}}{\mathrm{2}}\mathrm{cos}\:\mathrm{2}\theta+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{6}\theta \\ $$$$\mathrm{cos}\:^{\mathrm{6}} \theta=\frac{\mathrm{1}}{\mathrm{16}}\left(\mathrm{5}+\frac{\mathrm{15}}{\mathrm{2}}\mathrm{cos}\:\mathrm{2}\theta+\mathrm{3cos}\:\mathrm{4}\theta+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{6}\theta\right) \\ $$$$\mathrm{cos}\:^{\mathrm{6}} \theta=\frac{\mathrm{1}}{\mathrm{32}}\left(\mathrm{10}+\mathrm{15cos}\:\mathrm{2}\theta+\mathrm{6cos}\:\mathrm{4}\theta+\mathrm{cos}\:\mathrm{6}\theta\right) \\ $$
Answered by mathmax by abdo last updated on 18/Nov/21
f(a)=∫_0 ^a (a^2 −x^2 )^(5/2) dx changement x=asint give f(a)=∫_0 ^(π/2) (a^2 −a^2 sin^2 t)^(5/2) acost dt =a∫_0 ^(π/2) a^5 (cos^2 t)^(5/2) cost dt =a^6 ∫_0 ^(π/2) cos^6 t dt we have the formulae 2∫_0 ^(π/2) cos^(2p−1) x sin^(2q−1) x dx=B(p,q)=((Γ(p).Γ(q))/(Γ(p+q))) 2p−1=6 and 2q−1=0 ⇒p=(7/2) and q=(1/2) ⇒ f(a)=(a^6 /2)×2∫_0 ^(π/2) cos^(2((7/2))−1) t sin^(2((1/2))−1) tdt =(a^6 /2)B((7/2),(1/2))=(a^6 /2)×((Γ((7/2)).Γ((1/2)))/(Γ((7/2)+(1/2))))=((√π)/2)a^6 .((Γ((7/2)))/(3!)) =((√π)/(12))a^6 Γ((7/2)) we know Γ(x+1)=xΓ(x)⇒ Γ((7/2))=Γ((5/2)+1)=(5/2)Γ((5/2))=(5/2)Γ((3/2)+1)=(5/2).(3/2)Γ((3/2)) =((15)/4)Γ((1/2)+1)=((15)/4).(1/2)Γ((1/2))=((15)/8)(√π)⇒ f(a)=((√π)/(12)) a^6 .((15(√π))/8) =((15πa^6 )/(12.8)) =((3.5πa^6 )/(3.4.8)) =((5πa^6 )/(32))
$$\mathrm{f}\left(\mathrm{a}\right)=\int_{\mathrm{0}} ^{\mathrm{a}} \left(\mathrm{a}^{\mathrm{2}} −\mathrm{x}^{\mathrm{2}} \right)^{\frac{\mathrm{5}}{\mathrm{2}}} \:\mathrm{dx}\:\:\mathrm{changement}\:\mathrm{x}=\mathrm{asint}\:\mathrm{give}\: \\ $$$$\mathrm{f}\left(\mathrm{a}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{a}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \mathrm{t}\right)^{\frac{\mathrm{5}}{\mathrm{2}}} \mathrm{acost}\:\mathrm{dt}\:=\mathrm{a}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{a}^{\mathrm{5}} \left(\mathrm{cos}^{\mathrm{2}} \mathrm{t}\right)^{\frac{\mathrm{5}}{\mathrm{2}}} \:\mathrm{cost}\:\mathrm{dt} \\ $$$$=\mathrm{a}^{\mathrm{6}} \:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cos}^{\mathrm{6}} \:\mathrm{t}\:\mathrm{dt}\:\:\mathrm{we}\:\mathrm{have}\:\mathrm{the}\:\mathrm{formulae} \\ $$$$\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{cos}^{\mathrm{2p}−\mathrm{1}} \mathrm{x}\:\mathrm{sin}^{\mathrm{2q}−\mathrm{1}} \mathrm{x}\:\mathrm{dx}=\mathrm{B}\left(\mathrm{p},\mathrm{q}\right)=\frac{\Gamma\left(\mathrm{p}\right).\Gamma\left(\mathrm{q}\right)}{\Gamma\left(\mathrm{p}+\mathrm{q}\right)} \\ $$$$\mathrm{2p}−\mathrm{1}=\mathrm{6}\:\mathrm{and}\:\mathrm{2q}−\mathrm{1}=\mathrm{0}\:\Rightarrow\mathrm{p}=\frac{\mathrm{7}}{\mathrm{2}}\:\mathrm{and}\:\mathrm{q}=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{a}\right)=\frac{\mathrm{a}^{\mathrm{6}} }{\mathrm{2}}×\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{cos}^{\mathrm{2}\left(\frac{\mathrm{7}}{\mathrm{2}}\right)−\mathrm{1}} \mathrm{t}\:\mathrm{sin}^{\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\mathrm{1}} \mathrm{tdt} \\ $$$$=\frac{\mathrm{a}^{\mathrm{6}} }{\mathrm{2}}\mathrm{B}\left(\frac{\mathrm{7}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\mathrm{a}^{\mathrm{6}} }{\mathrm{2}}×\frac{\Gamma\left(\frac{\mathrm{7}}{\mathrm{2}}\right).\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left(\frac{\mathrm{7}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)}=\frac{\sqrt{\pi}}{\mathrm{2}}\mathrm{a}^{\mathrm{6}} \:.\frac{\Gamma\left(\frac{\mathrm{7}}{\mathrm{2}}\right)}{\mathrm{3}!} \\ $$$$=\frac{\sqrt{\pi}}{\mathrm{12}}\mathrm{a}^{\mathrm{6}} \:\Gamma\left(\frac{\mathrm{7}}{\mathrm{2}}\right)\:\:\mathrm{we}\:\mathrm{know}\:\:\Gamma\left(\mathrm{x}+\mathrm{1}\right)=\mathrm{x}\Gamma\left(\mathrm{x}\right)\Rightarrow \\ $$$$\Gamma\left(\frac{\mathrm{7}}{\mathrm{2}}\right)=\Gamma\left(\frac{\mathrm{5}}{\mathrm{2}}+\mathrm{1}\right)=\frac{\mathrm{5}}{\mathrm{2}}\Gamma\left(\frac{\mathrm{5}}{\mathrm{2}}\right)=\frac{\mathrm{5}}{\mathrm{2}}\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{1}\right)=\frac{\mathrm{5}}{\mathrm{2}}.\frac{\mathrm{3}}{\mathrm{2}}\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right) \\ $$$$=\frac{\mathrm{15}}{\mathrm{4}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{1}\right)=\frac{\mathrm{15}}{\mathrm{4}}.\frac{\mathrm{1}}{\mathrm{2}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\mathrm{15}}{\mathrm{8}}\sqrt{\pi}\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{a}\right)=\frac{\sqrt{\pi}}{\mathrm{12}}\:\mathrm{a}^{\mathrm{6}} .\frac{\mathrm{15}\sqrt{\pi}}{\mathrm{8}}\:=\frac{\mathrm{15}\pi\mathrm{a}^{\mathrm{6}} }{\mathrm{12}.\mathrm{8}}\:=\frac{\mathrm{3}.\mathrm{5}\pi\mathrm{a}^{\mathrm{6}} }{\mathrm{3}.\mathrm{4}.\mathrm{8}}\:=\frac{\mathrm{5}\pi\mathrm{a}^{\mathrm{6}} }{\mathrm{32}} \\ $$

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