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Question Number 136295 by mnjuly1970 last updated on 20/Mar/21

      ....nice  calculus...  prove ::  1 ::𝛗=∫_0 ^( (Ο€/2)) ((tan^(βˆ’1) ((√(tan(x))) ))/(tan(x)))dx=(Ο€/2)log(2+(√2) )  2::Ξ©=∫_(βˆ’Ο€) ^( Ο€) ((e^((sin(x)+cos(x))) cos(sin(x)))/(e^x +e^(sin(x)) ))dx=Ο€

$$\:\:\:\:\:\:....{nice}\:\:{calculus}... \\ $$$${prove}\::: \\ $$$$\mathrm{1}\:::\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \frac{{tan}^{βˆ’\mathrm{1}} \left(\sqrt{{tan}\left({x}\right)}\:\right)}{{tan}\left({x}\right)}{dx}=\frac{\pi}{\mathrm{2}}{log}\left(\mathrm{2}+\sqrt{\mathrm{2}}\:\right) \\ $$$$\mathrm{2}::\Omega=\int_{βˆ’\pi} ^{\:\pi} \frac{{e}^{\left({sin}\left({x}\right)+{cos}\left({x}\right)\right)} {cos}\left({sin}\left({x}\right)\right)}{{e}^{{x}} +{e}^{{sin}\left({x}\right)} }{dx}=\pi \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Answered by mindispower last updated on 20/Mar/21

Ξ©=Re∫_(βˆ’Ο€) ^Ο€ (e^(cos(x)+sin(x)(+isin(x)) /(e^x +sin(x)))dx=Re(W)  W=∫_(βˆ’Ο€) ^Ο€ (e^(cos(x)+isin(x)) /(e^(xβˆ’sin(x)) +1))dx,z=cos(x)+isin(x),z^βˆ’ ,conj(z)  by semetry  xβ†’βˆ’x  2w=∫_(βˆ’Ο€) ^Ο€ (e^z^βˆ’  /(e^(βˆ’x+sin(x)) +1))+(e^z /(e^(xβˆ’sin(x)) +1))dx  =∫_(βˆ’Ο€) ^Ο€ ((e^z^βˆ’  (e^(xβˆ’sin(x)) )+e^z )/(e^(xβˆ’sin(x)) +1))dx  =∫_(βˆ’Ο€) ^Ο€ e^z^βˆ’  +∫_(βˆ’Ο€) ^Ο€ ((e^z βˆ’e^z^βˆ’  )/(e^(xβˆ’sin(x)) +1))dx  2w=A+B  B∈{ia,a∈R}  Ξ©=Re(w)=(1/2)Re∫_(βˆ’Ο€) ^Ο€ e^z^βˆ’  dx,xβ†’βˆ’x,sin(βˆ’x)=βˆ’sin(x)β‡’  =(1/2)Re∫_(βˆ’Ο€) ^Ο€ e^z dx=(1/2)ReΞ£_(nβ‰₯0) ∫_(βˆ’Ο€) ^Ο€ (z^n /(n!))dx  z=e^(ix)   ⇔Ω=(1/2).ReΞ£_(nβ‰₯0) ∫_(βˆ’Ο€) ^Ο€ (e^(inx) /(n!))dx=(1/2)∫_(βˆ’Ο€) ^Ο€ dx_(=C) +Ξ£_(nβ‰₯1) [_(βˆ’Ο€) ^Ο€ (e^(inx) /(n!.n))]_(=D)   xβ†’e^(inx) is 2Ο€ periodicβ‡’D=0  Ξ©=(1/2).2Ο€+0=Ο€  ∫_(βˆ’Ο€) ^Ο€ ((e^(cos(x)+sin(x)) cos(sin(x))/(e^x +e^(sin(x)) ))dx=𝛑

