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Question Number 133016 by metamorfose last updated on 18/Feb/21

∫_0 ^(π/2) (tan(x))^(1/n) dx ...

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left({tan}\left({x}\right)\right)^{\frac{\mathrm{1}}{{n}}} {dx}\:... \\ $$

Answered by Ar Brandon last updated on 18/Feb/21

I=∫_0 ^(π/2) (tanx)^(1/n) dx=∫_0 ^(π/2) (sinx)^(1/n) (cosx)^(−(1/n)) dx =(1/2)Γ(((n+1)/(2n)))Γ(((n−1)/(2n)))=(1/2)∙(π/(sin(((n+1)/(2n))π))) , n>1

$$\mathcal{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{tanx}\right)^{\frac{\mathrm{1}}{\mathrm{n}}} \mathrm{dx}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{sinx}\right)^{\frac{\mathrm{1}}{\mathrm{n}}} \left(\mathrm{cosx}\right)^{−\frac{\mathrm{1}}{\mathrm{n}}} \mathrm{dx} \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\Gamma\left(\frac{\mathrm{n}+\mathrm{1}}{\mathrm{2n}}\right)\Gamma\left(\frac{\mathrm{n}−\mathrm{1}}{\mathrm{2n}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\centerdot\frac{\pi}{\mathrm{sin}\left(\frac{\mathrm{n}+\mathrm{1}}{\mathrm{2n}}\pi\right)}\:,\:\mathrm{n}>\mathrm{1} \\ $$

Answered by Dwaipayan Shikari last updated on 18/Feb/21

∫_0 ^(π/2) ((tanx))^(1/n) dx=((Γ((1/2)−(1/(2n)))Γ((1/2)+(1/(2n))))/2)=(π/(2sin((π/2)+(π/(2n)))))

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt[{{n}}]{{tanx}}\:{dx}=\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}{n}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}{n}}\right)}{\mathrm{2}}=\frac{\pi}{\mathrm{2}{sin}\left(\frac{\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{2}{n}}\right)} \\ $$

Answered by mathmax by abdo last updated on 20/Feb/21

I =∫_0 ^(π/2) (tanx)^(1/n) dx changement tanx =t give I =∫_0 ^∞ (t^(1/n) /(1+t^2 ))dt =_(t=z^(1/2) ) ∫_0 ^∞ (z^(1/(2n)) /(1+z))(1/2)z^(−(1/2)) dz =(1/2)∫_0 ^∞ (z^((1/(2n))+(1/2)−1) /(1+z))dz =(1/2)(π/(sin(π((1/(2n))+(1/2))))) =(π/(2sin((π/(2n))+(π/2)))) =(π/(2cos((π/(2n)))))(with n>1)

$$\mathrm{I}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{tanx}\right)^{\frac{\mathrm{1}}{\mathrm{n}}} \:\mathrm{dx}\:\mathrm{changement}\:\:\mathrm{tanx}\:=\mathrm{t}\:\mathrm{give} \\ $$$$\mathrm{I}\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\mathrm{t}^{\frac{\mathrm{1}}{\mathrm{n}}} }{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\mathrm{dt}\:\:=_{\mathrm{t}=\mathrm{z}^{\frac{\mathrm{1}}{\mathrm{2}}} } \:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\mathrm{z}^{\frac{\mathrm{1}}{\mathrm{2n}}} }{\mathrm{1}+\mathrm{z}}\frac{\mathrm{1}}{\mathrm{2}}\mathrm{z}^{−\frac{\mathrm{1}}{\mathrm{2}}} \mathrm{dz}\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{z}^{\frac{\mathrm{1}}{\mathrm{2n}}+\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} }{\mathrm{1}+\mathrm{z}}\mathrm{dz} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\frac{\pi}{\mathrm{sin}\left(\pi\left(\frac{\mathrm{1}}{\mathrm{2n}}+\frac{\mathrm{1}}{\mathrm{2}}\right)\right)}\:=\frac{\pi}{\mathrm{2sin}\left(\frac{\pi}{\mathrm{2n}}+\frac{\pi}{\mathrm{2}}\right)}\:=\frac{\pi}{\mathrm{2cos}\left(\frac{\pi}{\mathrm{2n}}\right)}\left(\mathrm{with}\:\mathrm{n}>\mathrm{1}\right) \\ $$

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