Menu Close

Question-108118

Tinku Tara 0 Comments
Pin




Question Number 108118 by mathdave last updated on 14/Aug/20
Answered by abdomathmax last updated on 14/Aug/20
I =∫_0 ^1 (1−x^(18) )^(1/(20)) dx−∫_0 ^1 (1−x^(20) )^(1/(18)) dx =H−K J =∫_0 ^1 (1−x^(18) )^(1/(20)) dx =_(x^(18) =sin^2 t →x=sin^(1/9) t) (1/9)∫_0 ^(π/2) cos^(1/(10)) tcostsin^((1/9)−1) t dt =(1/9)∫_0 ^(π/2) cos^((1/(10))+1) t sin^((1/9)−1) t dt we know 2 ∫_0 ^(π/2) cos^(2p−1) t sin^(2q−1) t dt =B(p,q)=((Γ(p).Γ(q))/(Γ(p+q))) 2p−1=(1/(10))+1 ⇒2p=(1/(10)) +2 ⇒p=(1/(20))+1 =((21)/(20)) 2q−1=(1/9)−1 ⇒q=(1/(18)) ⇒ H =(1/(18))B(((21)/(20)),(1/(18))) =(1/(18))((Γ(((21)/(20))).Γ((1/(18))))/(Γ(((21)/(20))+(1/(18))))) K = ∫_0 ^1 (1−x^(20) )^(1/(18)) dx =_(x^(20) =sin^2 t→x=sin^(1/(10)) t) (1/(10))∫_0 ^(π/2) cos^(1/9) t cost sin^((1/(10))−1) t dt =(1/(10)) ∫_0 ^(π/2) cos^((1/9)+1) t sin^((1/(10))−1) t dt 2p−1=(1/9)+1 ⇒2p =(1/9)+2 ⇒p =(1/(18)) +1 =((19)/(18)) 2q−1 =(1/(10))−1 ⇒q=(1/(20)) ⇒ K =(1/(20)) B(((19)/(18)),(1/(20))) =(1/(20))((Γ(((19)/(18))).Γ((1/(20))))/(Γ(((19)/(18))+(1/(20))))) I = H−K
$$\mathrm{I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−\mathrm{x}^{\mathrm{18}} \right)^{\frac{\mathrm{1}}{\mathrm{20}}} \mathrm{dx}−\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−\mathrm{x}^{\mathrm{20}} \right)^{\frac{\mathrm{1}}{\mathrm{18}}} \mathrm{dx}\:=\mathrm{H}−\mathrm{K} \\ $$$$\mathrm{J}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−\mathrm{x}^{\mathrm{18}} \right)^{\frac{\mathrm{1}}{\mathrm{20}}} \mathrm{dx}\:=_{\mathrm{x}^{\mathrm{18}} \:=\mathrm{sin}^{\mathrm{2}} \mathrm{t}\:\rightarrow\mathrm{x}=\mathrm{sin}^{\frac{\mathrm{1}}{\mathrm{9}}} \mathrm{t}} \frac{\mathrm{1}}{\mathrm{9}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cos}^{\frac{\mathrm{1}}{\mathrm{10}}} \mathrm{tcostsin}^{\frac{\mathrm{1}}{\mathrm{9}}−\mathrm{1}} \mathrm{t}\:\mathrm{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{9}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{cos}^{\frac{\mathrm{1}}{\mathrm{10}}+\mathrm{1}} \mathrm{t}\:\mathrm{sin}^{\frac{\mathrm{1}}{\mathrm{9}}−\mathrm{1}} \mathrm{t}\:\mathrm{dt}\:\mathrm{we}\:\mathrm{know} \\ $$$$\mathrm{2}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{cos}^{\mathrm{2p}−\mathrm{1}} \mathrm{t}\:\mathrm{sin}^{\mathrm{2q}−\mathrm{1}} \mathrm{t}\:\mathrm{dt}\:=\mathrm{B}\left(\mathrm{p},\mathrm{q}\right)=\frac{\Gamma\left(\mathrm{p}\right).\Gamma\left(\mathrm{q}\right)}{\Gamma\left(\mathrm{p}+\mathrm{q}\right)} \\ $$$$\mathrm{2p}−\mathrm{1}=\frac{\mathrm{1}}{\mathrm{10}}+\mathrm{1}\:\Rightarrow\mathrm{2p}=\frac{\mathrm{1}}{\mathrm{10}}\:+\mathrm{2}\:\Rightarrow\mathrm{p}=\frac{\mathrm{1}}{\mathrm{20}}+\mathrm{1}\:=\frac{\mathrm{21}}{\mathrm{20}} \\ $$$$\mathrm{2q}−\mathrm{1}=\frac{\mathrm{1}}{\mathrm{9}}−\mathrm{1}\:\Rightarrow\mathrm{q}=\frac{\mathrm{1}}{\mathrm{18}}\:\Rightarrow \\ $$$$\mathrm{H}\:=\frac{\mathrm{1}}{\mathrm{18}}\mathrm{B}\left(\frac{\mathrm{21}}{\mathrm{20}},\frac{\mathrm{1}}{\mathrm{18}}\right)\:=\frac{\mathrm{1}}{\mathrm{18}}\frac{\Gamma\left(\frac{\mathrm{21}}{\mathrm{20}}\right).