$$\Omega={Re}\int_{βˆ’\pi} ^{\pi} \frac{{e}^{{cos}\left({x}\right)+{sin}\left({x}\right)\left(+{isin}\left({x}\right)\right.} }{{e}^{{x}} +{sin}\left({x}\right)}{dx}={Re}\left({W}\right) \\ $$$${W}=\int_{βˆ’\pi} ^{\pi} \frac{{e}^{{cos}\left({x}\right)+{isin}\left({x}\right)} }{{e}^{{x}βˆ’{sin}\left({x}\right)} +\mathrm{1}}{dx},{z}={cos}\left({x}\right)+{isin}\left({x}\right),\overset{βˆ’} {{z}},{conj}\left({z}\right) \\ $$$${by}\:{semetry}\:\:{x}\rightarrowβˆ’{x} \\ $$$$\mathrm{2}{w}=\int_{βˆ’\pi} ^{\pi} \frac{{e}^{\overset{βˆ’} {{z}}} }{{e}^{βˆ’{x}+{sin}\left({x}\right)} +\mathrm{1}}+\frac{{e}^{{z}} }{{e}^{{x}βˆ’{sin}\left({x}\right)} +\mathrm{1}}{dx} \\ $$$$=\int_{βˆ’\pi} ^{\pi} \frac{{e}^{\overset{βˆ’} {{z}}} \left({e}^{{x}βˆ’{sin}\left({x}\right)} \right)+{e}^{{z}} }{{e}^{{x}βˆ’{sin}\left({x}\right)} +\mathrm{1}}{dx} \\ $$$$=\int_{βˆ’\pi} ^{\pi} {e}^{\overset{βˆ’} {{z}}} +\int_{βˆ’\pi} ^{\pi} \frac{{e}^{{z}} βˆ’{e}^{\overset{βˆ’} {{z}}} }{{e}^{{x}βˆ’{sin}\left({x}\right)} +\mathrm{1}}{dx} \\ $$$$\mathrm{2}{w}={A}+{B} \\ $$$${B}\in\left\{{ia},{a}\in\mathbb{R}\right\} \\ $$$$\Omega={Re}\left({w}\right)=\frac{\mathrm{1}}{\mathrm{2}}{Re}\int_{βˆ’\pi} ^{\pi} {e}^{\overset{βˆ’} {{z}}} {dx},{x}\rightarrowβˆ’{x},{sin}\left(βˆ’{x}\right)=βˆ’{sin}\left({x}\right)\Rightarrow \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{Re}\int_{βˆ’\pi} ^{\pi} {e}^{{z}} {dx}=\frac{\mathrm{1}}{\mathrm{2}}{Re}\underset{{n}\geqslant\mathrm{0}} {\sum}\int_{βˆ’\pi} ^{\pi} \frac{{z}^{{n}} }{{n}!}{dx} \\ $$$${z}={e}^{{ix}} \\ $$$$\Leftrightarrow\Omega=\frac{\mathrm{1}}{\mathrm{2}}.{Re}\underset{{n}\geqslant\mathrm{0}} {\sum}\int_{βˆ’\pi} ^{\pi} \frac{{e}^{{inx}} }{{n}!}{dx}=\frac{\mathrm{1}}{\mathrm{2}}\int_{βˆ’\pi} ^{\pi} {d}\underset{={C}} {{x}}+\underset{{n}\geqslant\mathrm{1}} {\sum}\left[_{βˆ’\pi} ^{\pi} \frac{{e}^{{inx}} }{{n}!.{n}}\underset{={D}} {\right]} \\ $$$${x}\rightarrow{e}^{{inx}} {is}\:\mathrm{2}\pi\:{periodic}\Rightarrow{D}=\mathrm{0} \\ $$$$\Omega=\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{2}\pi+\mathrm{0}=\pi \\ $$$$\int_{βˆ’\pi} ^{\pi} \frac{{e}^{{cos}\left({x}\right)+{sin}\left({x}\right)} {cos}\left({sin}\left({x}\right)\right.}{{e}^{{x}} +{e}^{{sin}\left({x}\right)} }{dx}=\boldsymbol{\pi} \\ $$$$ \\ $$$$ \\ $$

Commented by mnjuly1970 last updated on 20/Mar/21

very nice solution  thanks alot...

$${very}\:{nice}\:{solution} \\ $$$${thanks}\:{alot}... \\ $$

Commented by mindispower last updated on 20/Mar/21

thanx always pleasur

$${thanx}\:{always}\:{pleasur} \\ $$

Answered by Ar Brandon last updated on 20/Mar/21

𝛗=∫_0 ^(Ο€/2) ((tan^(βˆ’1) ((√(tanx))))/(tanx))dx=_(u=tanx) ∫_0 ^∞ ((tan^(βˆ’1) ((√u)))/(u(1+u^2 )))du      =∫_0 ^∞ Ξ£_(n=0) ^∞ (((βˆ’1)^n u^n )/((n+1)(1+u^2 )))du=Ξ£_(n=0) ^∞ (((βˆ’1)^n )/(n+1))∫_0 ^∞ (u^n /((1+u^2 )))du       =_(t=u^2 ) (1/2)Ξ£_(n=0) ^∞ (((βˆ’1)^n )/(n+1))∫_0 ^∞ (t^((n/2)βˆ’(1/2)) /((1+t)))dt=(1/2)Ξ£_(n=0) ^∞ (((βˆ’1)^n )/(n+1))Ξ²((n/2)+(1/2) , (1/2)βˆ’(n/2))       =(1/2)Ξ£_(n=0) ^∞ (((βˆ’1)^n )/(n+1))βˆ™(Ο€/(cos(((Ο€n)/2))))  😷

$$\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{tan}^{βˆ’\mathrm{1}} \left(\sqrt{\mathrm{tanx}}\right)}{\mathrm{tanx}}\mathrm{dx}\underset{\mathrm{u}=\mathrm{tanx}} {=}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{tan}^{βˆ’\mathrm{1}} \left(\sqrt{\mathrm{u}}\right)}{\mathrm{u}\left(\mathrm{1}+\mathrm{u}^{\mathrm{2}} \right)}\mathrm{du} \\ $$$$\:\:\:\:=\int_{\mathrm{0}} ^{\infty} \underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(βˆ’\mathrm{1}\right)^{\mathrm{n}} \mathrm{u}^{\mathrm{n}} }{\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{1}+\mathrm{u}^{\mathrm{2}} \right)}\mathrm{du}=\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(βˆ’\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}+\mathrm{1}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{u}^{\mathrm{n}} }{\left(\mathrm{1}+\mathrm{u}^{\mathrm{2}} \right)}\mathrm{du} \\ $$$$\:\:\:\:\:\underset{\mathrm{t}=\mathrm{u}^{\mathrm{2}} } {=}\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(βˆ’\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}+\mathrm{1}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{t}^{\frac{\mathrm{n}}{\mathrm{2}}βˆ’\frac{\mathrm{1}}{\mathrm{2}}} }{\left(\mathrm{1}+\mathrm{t}\right)}\mathrm{dt}=\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(βˆ’\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}+\mathrm{1}}\beta\left(\frac{\mathrm{n}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\:,\:\frac{\mathrm{1}}{\mathrm{2}}βˆ’\frac{\mathrm{n}}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(βˆ’\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}+\mathrm{1}}\centerdot\frac{\pi}{\mathrm{cos}\left(\frac{\pi\mathrm{n}}{\mathrm{2}}\right)} \\ $$😷

Commented by mindispower last updated on 20/Mar/21

you cant[switch Σ and ∫

$${you}\:{cant}\left[{switch}\:\Sigma\:{and}\:\int\right. \\ $$

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