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Question Number 108401 by bobhans last updated on 16/Aug/20

$$\frac{\mathit{b}\mathit{o}\mathit{b}\mathit{h}\mathit{a}\mathit{n}\mathit{s}}{\ddots \text{iddots}}$$$$I=\underset{0}{\stackrel{\pi /2}{\int }}\ln (a^2{\cos }^2\theta +b^2{\sin }^2\theta )d\theta ?$$

Commented by john santu last updated on 17/Aug/20

$$\frac{⊸JS⊸}{⊛}$$$$I=\pi \ln \left(\frac{a+b}{2}\right)$$

Commented by bobhans last updated on 17/Aug/20

$$jooss..thankyouall$$

Answered by Dwaipayan Shikari last updated on 16/Aug/20

$$I={\int }_0^{\frac{\pi }{2}}log({sin}^2\theta +\frac{a^2}{b^2}{cos}^2\theta )+2logb=I\left(k\right)+\pi logb$$$$I\left(k\right)={\int }_0^{\frac{\pi }{2}}log({sin}^2\theta +k^2{cos}^2\theta )(k=\frac{a}{b})$$$$I^{\prime }\left(k\right)={\int }_0^{\frac{\pi }{2}}\frac{2{kcos}^2\theta }{{sin}^2\theta +k^2{cos}^2\theta }d\theta$$$$I^{{}^{\prime }}\left(k\right)={\int }_0^{\frac{\pi }{2}}\frac{2k}{{tan}^2\theta +k^2}d\theta$$$$I^{{}^{\prime }}\left(k\right)={\int }_0^{\frac{\pi }{2}}\frac{2k}{({tan}^2\theta +k^2)({tan}^2\theta +1)}dttan\theta =t{sec}^2\theta =\frac{dt}{d\theta }$$$$I^{{}^{\prime }}\left(k\right)=2k{\int }_0^{\mathrm{\infty }}\frac{1}{t^2+k^2}-\frac{1}{t^2+1}dt$$$$I^{\prime }\left(k\right)=\frac{2k}{k^2-1}{\int }_0^{\mathrm{\infty }}\frac{1}{t^2+1}-\frac{1}{t^2+k^2}dt$$$$I^{\prime }\left(k\right)=\frac{2k}{k^2-1}{[{tan}^{-1}t-\frac{1}{k}{tan}^{-1}\frac{t}{k}]}_0^{\mathrm{\infty }}$$$$I^{{}^{\prime }}\left(k\right)=\frac{2k}{k^2-1}.\frac{k-1}{k}.\frac{\pi }{2}=\frac{\pi }{k+1}$$$$I\left(k\right)=\int \frac{\pi }{k+1}=\pi log(k+1)+C={\int }_0^{\frac{\pi }{2}}log({sin}^2\theta +k^2{cos}^2\theta )$$$$Ifk=1,\pi log\left(2\right)+C=0,C=-\pi log\left(2\right)$$$$I\left(k\right)=\pi log(k+1)-\pi log\left(2\right)=\pi log(a+b)-\pi log\left(2\right)-\pi log\left(b\right)$$$$So$$$$AnswerisI\left(k\right)+\pi logb$$$$\pi log\left(\frac{a+b}{2}\right)$$$$$$$$$$

Commented by mnjuly1970 last updated on 16/Aug/20

$$I={\int }_0^{\frac{\pi }{2}}log({sin}^2\theta +\frac{a^2}{b^2}{cos}^2\theta )+2logb=I\left(k\right)+\pi logb$$$$I\left(k\right)={\int }_0^{\frac{\pi }{2}}log({sin}^2\theta +k^2{cos}^2\theta )(k=\frac{a}{b})$$$$I^{\prime }\left(k\right)={\int }_0^{\frac{\pi }{2}}\frac{2{kcos}^2\theta }{{sin}^2\theta +k^2{cos}^2\theta }d\theta$$$$I^{{}^{\prime }}\left(k\right)={\int }_0^{\frac{\pi }{2}}\frac{2k}{{tan}^2\theta +k^2}d\theta$$$$I^{{}^{\prime }}\left(k\right)={\int }_0^{\frac{\pi }{2}}\frac{2k}{({tan}^2\theta +k^2)({tan}^2\theta +1)}dttan\theta =t{sec}^2\theta =\frac{dt}{d\theta }$$$$I^{{}^{\prime }}\left(k\right)=2k{\int }_0^{\mathrm{\infty }}\frac{1}{t^2+k^2}-\frac{1}{t^2+1}dt$$$$I^{\prime }\left(k\right)=\frac{2k}{k^2-1}{\int }_0^{\mathrm{\infty }}\frac{1}{t^2+1}-\frac{1}{t^2+k^2}dt$$$$I^{\prime }\left(k\right)=\frac{2k}{k^2-1}{[{tan}^{-1}t-\frac{1}{k}{tan}^{-1}\frac{t}{k}]}_0^{\mathrm{\infty }}$$$$I^{{}^{\prime }}\left(k\right)=\frac{2k}{k^2-1}.\frac{k-1}{k}.\frac{\pi }{2}=\frac{\pi }{k+1}$$$$I\left(k\right)=\int \frac{\pi }{k+1}=\pi log(k+1)+C={\int }_0^{\frac{\pi }{2}}log({sin}^2\theta +k^2{cos}^2\theta )$$$$Ifk=1,\pi log\left(2\right)+C=0,C=-\pi log\left(2\right)$$$$I\left(k\right)=\pi log(k+1)-\pi log\left(2\right)=\pi log(a+b)-\pi log\left(2\right)-\pi log\left(b\right)$$$$So$$$$AnswerisI\left(k\right)+\pi logb$$$$\mathrm{♣}\pi log\left(\frac{a+b}{2}\right)...\Rightarrow {\int }_0^{\frac{\pi }{2}}\left\{log(a^2{sin}^2\left(x\right)+b^2{cos}^2\left(x\right))\right\}dx=\pi log\left(\frac{a+b}{2}\right)\mathrm{♣}$$$$$$$$\mathrm{peace}\mathrm{upon}\mathrm{you}.\mathrm{thank}\mathrm{you}$$$$\mathrm{very}\mathrm{excellent}.$$