\Gamma\left(\frac{\mathrm{1}}{\mathrm{18}}\right)}{\Gamma\left(\frac{\mathrm{21}}{\mathrm{20}}+\frac{\mathrm{1}}{\mathrm{18}}\right)} \\ $$$$\mathrm{K}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−\mathrm{x}^{\mathrm{20}} \right)^{\frac{\mathrm{1}}{\mathrm{18}}} \mathrm{dx}\:=_{\mathrm{x}^{\mathrm{20}} =\mathrm{sin}^{\mathrm{2}} \mathrm{t}\rightarrow\mathrm{x}=\mathrm{sin}^{\frac{\mathrm{1}}{\mathrm{10}}} \mathrm{t}} \frac{\mathrm{1}}{\mathrm{10}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cos}^{\frac{\mathrm{1}}{\mathrm{9}}} \mathrm{t}\:\mathrm{cost}\:\mathrm{sin}^{\frac{\mathrm{1}}{\mathrm{10}}−\mathrm{1}} \:\mathrm{t}\:\mathrm{dt}\:\: \\ $$$$=\frac{\mathrm{1}}{\mathrm{10}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{cos}^{\frac{\mathrm{1}}{\mathrm{9}}+\mathrm{1}} \mathrm{t}\:\mathrm{sin}^{\frac{\mathrm{1}}{\mathrm{10}}−\mathrm{1}} \mathrm{t}\:\mathrm{dt} \\ $$$$\mathrm{2p}−\mathrm{1}=\frac{\mathrm{1}}{\mathrm{9}}+\mathrm{1}\:\Rightarrow\mathrm{2p}\:=\frac{\mathrm{1}}{\mathrm{9}}+\mathrm{2}\:\Rightarrow\mathrm{p}\:=\frac{\mathrm{1}}{\mathrm{18}}\:+\mathrm{1}\:=\frac{\mathrm{19}}{\mathrm{18}} \\ $$$$\mathrm{2q}−\mathrm{1}\:=\frac{\mathrm{1}}{\mathrm{10}}−\mathrm{1}\:\Rightarrow\mathrm{q}=\frac{\mathrm{1}}{\mathrm{20}}\:\Rightarrow \\ $$$$\mathrm{K}\:=\frac{\mathrm{1}}{\mathrm{20}}\:\mathrm{B}\left(\frac{\mathrm{19}}{\mathrm{18}},\frac{\mathrm{1}}{\mathrm{20}}\right)\:=\frac{\mathrm{1}}{\mathrm{20}}\frac{\Gamma\left(\frac{\mathrm{19}}{\mathrm{18}}\right).\Gamma\left(\frac{\mathrm{1}}{\mathrm{20}}\right)}{\Gamma\left(\frac{\mathrm{19}}{\mathrm{18}}+\frac{\mathrm{1}}{\mathrm{20}}\right)} \\ $$$$\mathrm{I}\:=\:\mathrm{H}−\mathrm{K} \\ $$
Commented by Her_Majesty last updated on 14/Aug/20
Γ(((21)/(20)))=(1/(20))Γ((1/(20))) Γ(((19)/(18)))=(1/(18))Γ((1/(18))) ((21)/(20))+(1/(18))=((19)/(18))+(1/(20)) ⇒ H=K ⇒ I=0
$$\Gamma\left(\frac{\mathrm{21}}{\mathrm{20}}\right)=\frac{\mathrm{1}}{\mathrm{20}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{20}}\right) \\ $$$$\Gamma\left(\frac{\mathrm{19}}{\mathrm{18}}\right)=\frac{\mathrm{1}}{\mathrm{18}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{18}}\right) \\ $$$$\frac{\mathrm{21}}{\mathrm{20}}+\frac{\mathrm{1}}{\mathrm{18}}=\frac{\mathrm{19}}{\mathrm{18}}+\frac{\mathrm{1}}{\mathrm{20}} \\ $$$$\Rightarrow\:{H}={K} \\ $$$$\Rightarrow\:{I}=\mathrm{0} \\ $$
Commented by abdomathmax last updated on 15/Aug/20
thank you for completing
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{for}\:\mathrm{completing} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *