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Question Number 141218 by mathmax by abdo last updated on 16/May/21

calculate I =∫_0 ^(π/2) ^4 (√(tanx))log(tanx)dx and J =∫_0 ^(π/2) ((log(tanx))/((^3 (√(tanx)))))dx

$$\mathrm{calculate}\:\:\mathrm{I}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:^{\mathrm{4}} \sqrt{\mathrm{tanx}}\mathrm{log}\left(\mathrm{tanx}\right)\mathrm{dx}\:\:\mathrm{and} \\ $$$$\mathrm{J}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{log}\left(\mathrm{tanx}\right)}{\left(^{\mathrm{3}} \sqrt{\mathrm{tanx}}\right)}\mathrm{dx} \\ $$

Answered by Ar Brandon last updated on 16/May/21

f(α)=∫_0 ^(π/2) tan^α xdx=(1/2)Γ(((α+1)/2))Γ(((1−α)/2))=(π/(2sin(((1−α)/2)π))) f((1/4))=((Γ((5/8))Γ((3/8)))/2)=(π/(2sin(((3π)/8))))★ ln(f(α))=ln((π/2))−lnsin(((1−α)/2)π) ((f ′(α))/(f(α)))=(π/2)∙tan(((1−α)/2)π)⇒f ′(α)=f(α){(π/2)∙tan(((1−α)/2)π)} f ′((1/4))=f((1/4)){(π/2)∙tan(((3π)/( 8)))} I=(π^2 /4)tan(((3π)/8))cosec(((3π)/8))

$$\mathrm{f}\left(\alpha\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{tan}^{\alpha} \mathrm{xdx}=\frac{\mathrm{1}}{\mathrm{2}}\Gamma\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{1}−\alpha}{\mathrm{2}}\right)=\frac{\pi}{\mathrm{2sin}\left(\frac{\mathrm{1}−\alpha}{\mathrm{2}}\pi\right)} \\ $$$$\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)=\frac{\Gamma\left(\frac{\mathrm{5}}{\mathrm{8}}\right)\Gamma\left(\frac{\mathrm{3}}{\mathrm{8}}\right)}{\mathrm{2}}=\frac{\pi}{\mathrm{2sin}\left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)}\bigstar \\ $$$$\mathrm{ln}\left(\mathrm{f}\left(\alpha\right)\right)=\mathrm{ln}\left(\frac{\pi}{\mathrm{2}}\right)−\mathrm{lnsin}\left(\frac{\mathrm{1}−\alpha}{\mathrm{2}}\pi\right) \\ $$$$\frac{\mathrm{f}\:'\left(\alpha\right)}{\mathrm{f}\left(\alpha\right)}=\frac{\pi}{\mathrm{2}}\centerdot\mathrm{tan}\left(\frac{\mathrm{1}−\alpha}{\mathrm{2}}\pi\right)\Rightarrow\mathrm{f}\:'\left(\alpha\right)=\mathrm{f}\left(\alpha\right)\left\{\frac{\pi}{\mathrm{2}}\centerdot\mathrm{tan}\left(\frac{\mathrm{1}−\alpha}{\mathrm{2}}\pi\right)\right\} \\ $$$$\mathrm{f}\:'\left(\frac{\mathrm{1}}{\mathrm{4}}\right)=\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\left\{\frac{\pi}{\mathrm{2}}\centerdot\mathrm{tan}\left(\frac{\mathrm{3}\pi}{\:\mathrm{8}}\right)\right\} \\ $$$$\mathrm{I}=\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\mathrm{tan}\left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)\mathrm{cosec}\left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right) \\ $$

Answered by Ar Brandon last updated on 16/May/21

f(α)=∫_0 ^(π/2) tan^α xdx=(1/2)Γ(((α+1)/2))Γ(((1−α)/2))=(π/(2sin(((1+α)/2)π))) f(−(1/3))=(π/(2sin((π/3))))★ ln(f(α))=ln((π/2))−lnsin(((1+α)/2)π) ((f ′(α))/(f(α)))=(π/2)∙tan(((1+α)/2)π)⇒f ′(α)=f(α){(π/2)∙tan(((1+α)/2)π)} f ′(−(1/3))=f(−(1/3)){(π/2)∙tan((π/( 3)))} J=(π^2 /4)tan((π/3))cosec((π/3))

$$\mathrm{f}\left(\alpha\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{tan}^{\alpha} \mathrm{xdx}=\frac{\mathrm{1}}{\mathrm{2}}\Gamma\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{1}−\alpha}{\mathrm{2}}\right)=\frac{\pi}{\mathrm{2sin}\left(\frac{\mathrm{1}+\alpha}{\mathrm{2}}\pi\right)} \\ $$$$\mathrm{f}\left(−\frac{\mathrm{1}}{\mathrm{3}}\right)=\frac{\pi}{\mathrm{2sin}\left(\frac{\pi}{\mathrm{3}}\right)}\bigstar \\ $$$$\mathrm{ln}\left(\mathrm{f}\left(\alpha\right)\right)=\mathrm{ln}\left(\frac{\pi}{\mathrm{2}}\right)−\mathrm{lnsin}\left(\frac{\mathrm{1}+\alpha}{\mathrm{2}}\pi\right) \\ $$$$\frac{\mathrm{f}\:'\left(\alpha\right)}{\mathrm{f}\left(\alpha\right)}=\frac{\pi}{\mathrm{2}}\centerdot\mathrm{tan}\left(\frac{\mathrm{1}+\alpha}{\mathrm{2}}\pi\right)\Rightarrow\mathrm{f}\:'\left(\alpha\right)=\mathrm{f}\left(\alpha\right)\left\{\frac{\pi}{\mathrm{2}}\centerdot\mathrm{tan}\left(\frac{\mathrm{1}+\alpha}{\mathrm{2}}\pi\right)\right\} \\ $$$$\mathrm{f}\:'\left(−\frac{\mathrm{1}}{\mathrm{3}}\right)=\mathrm{f}\left(−\frac{\mathrm{1}}{\mathrm{3}}\right)\left\{\frac{\pi}{\mathrm{2}}\centerdot\mathrm{tan}\left(\frac{\pi}{\:\mathrm{3}}\right)\right\} \\ $$$$\mathrm{J}=\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\mathrm{tan}\left(\frac{\pi}{\mathrm{3}}\right)\mathrm{cosec}\left(\frac{\pi}{\mathrm{3}}\right) \\ $$

Commented by Ar Brandon last updated on 16/May/21

f ′(α)=(π^2 /4)tan(((1−α)/2)π)cosec(((1−α)/2)π)

$$\mathrm{f}\:'\left(\alpha\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\mathrm{tan}\left(\frac{\mathrm{1}−\alpha}{\mathrm{2}}\pi\right)\mathrm{cosec}\left(\frac{\mathrm{1}−\alpha}{\mathrm{2}}\pi\right) \\ $$

Answered by mathmax by abdo last updated on 16/May/21

let f(a) =∫_0 ^(π/2) (tanx)^a dx ⇒f(a)=∫_0 ^(π/2) e^(aln(tanx)) dx ⇒ f^′ (a)=∫_0 ^(π/2) (tanx)^a log(tanx)dx ⇒f^′ ((1/4))=∫_0 ^(π/2) (tanx)^(1/4) log(tanx)dx=I changement tanx=t give f(a)=∫_0 ^∞ (t^a /(1+t^2 ))dt =_(t=(√z)) ∫_0 ^∞ (z^(a/2) /(1+z))(dz/(2(√z))) =(1/2)∫_0 ^∞ (z^((a/2)−(1/2)) /(1+z))dz =(1/2)∫_0 ^∞ (z^(((a+1)/2)−1) /(1+z))dz =(π/(2sin(π(((a+1))/2))))) =(π/(2cos(((πa)/2)))) cv of f(a) we must have0 <((a+1)/2)<1 ⇒−1<a<1 f^′ (a) =−(π/2)×((−(π/2)sin(((πa)/2)))/(cos^2 (((πa)/2)))) =(π^2 /4).((sin(((πa)/2)))/(cos^2 (((πa)/2)))) I =f^′ ((1/4))=(π^2 /4).((sin((π/8)))/(cos^2 ((π/8)))) =(π^2 /4).(((√(2−(√2)))/2)/((2+(√2))/4)) =(π^2 /2)×((√(2−(√2)))/(2+(√2))) J = ∫_0 ^(π/2) (tanx)^(−(1/3)) log(tanx)dx =f^′ (−(1/3)) =−(π^2 /4).((sin((π/6)))/(cos^2 ((π/6)))) =−(π^2 /4)×((1/2)/((((√3)/2))^2 )) =−(π^2 /8)×(4/3) =−(π^2 /6)

$$\mathrm{let}\:\mathrm{f}\left(\mathrm{a}\right)\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{tanx}\right)^{\mathrm{a}} \:\mathrm{dx}\:\:\Rightarrow\mathrm{f}\left(\mathrm{a}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{e}^{\mathrm{aln}\left(\mathrm{tanx}\right)} \mathrm{dx}\:\Rightarrow \\ $$$$\mathrm{f}^{'} \left(\mathrm{a}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{tanx}\right)^{\mathrm{a}} \mathrm{log}\left(\mathrm{tanx}\right)\mathrm{dx}\:\:\Rightarrow\mathrm{f}^{'} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\left(\mathrm{tanx}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} \:\mathrm{log}\left(\mathrm{tanx}\right)\mathrm{dx}=\mathrm{I} \\ $$$$\mathrm{changement}\:\mathrm{tanx}=\mathrm{t}\:\mathrm{give}\:\mathrm{f}\left(\mathrm{a}\right)=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{t}^{\mathrm{a}} }{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\mathrm{dt}\:=_{\mathrm{t}=\sqrt{\mathrm{z}}} \:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{z}^{\frac{\mathrm{a}}{\mathrm{2}}} }{\mathrm{1}+\mathrm{z}}\frac{\mathrm{dz}}{\mathrm{2}\sqrt{\mathrm{z}}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{z}^{\frac{\mathrm{a}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}} }{\mathrm{1}+\mathrm{z}}\mathrm{dz}\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{z}^{\frac{\mathrm{a}+\mathrm{1}}{\mathrm{2}}−\mathrm{1}} }{\mathrm{1}+\mathrm{z}}\mathrm{dz}\:=\frac{\pi}{\mathrm{2sin}\left(\pi\left(\frac{\left.\mathrm{a}+\mathrm{1}\right)}{\mathrm{2}}\right)\right)} \\ $$$$=\frac{\pi}{\mathrm{2cos}\left(\frac{\pi\mathrm{a}}{\mathrm{2}}\right)}\:\:\:\mathrm{cv}\:\mathrm{of}\:\mathrm{f}\left(\mathrm{a}\right)\:\:\:\mathrm{we}\:\mathrm{must}\:\mathrm{have0}\:<\frac{\mathrm{a}+\mathrm{1}}{\mathrm{2}}<\mathrm{1}\:\Rightarrow−\mathrm{1}<\mathrm{a}<\mathrm{1} \\ $$$$\mathrm{f}^{'} \left(\mathrm{a}\right)\:=−\frac{\pi}{\mathrm{2}}×\frac{−\frac{\pi}{\mathrm{2}}\mathrm{sin}\left(\frac{\pi\mathrm{a}}{\mathrm{2}}\right)}{\mathrm{cos}^{\mathrm{2}} \left(\frac{\pi\mathrm{a}}{\mathrm{2}}\right)}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{4}}.\frac{\mathrm{sin}\left(\frac{\pi\mathrm{a}}{\mathrm{2}}\right)}{\mathrm{cos}^{\mathrm{2}} \left(\frac{\pi\mathrm{a}}{\mathrm{2}}\right)} \\ $$$$\mathrm{I}\:=\mathrm{f}^{'} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{4}}.\frac{\mathrm{sin}\left(\frac{\pi}{\mathrm{8}}\right)}{\mathrm{cos}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{8}}\right)}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{4}}.\frac{\frac{\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}}{\mathrm{2}}}{\frac{\mathrm{2}+\sqrt{\mathrm{2}}}{\mathrm{4}}}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{2}}×\frac{\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}}{\mathrm{2}+\sqrt{\mathrm{2}}} \\ $$$$\mathrm{J}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{tanx}\right)^{−\frac{\mathrm{1}}{\mathrm{3}}} \:\mathrm{log}\left(\mathrm{tanx}\right)\mathrm{dx}\:=\mathrm{f}^{'} \left(−\frac{\mathrm{1}}{\mathrm{3}}\right) \\ $$$$=−\frac{\pi^{\mathrm{2}} }{\mathrm{4}}.\frac{\mathrm{sin}\left(\frac{\pi}{\mathrm{6}}\right)}{\mathrm{cos}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{6}}\right)}\:=−\frac{\pi^{\mathrm{2}} }{\mathrm{4}}×\frac{\frac{\mathrm{1}}{\mathrm{2}}}{\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} }\:=−\frac{\pi^{\mathrm{2}} }{\mathrm{8}}×\frac{\mathrm{4}}{\mathrm{3}} \\ $$$$=−\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$

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