Answered by mathmax by abdo last updated on 16/Aug/20

$$\mathrm{I}={\int }_0^{\frac{\pi }{2}}\ln ({\mathrm{a}}^2\frac{1+\cos \left(2\theta \right)}{2}+{\mathrm{b}}^2\frac{1-\cos \left(2\theta \right)}{2})\mathrm{d}\theta$$$$={\int }_0^{\frac{\pi }{2}}\ln (\frac{{\mathrm{a}}^2+{\mathrm{b}}^2}{2}+\frac{1}{2}({\mathrm{a}}^2-{\mathrm{b}}^2)\cos \left(2\theta \right))\mathrm{d}\theta$$$$={\int }_0^{\frac{\pi }{2}}\ln \{\left(\frac{{\mathrm{a}}^2+{\mathrm{b}}^2}{2}\right)(1+\frac{{\mathrm{a}}^2-{\mathrm{b}}^2}{{\mathrm{a}}^2+{\mathrm{b}}^2}\cos \left(2\theta \right)\}\mathrm{d}\theta$$$$=\frac{\pi }{2}\ln \left(\frac{{\mathrm{a}}^2+{\mathrm{b}}^2}{2}\right)+{\int }_0^{\frac{\pi }{2}}\ln \{1+\frac{{\mathrm{a}}^2-{\mathrm{b}}^2}{{\mathrm{a}}^2+{\mathrm{b}}^2}\cos \left(2\theta \right)\}\mathrm{d}\theta (2\theta =\mathrm{x})(\alpha =\frac{{\mathrm{a}}^2-{\mathrm{b}}^2}{{\mathrm{a}}^2+{\mathrm{b}}^2})$$$$=\frac{\pi }{2}\ln \left(\frac{{\mathrm{a}}^2+{\mathrm{b}}^2}{2}\right)+\frac{1}{2}{\int }_0^{\pi }\ln (1+\alpha \mathrm{cosx})\mathrm{dx}\mathrm{let}\mathrm{explicit}\mathrm{for}\mathrm{o}<\alpha <1$$$$\mathrm{f}\left(\alpha \right)={\int }_0^{\pi }\ln (1+\alpha \mathrm{cosx})\mathrm{dx}\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(\alpha \right)={\int }_0^{\pi }\frac{\mathrm{cosx}}{1+\alpha \mathrm{cosx}}\mathrm{dx}$$$$=\frac{1}{\alpha }{\int }_0^{\pi }\frac{1+\alpha \mathrm{cosx}-1}{1+\alpha \mathrm{cosx}}\mathrm{dx}=\frac{\pi }{\alpha }-\frac{1}{\alpha }{\int }_0^{\pi }\frac{\mathrm{dx}}{1+\alpha \mathrm{cosx}}$$$$\mathrm{and}{\int }_0^{\pi }\frac{\mathrm{dx}}{1+\alpha \mathrm{cosx}}{=}_{\tan \left(\frac{\mathrm{x}}{2}\right)=\mathrm{t}}{\int }_0^{\mathrm{\infty }}\frac{2\mathrm{dt}}{(1+{\mathrm{t}}^2)(1+\alpha \frac{1-{\mathrm{t}}^2}{1+{\mathrm{t}}^2})}$$$$={\int }_0^{\mathrm{\infty }}\frac{2\mathrm{dt}}{1+{\mathrm{t}}^2+\alpha -\alpha {\mathrm{t}}^2}=2{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dt}}{(1-\alpha ){\mathrm{t}}^2+1+\alpha }=\frac{2}{1-\alpha }{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dt}}{{\mathrm{t}}^2+\frac{1+\alpha }{1-\alpha }}$$$${=}_{\mathrm{t}=\sqrt{\frac{1+\alpha }{1-\alpha }}\mathrm{u}}\frac{2}{1-\alpha }.\frac{1-\alpha }{1+\alpha }{\int }_0^{\mathrm{\infty }}\frac{1}{1+{\mathrm{u}}^2}\times \frac{\sqrt{1+\alpha }}{\sqrt{1-\alpha }}$$$$=\frac{2}{\sqrt{1-{\alpha }^2}}\times \frac{\pi }{2}=\frac{\pi }{\sqrt{1-{\alpha }^2}}\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(\alpha \right)=\frac{\pi }{\alpha }-\frac{\pi }{\alpha \sqrt{1-{\alpha }^2}}\Rightarrow$$$$\mathrm{f}\left(\alpha \right)=\pi {\int }_1^{\alpha }\frac{\mathrm{dt}}{\mathrm{t}}-\pi {\int }_1^{\alpha }\frac{\mathrm{dt}}{\mathrm{t}\sqrt{1-{\mathrm{t}}^2}}+\mathrm{C}=\pi \ln \alpha -\pi {\int }_1^{\alpha }\frac{\mathrm{dt}}{\mathrm{t}\sqrt{1-{\mathrm{t}}^2}}+\mathrm{C}$$$$\mathrm{C}=\mathrm{f}\left(1\right)={\int }_0^{\pi }\ln (1+\mathrm{cosx})\mathrm{dx}\mathrm{we}\mathrm{have}$$$${\int }_1^{\alpha }\frac{\mathrm{dt}}{\mathrm{t}\sqrt{1-{\mathrm{t}}^2}}{=}_{\mathrm{t}=\mathrm{sinz}}{\int }_{\frac{\pi }{2}}^{\arcsin \alpha }\frac{\mathrm{cosz}\mathrm{dz}}{\mathrm{sinz}\mathrm{cosz}}={\int }_{\frac{\pi }{2}}^{\arcsin \alpha }\frac{\mathrm{dz}}{\mathrm{sinz}}$$$${=}_{\tan \left(\frac{\mathrm{z}}{2}\right)=\mathrm{m}}{\int }_1^{\tan \left(\frac{\arcsin \alpha }{2}\right)}\frac{2\mathrm{dm}}{(1+{\mathrm{m}}^2)\frac{2\mathrm{m}}{1+{\mathrm{m}}^2}}={\int }_1^{\tan \left(\frac{\arcsin \alpha }{2}\right)}\frac{\mathrm{dm}}{\mathrm{m}}$$$$=\ln \mid \tan \left(\frac{\arcsin \alpha }{2}\right)\mid \Rightarrow \mathrm{f}\left(\alpha \right)=\pi \ln \left(\alpha \right)-\pi \ln \mid \tan \left(\frac{\arcsin \alpha }{2}\right)\mid \Rightarrow$$$$\mathrm{I}=\frac{\pi }{2}\ln \left(\frac{{\mathrm{a}}^2+{\mathrm{b}}^2}{2}\right)+\frac{1}{2}\mathrm{f}\left(\frac{{\mathrm{a}}^2-{\mathrm{b}}^2}{{\mathrm{a}}^2+{\mathrm{b}}^2}\right)$$

Commented by mathmax by abdo last updated on 16/Aug/20

$$\mathrm{sorry}\mathrm{f}\left(\alpha \right)=\pi \ln \left(\alpha \right)-\pi \ln \mid \tan \left(\frac{\arcsin \alpha }{2}\right)\mid +{\int }_0^{\pi }\ln (1+\mathrm{cosx})\mathrm{dx}$$

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