Question Number 133688 by mnjuly1970 last updated on 23/Feb/21
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:#\:\:\:{calculus}\:# \\ $$$$\:\:\:\:\:\:\:{find}:\: \\ $$$$\:\:\:\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \frac{{dx}}{{sin}^{\mathrm{10}} \left({x}\right)+{cos}^{\mathrm{10}} \left({x}\right)}\:=??? \\ $$$$ \\ $$
Answered by Ar Brandon last updated on 23/Feb/21
$$\phi=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{d}{x}}{\mathrm{sin}^{\mathrm{10}} {x}+\mathrm{cos}^{\mathrm{10}} {x}}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{sec}^{\mathrm{10}} {x}}{\mathrm{tan}^{\mathrm{10}} {x}+\mathrm{1}}\mathrm{d}{x}=\int_{\mathrm{0}} ^{\infty} \frac{\left(\mathrm{t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{4}} }{\mathrm{t}^{\mathrm{10}} +\mathrm{1}}\mathrm{dt} \\ $$$$\:\:\:=\int_{\mathrm{0}} ^{\infty} \left\{\frac{\mathrm{t}^{\mathrm{8}} +\mathrm{4t}^{\mathrm{6}} +\mathrm{6t}^{\mathrm{4}} +\mathrm{4t}^{\mathrm{2}} +\mathrm{1}}{\mathrm{t}^{\mathrm{10}} +\mathrm{1}}\right\}\mathrm{dt}\:,\:\mathrm{u}=\mathrm{t}^{\mathrm{10}} \:\Rightarrow\mathrm{du}=\mathrm{10t}^{\mathrm{9}} \mathrm{dt}=\mathrm{10u}^{\frac{\mathrm{9}}{\mathrm{10}}} \mathrm{dt} \\ $$$$\phi=\frac{\mathrm{1}}{\mathrm{10}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{u}^{−\frac{\mathrm{1}}{\mathrm{10}}} }{\mathrm{u}+\mathrm{1}}\mathrm{du}+\frac{\mathrm{2}}{\mathrm{5}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{u}^{−\frac{\mathrm{3}}{\mathrm{10}}} }{\mathrm{u}+\mathrm{1}}\mathrm{du}+\frac{\mathrm{3}}{\mathrm{5}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{u}^{−\frac{\mathrm{5}}{\mathrm{10}}} }{\mathrm{u}+\mathrm{1}}\mathrm{du}+\frac{\mathrm{2}}{\mathrm{5}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{u}^{−\frac{\mathrm{7}}{\mathrm{10}}} }{\mathrm{u}+\mathrm{1}}\mathrm{du}+\frac{\mathrm{1}}{\mathrm{10}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{u}^{−\frac{\mathrm{9}}{\mathrm{10}}} }{\mathrm{u}+\mathrm{1}}\mathrm{du} \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{10}}\beta\left(\frac{\mathrm{9}}{\mathrm{10}},\frac{\mathrm{1}}{\mathrm{10}}\right)+\frac{\mathrm{2}}{\mathrm{5}}\beta\left(\frac{\mathrm{7}}{\mathrm{10}},\frac{\mathrm{3}}{\mathrm{10}}\right)+\frac{\mathrm{3}}{\mathrm{5}}\beta\left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}}\right)+\frac{\mathrm{2}}{\mathrm{5}}\beta\left(\frac{\mathrm{3}}{\mathrm{10}},\frac{\mathrm{7}}{\mathrm{10}}\right)+\frac{\mathrm{1}}{\mathrm{10}}\beta\left(\frac{\mathrm{1}}{\mathrm{10}},\frac{\mathrm{9}}{\mathrm{10}}\right) \\ $$$$\:\:\:=\frac{\pi}{\mathrm{5sin}\left(\frac{\pi}{\mathrm{10}}\right)}+\frac{\mathrm{4}\pi}{\mathrm{5sin}\left(\frac{\mathrm{3}\pi}{\mathrm{10}}\right)}+\frac{\mathrm{3}\pi}{\mathrm{5sin}\left(\frac{\pi}{\mathrm{2}}\right)} \\ $$
Commented by Dwaipayan Shikari last updated on 23/Feb/21
$$\frac{\pi}{\mathrm{5}}\left(\frac{\mathrm{4}}{\:\sqrt{\mathrm{5}}−\mathrm{1}}+\frac{\mathrm{16}}{\:\sqrt{\mathrm{5}}+\mathrm{1}}+\mathrm{3}\right)=\frac{\pi}{\mathrm{5}}\left(\sqrt{\mathrm{5}}+\mathrm{1}+\mathrm{4}\sqrt{\mathrm{5}}−\mathrm{4}+\mathrm{3}\right) \\ $$$$=\sqrt{\mathrm{5}}\pi \\ $$
Commented by mnjuly1970 last updated on 23/Feb/21
$${thanks}\:\:{alot}\:\:{master}\:{brandon} \\ $$$$\:\:{nice}\:\:\:{very}\:\:{nice}\:{solution}… \\ $$
Commented by Ar Brandon last updated on 23/Feb/21
My pleasure Sir
Answered by Ajetunmobi last updated on 23/Feb/21
$$\boldsymbol{\mathrm{i}}\:\boldsymbol{\mathrm{have}}\:\boldsymbol{\mathrm{did}}\:\boldsymbol{\mathrm{this}}\:\boldsymbol{\mathrm{on}}\:\boldsymbol{\mathrm{facebook}} \\ $$
Answered by Ajetunmobi last updated on 23/Feb/21
Commented by Ar Brandon last updated on 23/Feb/21
Ambiguity on your name Sir. Ajetunmobi or Ajeutunmobi ?��
Commented by Ajetunmobi last updated on 24/Feb/21
$${it}\:{is}\:\boldsymbol{\mathrm{Ajetunmobi}}\:\boldsymbol{\mathrm{Abdulqoyyum}} \\ $$$$\boldsymbol{\mathrm{thank}}\:\boldsymbol{\mathrm{u}}\:\boldsymbol{\mathrm{that}}\:\boldsymbol{\mathrm{was}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{mistake}} \\ $$
Answered by Ajetunmobi last updated on 23/Feb/21
