Question Number 40661 by math khazana by abdo last updated on 25/Jul/18 | ||
$$\left.\mathrm{1}\right){find}\:\:{g}\left({x}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\mathrm{1}−{x}^{\mathrm{2}} {cos}^{\mathrm{2}} \theta\right){d}\theta\:\:{with}\:{x}\:{from}\:{R} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{value}\:{of}\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\mathrm{1}−\mathrm{2}\:{cos}^{\mathrm{2}} \theta\right){d}\theta\:{and} \\ $$$$\left.\mathrm{3}\right)\:{find}\:{the}\:{value}\:{of}\:\: \\ $$$${A}\left(\alpha\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\mathrm{1}−{cos}^{\mathrm{2}} \alpha\:{cos}^{\mathrm{2}} \theta\right){d}\theta\: \\ $$ | ||
Commented by math khazana by abdo last updated on 04/Aug/18 | ||
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{g}\left({x}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\mathrm{1}+{xcos}\theta\right){d}\theta \\ $$$$+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\mathrm{1}−{xcos}\theta\right){d}\theta\:={h}\left({x}\right)+{k}\left({x}\right) \\ $$$${h}\left({x}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\mathrm{1}+{xcos}\theta\right){d}\theta\:\Rightarrow \\ $$$${h}^{'} \left({x}\right)\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{cos}\theta}{\mathrm{1}+{xcos}\theta}\:{d}\theta \\ $$$$=\frac{\mathrm{1}}{{x}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{\mathrm{1}+{xcos}\theta−\mathrm{1}}{\mathrm{1}+{xcos}\theta}{d}\theta=\frac{\pi}{\mathrm{2}{x}}\:−\frac{\mathrm{1}}{{x}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{d}\theta}{\mathrm{1}+{xcos}\theta} \\ $$$${but}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{d}\theta}{\mathrm{1}+{xcos}\theta}\:=_{{tan}\left(\frac{\theta}{\mathrm{2}}\right)={t}} \:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{\mathrm{1}}{\mathrm{1}+{x}\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} \:+{x}−{xt}^{\mathrm{2}} }\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{x}\:+\left(\mathrm{1}−{x}\right){t}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{2}}{\left(\mathrm{1}+{x}\right)}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dt}}{\mathrm{1}+\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}{t}^{\mathrm{2}} }\:{if}\:\mid{x}\mid<\mathrm{1}\:{we}\:{do}\:{the}\: \\ $$$${changement}\:\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}{t}\:={u}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\frac{{d}\theta}{\mathrm{1}+{xcos}\theta}\:=\frac{\mathrm{2}}{\mathrm{1}+{x}}\:\int_{\mathrm{0}} ^{\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}} \:\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{2}} }\:\sqrt{\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}}{du} \\ $$$$=\:\frac{\mathrm{2}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:{arctan}\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}\:\Rightarrow \\ $$$${h}\left({x}\right)\:=\:\int\:\:\frac{\mathrm{2}{arctan}\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:{dx}\:+{c} \\ $$$${changement}\:{x}={cost}\:{give} \\ $$$${h}\left({x}\right)\:=\:\int\:\:\:\frac{\mathrm{2}{arctan}\left({tan}\left(\frac{{t}}{\mathrm{2}}\right)\right)}{{sint}}\:\left(−{sint}\right){dt}\:+{c} \\ $$$$=\:−\int\:{t}\:{dt}\:+{c}\:=−\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\:+{c}\:={c}−\frac{\mathrm{1}}{\mathrm{2}}\left({arccosx}\right)^{\mathrm{2}} \\ $$$${h}\left(\mathrm{0}\right)=\mathrm{0}\:={c}−\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow{c}=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\mathrm{1}+{xcos}\theta\right){d}\theta\:=\frac{\mathrm{1}}{\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{2}}\left({arcosx}\right)^{\mathrm{2}} \\ $$ | ||
