Question Number 122176 by Dwaipayan Shikari last updated on 14/Nov/20
$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt[{\mathrm{7}}]{{tanx}}\:{dx} \\ $$
Commented by Dwaipayan Shikari last updated on 14/Nov/20
$${I}\:{have}\:{found}\:\frac{\pi}{\mathrm{2}}{cosec}\left(\frac{\mathrm{3}\pi}{\mathrm{7}}\right) \\ $$
Commented by mnjuly1970 last updated on 14/Nov/20
$$\:{perfect}\:{mr}\:{dwaipayan}.. \\ $$
Answered by mnjuly1970 last updated on 14/Nov/20
$$\:\mathrm{I}=\frac{\mathrm{1}}{\mathrm{2}}\ast\mathrm{2}\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} {sin}^{\frac{\mathrm{1}}{\mathrm{7}}} \left({x}\right){cos}^{\frac{−\mathrm{1}}{\mathrm{7}}} {dx} \\ $$$$\:\:=\frac{\mathrm{1}}{\mathrm{2}}\beta\left(\frac{\mathrm{4}}{\mathrm{7}}\:,\:\frac{\mathrm{3}}{\mathrm{7}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\:\ast\frac{\Gamma\left(\frac{\mathrm{4}}{\mathrm{7}}\right)\Gamma\left(\mathrm{1}−\frac{\mathrm{4}}{\mathrm{7}}\right)}{\Gamma\left(\mathrm{1}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\ast\frac{\pi}{{sin}\left(\frac{\mathrm{4}\pi}{\mathrm{7}}\right)}=\frac{\pi}{\mathrm{2}}\:{csc}\left(\frac{\mathrm{4}\pi}{\mathrm{7}}\right)\checkmark \\ $$
Commented by Dwaipayan Shikari last updated on 15/Nov/20
$${With}\:{pleasure} \\ $$
Answered by mathmax by abdo last updated on 14/Nov/20
$$\mathrm{I}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{tanx}\right)^{\frac{\mathrm{1}}{\mathrm{7}}} \:\mathrm{dx}\:\Rightarrow\mathrm{I}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{\frac{\mathrm{1}}{\mathrm{7}}} \mathrm{x}\:\mathrm{cos}^{−\frac{\mathrm{1}}{\mathrm{7}}} \mathrm{xdx}\:\Rightarrow \\ $$$$\mathrm{2I}\:=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{cos}^{−\frac{\mathrm{1}}{\mathrm{7}}} \mathrm{x}\:\:\mathrm{sin}^{\frac{\mathrm{1}}{\mathrm{7}}} \mathrm{xdx}\:\mathrm{we}\:\mathrm{know}\:\mathrm{that} \\ $$$$\mathrm{2}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{cos}^{\mathrm{2p}−\mathrm{1}} \mathrm{x}\:\mathrm{sin}^{\mathrm{2q}−\mathrm{1}} \mathrm{xdx}\:=\mathrm{B}\left(\mathrm{p},\mathrm{q}\right)\:=\frac{\Gamma\left(\mathrm{p}\right)\Gamma\left(\mathrm{q}\right)}{\Gamma\left(\mathrm{p}+\mathrm{q}\right)} \\ $$$$\mathrm{2p}−\mathrm{1}=−\frac{\mathrm{1}}{\mathrm{7}}\:\Rightarrow\mathrm{2p}\:=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{7}}\:=\frac{\mathrm{6}}{\mathrm{7}}\:\Rightarrow\mathrm{p}\:=\frac{\mathrm{3}}{\mathrm{7}} \\ $$$$\mathrm{2q}−\mathrm{1}=\frac{\mathrm{1}}{\mathrm{7}}\:\Rightarrow\mathrm{2q}=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{7}}\:=\frac{\mathrm{8}}{\mathrm{7}}\:\Rightarrow\mathrm{q}=\frac{\mathrm{4}}{\mathrm{7}}\:\Rightarrow\: \\ $$$$\mathrm{2}\:\mathrm{I}\:=\frac{\Gamma\left(\frac{\mathrm{3}}{\mathrm{7}}\right).\Gamma\left(\frac{\mathrm{4}}{\mathrm{7}}\right)}{\Gamma\left(\mathrm{1}\right)}\:\Rightarrow\mathrm{I}\:=\frac{\mathrm{1}}{\mathrm{2}}\Gamma\left(\frac{\mathrm{3}}{\mathrm{7}}\right).\Gamma\left(\mathrm{1}−\frac{\mathrm{3}}{\mathrm{7}}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}.\frac{\pi}{\mathrm{sin}\left(\frac{\pi}{\mathrm{7}}\right)}\:\Rightarrow \\ $$$$\mathrm{I}\:=\frac{\pi}{\mathrm{2sin}\left(\frac{\pi}{\mathrm{7}}\right)} \\ $$
Commented by mathmax by abdo last updated on 14/Nov/20
$$\mathrm{sorry}\:\mathrm{I}\:=\frac{\pi}{\mathrm{2sin}\left(\frac{\mathrm{3}\pi}{\mathrm{7}}\right)} \\ $$
Commented by Dwaipayan Shikari last updated on 15/Nov/20
$${Thanking}\:{you} \\ $$
Commented by mathmax by abdo last updated on 15/Nov/20
$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcome} \\ $$