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Question Number 113630 by mathmax by abdo last updated on 14/Sep/20

$$\mathrm{explicit}\mathrm{g}\left(\mathrm{a}\right)={\int }_0^{\frac{\pi }{4}}\ln (1+{\mathrm{acos}}^2\theta )\mathrm{d}\theta$$

Answered by Dwaipayan Shikari last updated on 15/Sep/20

$$I\left(a\right)={\int }_0^{\frac{\pi }{4}}log(1+{acos}^2\theta )d\theta$$$$I^{\prime }\left(a\right)={\int }_0^{\frac{\pi }{4}}\frac{{cos}^2\theta }{1+{acos}^2\theta }$$$$I^{\prime }\left(a\right)=\frac{1}{a}{\int }_0^{\frac{\pi }{4}}1-\frac{1}{1+{acos}^2\theta }$$$$I^{{}^{\prime }}\left(a\right)=\frac{\pi }{4a}-{\int }^{\frac{\pi }{4}}\frac{{sec}^2\theta }{{sec}^2\theta +a}d\theta$$$$I^{\prime }\left(a\right)=\frac{\pi }{4a}-{\int }_0^{\frac{\pi }{4}}\frac{{sec}^2\theta }{{tan}^2\theta +1+a}$$$$I^{\prime }\left(a\right)=\frac{\pi }{4a}-{\int }_0^1\frac{dt}{t^2+{\left(\sqrt{1+a}\right)}^2}$$$$I^{\prime }\left(a\right)=\frac{\pi }{4a}-{\left[\frac{1}{\sqrt{1+a}}{tan}^{-1}\frac{t}{\sqrt{1+a}}\right]}_0^1$$$$I\left(a\right)=\int \frac{\pi }{4a}-\int \frac{1}{\sqrt{1+a}}{tan}^{-1}\frac{1}{\sqrt{1+a}}da$$$$I\left(a\right)=\frac{\pi }{4}log\left(a\right)-\int \frac{2u}{u}.{tan}^{-1}\frac{1}{u}du1+a=u^2,1=2u\frac{du}{da}$$$$I\left(a\right)=\frac{\pi }{4}log\left(a\right)-2\int {tan}^{-1}\frac{1}{u}du{tan}^{-1}\frac{1}{u}=\alpha$$$$I\left(a\right)=\frac{\pi }{4}log\left(a\right)-2{utan}^{-1}\frac{1}{u}-2\int \frac{1}{u\sqrt{1+\frac{1}{u^2}}}du$$$$I\left(a\right)=\frac{\pi }{4}log\left(a\right)-2\sqrt{1+a}{tan}^{-1}\frac{1}{\sqrt{1+a}}-2log(u+\sqrt{u^2+1})+C$$$$I\left(a\right)=\frac{\pi }{4}log\left(a\right)-2\sqrt{1+a}{tan}^{-1}\frac{1}{\sqrt{1+a}}-2log(\sqrt{1+a}+\sqrt{2+a})+C$$$$I(-1)=\frac{{\pi }^2}{4}i+C={\int }_0^{\frac{\pi }{4}}log(1-{cos}^{2}x)dx$$$${\int }_0^{\frac{\pi }{4}}log(1-{cos}^{2}x)dx={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}log\left(sinx\right)dx={\int }_0^{\frac{\pi }{2}}log\left(sinx\right)+{\int }_{-\frac{\pi }{2}}^{0}log\left(sinx\right)$$$$=-\frac{\pi }{2}log\left(2\right)+{\int }_{-\frac{\pi }{2}}^{0}log(-1)+log\left(cosx\right)=\frac{{\pi }^2}{2}i$$$$I(-1)=\frac{{\pi }^2}{4}i+C=\frac{{\pi }^{2}i}{2}\Rightarrow C=\frac{{\pi }^2}{4}i$$$$I\left(a\right)=\frac{\pi }{4}log\left(a\right)-2\sqrt{1+a}{tan}^{-1}\frac{1}{\sqrt{1+a}}-2log(\sqrt{1+a}+\sqrt{2+a})+\frac{{\pi }^2}{4}i$$

Commented by Tawa11 last updated on 06/Sep/21

$$\mathrm{great}\mathrm{sir}$$

Answered by mathmax by abdo last updated on 14/Sep/20

$$\mid \mathrm{a}\mid <1$$

Answered by Olaf last updated on 14/Sep/20

$$\frac{1}{1+a{\cos }^2\theta }=\underset{k=0}{\sum ^{\mathrm{\infty }}}{(-1)}^k{a}^k{\cos }^{2k}\theta$$$$-\frac{2a\cos \theta \sin \theta }{1+a{\cos }^2\theta }=-2a\sin \theta \cos \theta \underset{k=0}{\sum ^{\mathrm{\infty }}}{(-1)}^k{a}^k{\cos }^{2k}\theta$$$$-\frac{2a\cos \theta \sin \theta }{1+a{\cos }^2\theta }=-2a\sin \theta \underset{k=0}{\sum ^{\mathrm{\infty }}}{(-1)}^k{a}^k{\cos }^{2k+1}\theta$$$$\ln (1+a{\cos }^2\theta )=-2a\underset{k=0}{\sum ^{\mathrm{\infty }}}{(-1)}^k{a}^k\frac{{\cos }^{2k+2}\theta }{2k+2}$$$${\int }_0^{\frac{\pi }{4}}\ln (1+a{\cos }^2\theta )d\theta =-2a\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^k{a}^k}{2k+2}{\int }_0^{\frac{\pi }{4}}{\cos }^{2k+2}\theta d\theta$$$${\cos }^{2k}\theta ={\left(\frac{1+\cos 2\theta }{2}\right)}^k$$$${\cos }^{2k}\theta =\frac{1}{2^k}\underset{p=0}{\sum ^k}{\mathrm{C}}_k^p{\cos }^{p}2\theta$$$${\int }_0^{\frac{\pi }{4}}\ln (1+a{\cos }^2\theta )d\theta =-2a\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^k{a}^k}{2k+2}{\left[\frac{{\cos }^2\theta }{2^k}\underset{p=0}{\sum ^k}{\mathrm{C}}_k^p{\cos }^{p}2\theta \right]}_0^{\frac{\pi }{4}}$$$${\int }_0^{\frac{\pi }{4}}\ln (1+a{\cos }^2\theta )d\theta =2a\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^k{a}^k}{(2k+2)2^k}\underset{p=0}{\sum ^k}{\mathrm{C}}_k^p$$$${\int }_0^{\frac{\pi }{4}}\ln (1+a{\cos }^2\theta )d\theta =2a\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^k{a}^k}{(2k+2)}$$$$...$$

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