Commented by math khazana by abdo last updated on 04/Aug/18 | ||
$$ \\ $$$${error}\:{from}\:{line}\:\mathrm{12}\:{we}\:{have} \\ $$$${h}^{'} \left({x}\right)=\frac{\pi}{\mathrm{2}{x}}\:−\frac{\mathrm{1}}{{x}}\:\:\frac{\mathrm{2}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:{arctan}\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}\:\Rightarrow \\ $$$${h}\left({x}\right)\:=\frac{\pi}{\mathrm{2}}{ln}\mid{x}\mid\:−\:\int\:\:\:\:\frac{\mathrm{2}{arctan}\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}}{{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:{dx}\:+{c}\:{but} \\ $$$${changement}\:{x}\:={cost}\:{give} \\ $$$$\int\:\:\:\:\:\frac{\mathrm{2}\:{arctan}\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}}{{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:{dx}=\:−\int\:\:\:\frac{\mathrm{2}{arctan}\left({tan}\left(\frac{{t}}{\mathrm{2}}\right)\right)}{{costsint}}{sintdt} \\ $$$$=−\int\:\:\:\:\:\:\frac{{t}}{{cost}}\:{dt}\:\:\:=_{{tan}\left(\frac{{t}}{\mathrm{2}}\right)={u}} −\int\:\:\:\:\:\:\frac{\mathrm{2}{arctanu}}{\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }}\:\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$=−\int\:\:\:\frac{{arctanu}}{\mathrm{1}−{u}^{\mathrm{2}} }\:{du}\:\:{any}\:{way}\:{we}\:{have} \\ $$$${h}\left({x}\right)\:=\frac{\pi}{\mathrm{2}}{ln}\mid{x}\mid\:+\int_{\mathrm{1}} ^{{x}} \:\:\frac{{t}}{{cost}}\:{dt}\:+{c} \\ $$$${c}={h}\left(\mathrm{1}\right)\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\mathrm{1}+{cos}\theta\right){d}\theta\:\Rightarrow \\ $$$${h}\left({x}\right)=\frac{\pi}{\mathrm{2}}{ln}\mid{x}\mid\:+\int_{\mathrm{1}} ^{{x}} \:\frac{{t}}{{cost}}\:{dt}\:\:+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\mathrm{1}+{cos}\theta\right){d}\theta \\ $$$${if}\:\mid{x}\mid<\mathrm{1} \\ $$$${if}\:\mid{x}\mid>\mathrm{1}\:{we}\:{follow}\:{the}\:{same}\:{way}\: \\ $$ | ||
Commented by math khazana by abdo last updated on 04/Aug/18 | ||
$${let}\:{find}\:{k}\left({x}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\mathrm{1}−{xcos}\theta\right){d}\theta \\ $$$${we}\:{have}\:{g}\left({x}\right)={h}\left({x}\right)+{k}\left({x}\right)\:\Rightarrow{k}\left({x}\right)={g}\left({x}\right)−{h}\left({x}\right)\Rightarrow \\ $$$${k}\left(−{x}\right)={g}\left(−{x}\right)−{h}\left(−{x}\right)\:={g}\left({x}\right)−{h}\left({x}\right)={k}\left({x}\right) \\ $$$${because}\:{g}\:{and}\:{h}\:{are}\:{even}\:{so} \\ $$$${k}\left({x}\right)={k}\left(−{x}\right)={h}\left({x}\right)\:\Rightarrow \\ $$$${g}\left({x}\right)\:=\mathrm{2}{h}\left({x}\right) \\ $$$$ \\ $$ | ||
Commented by math khazana by abdo last updated on 04/Aug/18 | ||
$${g}\left({x}\right)=\pi{ln}\mid{x}\mid\:+\mathrm{2}\:\int_{\mathrm{1}} ^{{x}} \:\:\frac{{t}}{{cost}}\:{dt}\:+\mathrm{2}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\mathrm{1}+{cos}\theta\right){d}\theta\:. \\ $$ | ||