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Question Number 150325 by mnjuly1970 last updated on 11/Aug/21

            Solve.....              𝛗= ∫_0 ^( (Ο€/2)) cos^( 2) (x ). ln (cot( x ))dx=?                .....m.n.....

$$\:\:\:\: \\ $$$$\:\:\:\:\:\:\mathrm{Solve}..... \\ $$$$\:\:\:\:\:\:\:\:\: \\ $$$$\:\boldsymbol{\phi}=\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} {cos}^{\:\mathrm{2}} \left({x}\:\right).\:{ln}\:\left({cot}\left(\:{x}\:\right)\right){dx}=?\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:.....{m}.{n}..... \\ $$

Answered by Ar Brandon last updated on 11/Aug/21

𝛗=∫_0 ^(Ο€/2) cos^2 xβˆ™ln(cotx)dx      =∫_0 ^(Ο€/2) cos^2 xβˆ™ln(cosx)dxβˆ’βˆ«_0 ^(Ο€/2) cos^2 xβˆ™ln(sinx)dx      =(βˆ‚/βˆ‚Ξ±)∣_(Ξ±=2) ∫_0 ^(Ο€/2) cos^Ξ± xdxβˆ’(βˆ‚/βˆ‚Ξ±)∣_(Ξ±=0) ∫_0 ^(Ο€/2) cos^2 xβˆ™sin^Ξ± xdx      =(βˆ‚/βˆ‚Ξ±)∣_(Ξ±=2) Ξ²(((Ξ±+1)/2), (1/2))βˆ’(βˆ‚/βˆ‚Ξ±)∣_(Ξ±=0) Ξ²((3/2), ((Ξ±+1)/2))      =(βˆ‚/βˆ‚Ξ±)∣_(Ξ±=2) ((Ξ“(((Ξ±+1)/2))Ξ“((1/2)))/(Ξ“((Ξ±/2)+1)))βˆ’(βˆ‚/βˆ‚Ξ±)∣_(Ξ±=0) ((Ξ“((3/2))Ξ“(((Ξ±+1)/2)))/(Ξ“((Ξ±/2)+2)))       =((βˆšΟ€)/2)βˆ™βˆ£_(Ξ±=2) ((Ξ“(((Ξ±+1)/2))Ξ“((Ξ±/2)+1)[ψ(((Ξ±+1)/2))βˆ’Οˆ((Ξ±/2)+1)])/(Ξ“^2 ((Ξ±/2)+1)))             βˆ’((βˆšΟ€)/4)∣_(Ξ±=0) ((Ξ“(((Ξ±+1)/2))Ξ“((Ξ±/2)+2)[ψ(((Ξ±+1)/2))βˆ’Οˆ((Ξ±/2)+2)])/(Ξ“^2 ((Ξ±/2)+2)))       =((βˆšΟ€)/2)βˆ™((Ξ“((3/2))Ξ“(2)[ψ((3/2))βˆ’Οˆ(2)])/(Ξ“^2 (2)))βˆ’((βˆšΟ€)/4)βˆ™((Ξ“((1/2))Ξ“(2)[ψ((1/2))βˆ’Οˆ(2)])/(Ξ“^2 (2)))       =(Ο€/4)(2βˆ’Ξ³βˆ’2ln2βˆ’1+Ξ³)βˆ’(Ο€/4)(βˆ’Ξ³βˆ’2ln2βˆ’1+Ξ³)       =(Ο€/4)(1βˆ’2ln2+1+2ln2)=(Ο€/2)

