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:: 𝚽=∫_0 ^( (𝛑/4)) (((√(tan(x)+tan^2 (x)))/( (√(tan(x)−tan^2 (x))))) sin(x))dx =(1/2)+((√𝛑)/8) (((𝚪((1/4)))/(𝚪((3/4))))−((𝚪((3/4)))/(𝚪((5/4)))))

$$\:\:\:\:\:\:\:\:\:\:\:\:…{calculus}… \\ $$$$\:{prove}\:{that}:: \\ $$$$\:\boldsymbol{\Phi}=\int_{\mathrm{0}} ^{\:\frac{\boldsymbol{\pi}}{\mathrm{4}}} \left(\frac{\sqrt{\boldsymbol{{tan}}\left(\boldsymbol{{x}}\right)+\boldsymbol{{tan}}^{\mathrm{2}} \left(\boldsymbol{{x}}\right)}}{\:\sqrt{\boldsymbol{{tan}}\left(\boldsymbol{{x}}\right)−\boldsymbol{{tan}}^{\mathrm{2}} \left(\boldsymbol{{x}}\right)}}\:\boldsymbol{{sin}}\left(\boldsymbol{{x}}\right)\right)\boldsymbol{{dx}}\: \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\boldsymbol{\pi}}}{\mathrm{8}}\:\left(\frac{\boldsymbol{\Gamma}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)}{\boldsymbol{\Gamma}\left(\frac{\mathrm{3}}{\mathrm{4}}\right)}−\frac{\boldsymbol{\Gamma}\left(\frac{\mathrm{3}}{\mathrm{4}}\right)}{\boldsymbol{\Gamma}\left(\frac{\mathrm{5}}{\mathrm{4}}\right)}\right) \\ $$

Answered by Ar Brandon last updated on 02/Feb/21

Φ=∫_0 ^(π/4) (((√(tanx+tan^2 x))/( (√(tanx−tan^2 x))))sinx)dx=∫_0 ^(π/4) (√((1+tanx)/(1−tanx)))∙((tanx)/( (√(1+tan^2 x))))dx =∫_0 ^(π/4) (((1+tanx)tanx)/( (√((1−tan^2 x)(1+tan^2 x)))))dx, tanx=t ⇒dx=(dt/(1+t^2 )) =∫_0 ^1 ((t+t^2 )/( (√((1−t^2 )(1+t^2 )))))∙(dt/(1+t^2 )) =∫_0 ^1 {(t/( (1+t^2 )(√((1−t^2 )(1+t^2 )))))+(t^2 /((1+t^2 )(√((1−t^2 )(1+t^2 )))))}dt =(1/2)∫_0 ^1 (du/((1+u)(√(1−u^2 ))))+∫_0 ^1 ((u^(1/2) du)/((1+u)(√(1−u^2 )))) (1/(u+1))=v ⇒−(du/((u+1)^2 ))=dv ⇒du=−(dv/v^2 ) Φ=(1/2)∫_(1/2) ^1 (v/( (√((2/v)−(1/v^2 )))))∙(dv/v^2 )+∫_0 ^1 ((u^(1/2) du)/((1+u)(√(1−u^2 )))) =(1/2)∫_(1/2) ^1 (dv/( (√(2v−1))))+(1/2)∫_0 ^1 (p^(1/4) /((1+p^(1/2) )(√(1−p))))∙(dp/p^(1/2) ) =1+(1/2)∫_0 ^1 (p^(−1/4) /((1+p^(1/2) )(√(1−p))))dp ...

$$\Phi=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \left(\frac{\sqrt{\mathrm{tanx}+\mathrm{tan}^{\mathrm{2}} \mathrm{x}}}{\:\sqrt{\mathrm{tanx}−\mathrm{tan}^{\mathrm{2}} \mathrm{x}}}\mathrm{sinx}\right)\mathrm{dx}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \sqrt{\frac{\mathrm{1}+\mathrm{tanx}}{\mathrm{1}−\mathrm{tanx}}}\centerdot\frac{\mathrm{tanx}}{\:\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \mathrm{x}}}\mathrm{dx} \\ $$$$\:\:\:\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\left(\mathrm{1}+\mathrm{tanx}\right)\mathrm{tanx}}{\:\sqrt{\left(\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \mathrm{x}\right)\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \mathrm{x}\right)}}\mathrm{dx},\:\mathrm{tanx}=\mathrm{t}\:\Rightarrow\mathrm{dx}=\frac{\mathrm{dt}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} } \\ $$$$\:\:\:\:=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{t}+\mathrm{t}^{\mathrm{2}} }{\:\sqrt{\left(\mathrm{1}−\mathrm{t}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)}}\centerdot\frac{\mathrm{dt}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} } \\ $$$$\:\:\:\:=\int_{\mathrm{0}} ^{\mathrm{1}} \left\{\frac{\mathrm{t}}{\:\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)\sqrt{\left(\mathrm{1}−\mathrm{t}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)}}+\frac{\mathrm{t}^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)\sqrt{\left(\mathrm{1}−\mathrm{t}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)}}\right\}\mathrm{dt} \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{du}}{\left(\mathrm{1}+\mathrm{u}\right)\sqrt{\mathrm{1}−\mathrm{u}^{\mathrm{2}} }}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{u}^{\mathrm{1}/\mathrm{2}} \mathrm{du}}{\left(\mathrm{1}+\mathrm{u}\right)\sqrt{\mathrm{1}−\mathrm{u}^{\mathrm{2}} }} \\ $$$$\frac{\mathrm{1}}{\mathrm{u}+\mathrm{1}}=\mathrm{v}\:\Rightarrow−\frac{\mathrm{du}}{\left(\mathrm{u}+\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{dv}\:\Rightarrow\mathrm{du}=−\frac{\mathrm{dv}}{\mathrm{v}^{\mathrm{2}} } \\ $$$$\Phi=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{1}/\mathrm{2}} ^{\mathrm{1}} \frac{\mathrm{v}}{\:\sqrt{\frac{\mathrm{2}}{\mathrm{v}}−\frac{\mathrm{1}}{\mathrm{v}^{\mathrm{2}} }}}\centerdot\frac{\mathrm{dv}}{\mathrm{v}^{\mathrm{2}} }+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{u}^{\mathrm{1}/\mathrm{2}} \mathrm{du}}{\left(\mathrm{1}+\mathrm{u}\right)\sqrt{\mathrm{1}−\mathrm{u}^{\mathrm{2}} }} \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{1}/\mathrm{2}} ^{\mathrm{1}} \frac{\mathrm{dv}}{\:\sqrt{\mathrm{2v}−\mathrm{1}}}+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{p}^{\mathrm{1}/\mathrm{4}} }{\left(\mathrm{1}+\mathrm{p}^{\mathrm{1}/\mathrm{2}} \right)\sqrt{\mathrm{1}−\mathrm{p}}}\centerdot\frac{\mathrm{dp}}{\mathrm{p}^{\mathrm{1}/\mathrm{2}} } \\ $$$$\:\:\:\:=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{p}^{−\mathrm{1}/\mathrm{4}} }{\left(\mathrm{1}+\mathrm{p}^{\mathrm{1}/\mathrm{2}} \right)\sqrt{\mathrm{1}−\mathrm{p}}}\mathrm{dp} \\ $$$$… \\ $$

Commented by mnjuly1970 last updated on 02/Feb/21