# calculus-prove-that-0-4-tan-x-tan-2-x-tan-x-tan-2-x-sin-x-dx-1-2-8-1-4-3-4-3-4-5-4-

Question Number 131211 by mnjuly1970 last updated on 02/Feb/21
$$\:\:\:\:\:\:\:\:\:\:\:\:…{calculus}… \\$$$$\:{prove}\:{that}:: \\$$$$\:\boldsymbol{\Phi}=\int_{\mathrm{0}} ^{\:\frac{\boldsymbol{\pi}}{\mathrm{4}}} \left(\frac{\sqrt{\boldsymbol{{tan}}\left(\boldsymbol{{x}}\right)+\boldsymbol{{tan}}^{\mathrm{2}} \left(\boldsymbol{{x}}\right)}}{\:\sqrt{\boldsymbol{{tan}}\left(\boldsymbol{{x}}\right)β\boldsymbol{{tan}}^{\mathrm{2}} \left(\boldsymbol{{x}}\right)}}\:\boldsymbol{{sin}}\left(\boldsymbol{{x}}\right)\right)\boldsymbol{{dx}}\: \\$$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\boldsymbol{\pi}}}{\mathrm{8}}\:\left(\frac{\boldsymbol{\Gamma}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)}{\boldsymbol{\Gamma}\left(\frac{\mathrm{3}}{\mathrm{4}}\right)}β\frac{\boldsymbol{\Gamma}\left(\frac{\mathrm{3}}{\mathrm{4}}\right)}{\boldsymbol{\Gamma}\left(\frac{\mathrm{5}}{\mathrm{4}}\right)}\right) \\$$
Answered by Ar Brandon last updated on 02/Feb/21
$$\Phi=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \left(\frac{\sqrt{\mathrm{tanx}+\mathrm{tan}^{\mathrm{2}} \mathrm{x}}}{\:\sqrt{\mathrm{tanx}β\mathrm{tan}^{\mathrm{2}} \mathrm{x}}}\mathrm{sinx}\right)\mathrm{dx}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \sqrt{\frac{\mathrm{1}+\mathrm{tanx}}{\mathrm{1}β\mathrm{tanx}}}\centerdot\frac{\mathrm{tanx}}{\:\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \mathrm{x}}}\mathrm{dx} \\$$$$\:\:\:\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\left(\mathrm{1}+\mathrm{tanx}\right)\mathrm{tanx}}{\:\sqrt{\left(\mathrm{1}β\mathrm{tan}^{\mathrm{2}} \mathrm{x}\right)\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \mathrm{x}\right)}}\mathrm{dx},\:\mathrm{tanx}=\mathrm{t}\:\Rightarrow\mathrm{dx}=\frac{\mathrm{dt}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} } \\$$$$\:\:\:\:=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{t}+\mathrm{t}^{\mathrm{2}} }{\:\sqrt{\left(\mathrm{1}β\mathrm{t}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)}}\centerdot\frac{\mathrm{dt}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} } \\$$$$\:\:\:\:=\int_{\mathrm{0}} ^{\mathrm{1}} \left\{\frac{\mathrm{t}}{\:\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)\sqrt{\left(\mathrm{1}β\mathrm{t}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)}}+\frac{\mathrm{t}^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)\sqrt{\left(\mathrm{1}β\mathrm{t}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)}}\right\}\mathrm{dt} \\$$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{du}}{\left(\mathrm{1}+\mathrm{u}\right)\sqrt{\mathrm{1}β\mathrm{u}^{\mathrm{2}} }}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{u}^{\mathrm{1}/\mathrm{2}} \mathrm{du}}{\left(\mathrm{1}+\mathrm{u}\right)\sqrt{\mathrm{1}β\mathrm{u}^{\mathrm{2}} }} \\$$$$\frac{\mathrm{1}}{\mathrm{u}+\mathrm{1}}=\mathrm{v}\:\Rightarrowβ\frac{\mathrm{du}}{\left(\mathrm{u}+\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{dv}\:\Rightarrow\mathrm{du}=β\frac{\mathrm{dv}}{\mathrm{v}^{\mathrm{2}} } \\$$$$\Phi=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{1}/\mathrm{2}} ^{\mathrm{1}} \frac{\mathrm{v}}{\:\sqrt{\frac{\mathrm{2}}{\mathrm{v}}β\frac{\mathrm{1}}{\mathrm{v}^{\mathrm{2}} }}}\centerdot\frac{\mathrm{dv}}{\mathrm{v}^{\mathrm{2}} }+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{u}^{\mathrm{1}/\mathrm{2}} \mathrm{du}}{\left(\mathrm{1}+\mathrm{u}\right)\sqrt{\mathrm{1}β\mathrm{u}^{\mathrm{2}} }} \\$$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{1}/\mathrm{2}} ^{\mathrm{1}} \frac{\mathrm{dv}}{\:\sqrt{\mathrm{2v}β\mathrm{1}}}+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{p}^{\mathrm{1}/\mathrm{4}} }{\left(\mathrm{1}+\mathrm{p}^{\mathrm{1}/\mathrm{2}} \right)\sqrt{\mathrm{1}β\mathrm{p}}}\centerdot\frac{\mathrm{dp}}{\mathrm{p}^{\mathrm{1}/\mathrm{2}} } \\$$$$\:\:\:\:=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{p}^{β\mathrm{1}/\mathrm{4}} }{\left(\mathrm{1}+\mathrm{p}^{\mathrm{1}/\mathrm{2}} \right)\sqrt{\mathrm{1}β\mathrm{p}}}\mathrm{dp} \\$$$$… \\$$
Commented by mnjuly1970 last updated on 02/Feb/21
$${grateful}\:..{for}\:{your}\:\:{effort} \\$$$${mr}\:{brandon}… \\$$
Commented by Ar Brandon last updated on 02/Feb/21
Thank you Sir ���� Got stucked unfortunately
Commented by mnjuly1970 last updated on 02/Feb/21
Commented by Ar Brandon last updated on 02/Feb/21
Nice��
Commented by mnjuly1970 last updated on 02/Feb/21
$${thank}\:{you}\:{master}\:{brandon} \\$$$$\:{God}\:\:{keep}\:{you}… \\$$