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Question Number 197060 by universe last updated on 07/Sep/23

$$\mathrm{Prove}\mathrm{that}$$$${\underset{0}{\int }}^{\frac{\pi }{2}}\frac{\ln (1+\alpha \sin t)}{\sin t}dt=\frac{{\pi }^2}{8}-\frac{1}{2}{\left(\arccos \alpha \right)}^2$$

Commented by universe last updated on 07/Sep/23

$$question196950$$

Answered by universe last updated on 07/Sep/23

$$$$$$\mathit{I}={\int }_0^{\mathit{\pi }/2}\frac{\mathbf{log}(1+\mathit{\alpha }\mathbf{sin}\mathit{x})\mathit{d}\mathit{x}}{\mathbf{sin}\mathit{x}}$$$$\frac{\mathit{d}\mathit{I}}{\mathit{d}\mathit{\alpha }}={\int }_0^{\mathit{\pi }/2}\frac{1}{1+\mathit{\alpha }\mathbf{sin}\mathit{x}}\mathit{d}\mathit{x}$$$$\frac{\mathit{d}\mathit{I}}{\mathit{d}\mathit{\alpha }}={\int }_0^{\mathit{\pi }/2}\frac{{\mathbf{sec}}^2\mathit{x}/2}{1+{\mathbf{tan}}^2\mathit{x}/2+2\mathit{\alpha }\mathbf{tan}\mathit{x}/2}\mathit{d}\mathit{x}$$$$\frac{\mathit{d}\mathit{I}}{\mathit{d}\mathit{\alpha }}={\int }_0^1\frac{2\mathit{d}\mathit{y}}{{\mathit{y}}^2+2\mathit{\alpha }\mathit{y}+1}={\underset{0}{\int }}^1\frac{2\mathit{d}\mathit{y}}{{(\mathit{y}+\mathit{\alpha })}^2+(1-{\mathit{\alpha }}^2)}$$$$\frac{\mathit{d}\mathit{I}}{\mathit{d}\mathit{\alpha }}=\frac{2}{\sqrt{1-{\mathit{\alpha }}^2}}{\mathbf{tan}}^{-1}\frac{\mathit{y}+\mathit{\alpha }}{\sqrt{1-{\mathit{\alpha }}^2}}{\mid }_0^1$$$$\frac{\mathit{d}\mathit{I}}{\mathit{d}\mathit{\alpha }}=\frac{2}{\sqrt{1-{\mathit{\alpha }}^2}}[{\mathbf{tan}}^{-1}\frac{1-\mathit{\alpha }}{\sqrt{1-{\mathit{\alpha }}^2}}-{\mathbf{tan}}^{-1}\frac{\mathit{\alpha }}{\sqrt{1-{\mathit{\alpha }}^2}}]$$$$\frac{\mathit{d}\mathit{I}}{\mathit{d}\mathit{\alpha }}=\frac{2}{\sqrt{1-{\mathit{\alpha }}^2}}{\mathbf{tan}}^{-1}\left(\frac{1/\sqrt{1-{\mathit{\alpha }}^2}}{1+(1+\mathit{\alpha })\mathit{\alpha }/1-{\mathit{\alpha }}^2}\right)$$$$\int \frac{\mathit{d}\mathit{I}}{\mathit{d}\mathit{\alpha }}=\int \frac{2}{\sqrt{1-{\mathit{\alpha }}^2}}{\mathbf{tan}}^{-1}\sqrt{\frac{1-\mathit{\alpha }}{1+\mathit{\alpha }}}\mathit{d}\mathit{\alpha }$$$$\mathit{l}\mathit{e}\mathit{t}\mathit{\alpha }=\mathbf{cos}\mathit{\beta }\Rightarrow \mathit{d}\mathit{\alpha }=-\mathbf{sin}\mathit{\beta }\mathit{d}\mathit{\beta }$$$$\mathit{I}=-2\int {\mathbf{tan}}^{-1}\left(\sqrt{\frac{1-\mathbf{cos}\mathit{\beta }}{1+\mathbf{cos}\mathit{\beta }}}\right)\mathit{d}\mathit{\beta }$$$$\mathit{I}=-2\int \frac{\mathit{\beta }}{2}\mathit{d}\mathit{\beta }$$$$\mathit{I}=-\frac{{\mathit{\beta }}^2}{2}+\mathit{c}$$$$\mathit{I}=-\frac{{\left({\mathbf{cos}}^{-1}\mathit{\alpha }\right)}^2}{2}+\mathit{c}$$$$\mathit{l}\mathit{e}\mathit{t}\mathit{\alpha }=0\mathit{t}\mathit{h}\mathit{e}\mathit{n}\mathit{I}=0$$$$\mathit{c}=\frac{{\mathit{\pi }}^2}{8}$$$$\mathit{I}=\frac{{\mathit{\pi }}^2}{8}-\frac{{\left({\mathbf{cos}}^{-1}\mathit{\alpha }\right)}^2}{2}$$

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