$$\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cos}^{\mathrm{2}} {x}\centerdot\mathrm{ln}\left(\mathrm{cot}{x}\right){dx} \\ $$$$\:\:\:\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cos}^{\mathrm{2}} {x}\centerdot\mathrm{ln}\left(\mathrm{cos}{x}\right){dx}βˆ’\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cos}^{\mathrm{2}} {x}\centerdot\mathrm{ln}\left(\mathrm{sin}{x}\right){dx} \\ $$$$\:\:\:\:=\frac{\partial}{\partial\alpha}\mid_{\alpha=\mathrm{2}} \int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cos}^{\alpha} {xdx}βˆ’\frac{\partial}{\partial\alpha}\mid_{\alpha=\mathrm{0}} \int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cos}^{\mathrm{2}} {x}\centerdot\mathrm{sin}^{\alpha} {xdx} \\ $$$$\:\:\:\:=\frac{\partial}{\partial\alpha}\mid_{\alpha=\mathrm{2}} \beta\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}},\:\frac{\mathrm{1}}{\mathrm{2}}\right)βˆ’\frac{\partial}{\partial\alpha}\mid_{\alpha=\mathrm{0}} \beta\left(\frac{\mathrm{3}}{\mathrm{2}},\:\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\:\:\:\:=\frac{\partial}{\partial\alpha}\mid_{\alpha=\mathrm{2}} \frac{\Gamma\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left(\frac{\alpha}{\mathrm{2}}+\mathrm{1}\right)}βˆ’\frac{\partial}{\partial\alpha}\mid_{\alpha=\mathrm{0}} \frac{\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\Gamma\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left(\frac{\alpha}{\mathrm{2}}+\mathrm{2}\right)} \\ $$$$\:\:\:\:\:=\frac{\sqrt{\pi}}{\mathrm{2}}\centerdot\mid_{\alpha=\mathrm{2}} \frac{\Gamma\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\alpha}{\mathrm{2}}+\mathrm{1}\right)\left[\psi\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)βˆ’\psi\left(\frac{\alpha}{\mathrm{2}}+\mathrm{1}\right)\right]}{\Gamma^{\mathrm{2}} \left(\frac{\alpha}{\mathrm{2}}+\mathrm{1}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:βˆ’\frac{\sqrt{\pi}}{\mathrm{4}}\mid_{\alpha=\mathrm{0}} \frac{\Gamma\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\alpha}{\mathrm{2}}+\mathrm{2}\right)\left[\psi\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)βˆ’\psi\left(\frac{\alpha}{\mathrm{2}}+\mathrm{2}\right)\right]}{\Gamma^{\mathrm{2}} \left(\frac{\alpha}{\mathrm{2}}+\mathrm{2}\right)} \\ $$$$\:\:\:\:\:=\frac{\sqrt{\pi}}{\mathrm{2}}\centerdot\frac{\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\Gamma\left(\mathrm{2}\right)\left[\psi\left(\frac{\mathrm{3}}{\mathrm{2}}\right)βˆ’\psi\left(\mathrm{2}\right)\right]}{\Gamma^{\mathrm{2}} \left(\mathrm{2}\right)}βˆ’\frac{\sqrt{\pi}}{\mathrm{4}}\centerdot\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\mathrm{2}\right)\left[\psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)βˆ’\psi\left(\mathrm{2}\right)\right]}{\Gamma^{\mathrm{2}} \left(\mathrm{2}\right)} \\ $$$$\:\:\:\:\:=\frac{\pi}{\mathrm{4}}\left(\mathrm{2}βˆ’\gammaβˆ’\mathrm{2ln2}βˆ’\mathrm{1}+\gamma\right)βˆ’\frac{\pi}{\mathrm{4}}\left(βˆ’\gammaβˆ’\mathrm{2ln2}βˆ’\mathrm{1}+\gamma\right) \\ $$$$\:\:\:\:\:=\frac{\pi}{\mathrm{4}}\left(\mathrm{1}βˆ’\mathrm{2ln2}+\mathrm{1}+\mathrm{2ln2}\right)=\frac{\pi}{\mathrm{2}} \\ $$$$ \\ $$

Commented by mnjuly1970 last updated on 11/Aug/21

   thanks a lot sir brandon..

$$\:\:\:{thanks}\:{a}\:{lot}\:{sir}\:{brandon}.. \\ $$

Answered by mnjuly1970 last updated on 11/Aug/21

    .....solution....   𝛗:= ∫_0 ^( ∞) cos^( 2) (x) ln (cot(x))dx=?      𝛗 :=^({tan(x)= t })  βˆ’βˆ«_0 ^( ∞) (((1βˆ’t^( 2) )ln (t ))/((1+t^( 2) )^2 )) dt       := βˆ’2∫_0 ^( ∞) ((l^  n(t ))/((1+t^( 2) )^( 2) ))  dt = βˆ’2(((βˆ’Ο€)/4))             ........ 𝛗 := (Ο€/2) .......              note (1) :  ∫_0 ^( ∞) ((ln (t ))/(1+ t^( 2) )) dt =0  ( easy )           note (2) : ∫_0 ^( ∞) (( ln(t ))/((1+t^( 2) )^( 2) )) dt =βˆ’(Ο€/4) (derived earlier )......β– 

$$\:\:\:\:.....{solution}....\:\:\:\boldsymbol{\phi}:=\:\int_{\mathrm{0}} ^{\:\infty} {cos}^{\:\mathrm{2}} \left({x}\right)\:{ln}\:\left({cot}\left({x}\right)\right){dx}=? \\ $$$$\:\:\:\:\boldsymbol{\phi}\::\overset{\left\{{tan}\left({x}\right)=\:{t}\:\right\}} {=}\:βˆ’\int_{\mathrm{0}} ^{\:\infty} \frac{\left(\mathrm{1}βˆ’{t}^{\:\mathrm{2}} \right){ln}\:\left({t}\:\right)}{\left(\mathrm{1}+{t}^{\:\mathrm{2}} \right)^{\mathrm{2}} }\:{dt} \\ $$$$\:\:\:\:\::=\:βˆ’\mathrm{2}\int_{\mathrm{0}} ^{\:\infty} \frac{{l}^{\:} {n}\left({t}\:\right)}{\left(\mathrm{1}+{t}^{\:\mathrm{2}} \right)^{\:\mathrm{2}} }\:\:{dt}\:=\:βˆ’\mathrm{2}\left(\frac{βˆ’\pi}{\mathrm{4}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:........\:\boldsymbol{\phi}\::=\:\frac{\pi}{\mathrm{2}}\:....... \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{note}\:\left(\mathrm{1}\right)\::\:\:\int_{\mathrm{0}} ^{\:\infty} \frac{{ln}\:\left({t}\:\right)}{\mathrm{1}+\:{t}^{\:\mathrm{2}} }\:{dt}\:=\mathrm{0}\:\:\left(\:{easy}\:\right) \\ $$$$\:\:\:\:\:\:\:\:\:{note}\:\left(\mathrm{2}\right)\::\:\int_{\mathrm{0}} ^{\:\infty} \frac{\:{ln}\left({t}\:\right)}{\left(\mathrm{1}+{t}^{\:\mathrm{2}} \right)^{\:\mathrm{2}} }\:{dt}\:=βˆ’\frac{\pi}{\mathrm{4}}\:\left({derived}\:{earlier}\:\right)......\blacksquare \\ $$$$ \\ $$

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