Question Number 196950 by Erico last updated on 05/Sep/23
$$\mathrm{Prove}\:\mathrm{that}\:\underset{\:\mathrm{0}} {\int}^{\:\frac{\pi}{\mathrm{2}}} \frac{\mathrm{ln}\left(\mathrm{1}+\alpha\mathrm{sin}{t}\right)}{\mathrm{sin}{t}}{dt}=\:\frac{\pi^{\mathrm{2}} }{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{arccos}\alpha\right)^{\mathrm{2}} \\ $$
Answered by Mathspace last updated on 06/Sep/23
$${f}\left({x}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{ln}\left(\mathrm{1}+{xsint}\right)}{{sint}}{dt} \\ $$$${f}^{'} \left({x}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{dt}}{\mathrm{1}+{xsint}}\:\left({tan}\left(\frac{{x}}{\mathrm{2}}\right)={y}\right) \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}{dy}}{\left(\mathrm{1}+{y}^{\mathrm{2}} \right)\left(\mathrm{1}+{x}\frac{\mathrm{2}{y}}{\mathrm{1}+{y}^{\mathrm{2}} }\right)} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}{dy}}{\mathrm{1}+{y}^{\mathrm{2}} +\mathrm{2}{xy}}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}{dy}}{{y}^{\mathrm{2}} +\mathrm{2}{xy}+{x}^{\mathrm{2}} +\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}{dy}}{\left({y}+{x}\right)^{\mathrm{2}} +\mathrm{1}−{x}^{\mathrm{2}} }{dy}\left(\:{y}+{x}=\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }{z}\right) \\ $$$$=\int_{\frac{{x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}} ^{\frac{\mathrm{1}+{x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}} \:\:\:\frac{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }{dz}}{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{z}^{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:\left({arctan}\left(\frac{\mathrm{1}+{x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\right)−{arctan}\left(\frac{{x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\right)\right\} \\ $$$${we}\:{have} \\ $$$${tan}\left(\:{arctana}−{arctanb}\right) \\ $$$$=\frac{{a}−{b}}{\mathrm{1}+{ab}}\:\Rightarrow \\ $$$${arctan}\left(\frac{\mathrm{1}+{x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\right)−{arctan}\left(\frac{{x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\right) \\ $$$$={arctan}\left(\frac{\mathrm{1}+{x}−{x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\right) \\ $$$$=\frac{\pi}{\mathrm{2}}−{arctan}\left(\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)\:\Rightarrow \\ $$$${f}^{'} \left({x}\right)=\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\left(\frac{\pi}{\mathrm{2}}−{arctan}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right) \\ $$$$=\frac{\pi}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}−\frac{\mathrm{2}{arctan}\left(\sqrt{\left.\mathrm{1}−{x}^{\mathrm{2}} \right)}\right.}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \\ $$$$\Rightarrow{f}\left({x}\right)=\pi{arctanx} \\ $$$$−\mathrm{2}\int_{.} ^{{x}} \frac{{arctan}\left(\sqrt{\mathrm{1}−{u}^{\mathrm{2}} }\right)}{\:\sqrt{\mathrm{1}−{u}^{\mathrm{2}} }}{du}\:+{c} \\ $$$${test}\:{to}\:{find} \\ $$$$\int_{.} ^{{x}} \frac{{arctan}\sqrt{\mathrm{1}−{u}^{\mathrm{2}} }}{\:\sqrt{\mathrm{1}−{u}^{\mathrm{2}} }}{du} \\ $$$$…{be}\:{continued}… \\ $$
Commented by Mathspace last updated on 07/Sep/23
$${sorry}\:{f}\left({x}\right)=\pi\:{arcsinx} \\ $$$$−\mathrm{2}\int^{{x}} \frac{{arctan}\sqrt{\mathrm{1}−{u}^{\mathrm{2}} }}{\:\sqrt{\mathrm{1}−{u}^{\mathrm{2}} }}{du}\:+{c} \\ $$