Question Number 49938 by turbo msup by abdo last updated on 12/Dec/18
$${find}\:\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left({cost}+{xsint}\right){dt} \\ $$
Commented by Abdo msup. last updated on 24/Dec/18
$${we}\:{have}\:{f}^{'} \left({x}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\frac{{sint}}{{cost}\:+{x}\:{sint}}{dt} \\ $$$$=_{{tan}\left(\frac{{t}}{\mathrm{2}}\right)={u}} \:\:\:\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\:\:\:\:\frac{\frac{\mathrm{2}{u}}{\mathrm{1}+{u}^{\mathrm{2}} }}{\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }\:+\frac{\mathrm{2}{xu}}{\mathrm{1}+{u}^{\mathrm{2}} }}\:\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$=\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\:\frac{\mathrm{4}{udu}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left(\mathrm{1}−{u}^{\mathrm{2}} \:+\mathrm{2}{xu}\right)} \\ $$$$=−\mathrm{4}\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\:\:\frac{{u}}{\left({u}^{\mathrm{2}} \:+\mathrm{1}\right)\left({u}^{\mathrm{2}} −\mathrm{2}{xu}\:−\mathrm{1}\right)}{du}\:{let}\:{decompose} \\ $$$${F}\left({u}\right)\:=\:\frac{{u}}{\left({u}^{\mathrm{2}} \:+\mathrm{1}\right)\left({u}^{\mathrm{2}} \:−\mathrm{2}{xu}\:−\mathrm{1}\right)} \\ $$$${let}\:{u}^{\mathrm{2}} −\mathrm{2}{xu}−\mathrm{1}\: \\ $$$$\Delta^{'} ={x}^{\mathrm{2}} +\mathrm{1}>\mathrm{0}\:\Rightarrow{u}_{\mathrm{1}} ={x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$${u}_{\mathrm{2}} ={x}−\sqrt{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$${F}\left({u}\right)=\frac{{au}+{b}}{{u}^{\mathrm{2}} \:+\mathrm{1}}\:+\:\frac{{c}}{{u}−{u}_{\mathrm{1}} }\:+\frac{{d}}{{u}−{u}_{\mathrm{2}} } \\ $$$${c}\:={lim}_{{u}\rightarrow{u}_{\mathrm{1}} } \left({u}−{u}_{\mathrm{1}} \right){F}\left({u}\right)\:=\frac{{u}_{\mathrm{1}} }{\left({u}_{\mathrm{1}} ^{\mathrm{2}} \:+\mathrm{1}\right)\left({u}_{\mathrm{1}} −{u}_{\mathrm{2}} \right)} \\ $$$$=\frac{{x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{\left(\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)^{\mathrm{2}} \:+\mathrm{1}\right)\mathrm{2}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} \\ $$$${d}\:={lim}_{{u}\rightarrow{u}_{\mathrm{2}} } \left({u}−{u}_{\mathrm{2}} \right){F}\left({u}\right)\:=\frac{{u}_{\mathrm{2}} }{\left({u}_{\mathrm{2}} ^{\mathrm{2}} \:+\mathrm{1}\right)\left(−\mathrm{2}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)} \\ $$$$=−\frac{{u}_{\mathrm{2}} }{\left(\left({x}−\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)^{\mathrm{2}} \:+\mathrm{1}\right)\mathrm{2}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:\:\:…{be}\:{continued}… \\ $$