Question Number 150325 by mnjuly1970 last updated on 11/Aug/21
$$\:\:\:\: \\ $$$$\:\:\:\:\:\:\mathrm{Solve}….. \\ $$$$\:\:\:\:\:\:\:\:\: \\ $$$$\:\boldsymbol{\phi}=\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} {cos}^{\:\mathrm{2}} \left({x}\:\right).\:{ln}\:\left({cot}\left(\:{x}\:\right)\right){dx}=?\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:…..{m}.{n}….. \\ $$
Answered by Ar Brandon last updated on 11/Aug/21
![π=β«_0 ^(Ο/2) cos^2 xβln(cotx)dx =β«_0 ^(Ο/2) cos^2 xβln(cosx)dxββ«_0 ^(Ο/2) cos^2 xβln(sinx)dx =(β/βΞ±)β£_(Ξ±=2) β«_0 ^(Ο/2) cos^Ξ± xdxβ(β/βΞ±)β£_(Ξ±=0) β«_0 ^(Ο/2) cos^2 xβsin^Ξ± xdx =(β/βΞ±)β£_(Ξ±=2) Ξ²(((Ξ±+1)/2), (1/2))β(β/βΞ±)β£_(Ξ±=0) Ξ²((3/2), ((Ξ±+1)/2)) =(β/βΞ±)β£_(Ξ±=2) ((Ξ(((Ξ±+1)/2))Ξ((1/2)))/(Ξ((Ξ±/2)+1)))β(β/βΞ±)β£_(Ξ±=0) ((Ξ((3/2))Ξ(((Ξ±+1)/2)))/(Ξ((Ξ±/2)+2))) =((βΟ)/2)ββ£_(Ξ±=2) ((Ξ(((Ξ±+1)/2))Ξ((Ξ±/2)+1)[Ο(((Ξ±+1)/2))βΟ((Ξ±/2)+1)])/(Ξ^2 ((Ξ±/2)+1))) β((βΟ)/4)β£_(Ξ±=0) ((Ξ(((Ξ±+1)/2))Ξ((Ξ±/2)+2)[Ο(((Ξ±+1)/2))βΟ((Ξ±/2)+2)])/(Ξ^2 ((Ξ±/2)+2))) =((βΟ)/2)β((Ξ((3/2))Ξ(2)[Ο((3/2))βΟ(2)])/(Ξ^2 (2)))β((βΟ)/4)β((Ξ((1/2))Ξ(2)[Ο((1/2))βΟ(2)])/(Ξ^2 (2))) =(Ο/4)(2βΞ³β2ln2β1+Ξ³)β(Ο/4)(βΞ³β2ln2β1+Ξ³) =(Ο/4)(1β2ln2+1+2ln2)=(Ο/2)](https://www.tinkutara.com/question/Q150336.png)
$$\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cos}^{\mathrm{2}} {x}\centerdot\mathrm{ln}\left(\mathrm{cot}{x}\right){dx} \\ $$$$\:\:\:\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cos}^{\mathrm{2}} {x}\centerdot\mathrm{ln}\left(\mathrm{cos}{x}\right){dx}β\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cos}^{\mathrm{2}} {x}\centerdot\mathrm{ln}\left(\mathrm{sin}{x}\right){dx} \\ $$$$\:\:\:\:=\frac{\partial}{\partial\alpha}\mid_{\alpha=\mathrm{2}} \int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cos}^{\alpha} {xdx}β\frac{\partial}{\partial\alpha}\mid_{\alpha=\mathrm{0}} \int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cos}^{\mathrm{2}} {x}\centerdot\mathrm{sin}^{\alpha} {xdx} \\ $$$$\:\:\:\:=\frac{\partial}{\partial\alpha}\mid_{\alpha=\mathrm{2}} \beta\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}},\:\frac{\mathrm{1}}{\mathrm{2}}\right)β\frac{\partial}{\partial\alpha}\mid_{\alpha=\mathrm{0}} \beta\left(\frac{\mathrm{3}}{\mathrm{2}},\:\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\:\:\:\:=\frac{\partial}{\partial\alpha}\mid_{\alpha=\mathrm{2}} \frac{\Gamma\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left(\frac{\alpha}{\mathrm{2}}+\mathrm{1}\right)}β\frac{\partial}{\partial\alpha}\mid_{\alpha=\mathrm{0}} \frac{\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\Gamma\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left(\frac{\alpha}{\mathrm{2}}+\mathrm{2}\right)} \\ $$$$\:\:\:\:\:=\frac{\sqrt{\pi}}{\mathrm{2}}\centerdot\mid_{\alpha=\mathrm{2}} \frac{\Gamma\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\alpha}{\mathrm{2}}+\mathrm{1}\right)\left[\psi\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)β\psi\left(\frac{\alpha}{\mathrm{2}}+\mathrm{1}\right)\right]}{\Gamma^{\mathrm{2}} \left(\frac{\alpha}{\mathrm{2}}+\mathrm{1}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:β\frac{\sqrt{\pi}}{\mathrm{4}}\mid_{\alpha=\mathrm{0}} \frac{\Gamma\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\alpha}{\mathrm{2}}+\mathrm{2}\right)\left[\psi\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)β\psi\left(\frac{\alpha}{\mathrm{2}}+\mathrm{2}\right)\right]}{\Gamma^{\mathrm{2}} \left(\frac{\alpha}{\mathrm{2}}+\mathrm{2}\right)} \\ $$$$\:\:\:\:\:=\frac{\sqrt{\pi}}{\mathrm{2}}\centerdot\frac{\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\Gamma\left(\mathrm{2}\right)\left[\psi\left(\frac{\mathrm{3}}{\mathrm{2}}\right)β\psi\left(\mathrm{2}\right)\right]}{\Gamma^{\mathrm{2}} \left(\mathrm{2}\right)}β\frac{\sqrt{\pi}}{\mathrm{4}}\centerdot\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\mathrm{2}\right)\left[\psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)β\psi\left(\mathrm{2}\right)\right]}{\Gamma^{\mathrm{2}} \left(\mathrm{2}\right)} \\ $$$$\:\:\:\:\:=\frac{\pi}{\mathrm{4}}\left(\mathrm{2}β\gammaβ\mathrm{2ln2}β\mathrm{1}+\gamma\right)β\frac{\pi}{\mathrm{4}}\left(β\gammaβ\mathrm{2ln2}β\mathrm{1}+\gamma\right) \\ $$$$\:\:\:\:\:=\frac{\pi}{\mathrm{4}}\left(\mathrm{1}β\mathrm{2ln2}+\mathrm{1}+\mathrm{2ln2}\right)=\frac{\pi}{\mathrm{2}} \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 11/Aug/21
$$\:\:\:{thanks}\:{a}\:{lot}\:{sir}\:{brandon}.. \\ $$
Answered by mnjuly1970 last updated on 11/Aug/21
$$\:\:\:\:…..{solution}….\:\:\:\boldsymbol{\phi}:=\:\int_{\mathrm{0}} ^{\:\infty} {cos}^{\:\mathrm{2}} \left({x}\right)\:{ln}\:\left({cot}\left({x}\right)\right){dx}=? \\ $$$$\:\:\:\:\boldsymbol{\phi}\::\overset{\left\{{tan}\left({x}\right)=\:{t}\:\right\}} {=}\:β\int_{\mathrm{0}} ^{\:\infty} \frac{\left(\mathrm{1}β{t}^{\:\mathrm{2}} \right){ln}\:\left({t}\:\right)}{\left(\mathrm{1}+{t}^{\:\mathrm{2}} \right)^{\mathrm{2}} }\:{dt} \\ $$$$\:\:\:\:\::=\:β\mathrm{2}\int_{\mathrm{0}} ^{\:\infty} \frac{{l}^{\:} {n}\left({t}\:\right)}{\left(\mathrm{1}+{t}^{\:\mathrm{2}} \right)^{\:\mathrm{2}} }\:\:{dt}\:=\:β\mathrm{2}\left(\frac{β\pi}{\mathrm{4}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:……..\:\boldsymbol{\phi}\::=\:\frac{\pi}{\mathrm{2}}\:……. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{note}\:\left(\mathrm{1}\right)\::\:\:\int_{\mathrm{0}} ^{\:\infty} \frac{{ln}\:\left({t}\:\right)}{\mathrm{1}+\:{t}^{\:\mathrm{2}} }\:{dt}\:=\mathrm{0}\:\:\left(\:{easy}\:\right) \\ $$$$\:\:\:\:\:\:\:\:\:{note}\:\left(\mathrm{2}\right)\::\:\int_{\mathrm{0}} ^{\:\infty} \frac{\:{ln}\left({t}\:\right)}{\left(\mathrm{1}+{t}^{\:\mathrm{2}} \right)^{\:\mathrm{2}} }\:{dt}\:=β\frac{\pi}{\mathrm{4}}\:\left({derived}\:{earlier}\:\right)……\blacksquare \\ $$$$ \\ $$