Question Number 48500 by peter frank last updated on 24/Nov/18
Commented by Abdo msup. last updated on 24/Nov/18
$${y}\left({x}\right)={e}^{{arctan}\left({x}\right)} \:\Rightarrow{y}^{'} \left({x}\right)=\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\:{y}\left({x}\right)\:\Rightarrow \\ $$$${y}^{''} \left({x}\right)\:=−\frac{\mathrm{2}{x}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{y}\left({x}\right)\:+\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }{y}^{'} \left({x}\right) \\ $$$$=\frac{−\mathrm{2}{x}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{y}\left({x}\right)\:+\frac{\mathrm{1}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{y}\left({x}\right)\:=\frac{−\mathrm{2}{x}+\mathrm{1}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{y}\left({x}\right)\:\Rightarrow \\ $$$$\left(\mathrm{1}+{x}^{\mathrm{2}} \right){y}^{''} \:−\left(\mathrm{2}{x}−\mathrm{1}\right){y}^{'} \\ $$$$=\frac{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(−\mathrm{2}{x}+\mathrm{1}\right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{y}\left({x}\right)−\left(\mathrm{2}{x}−\mathrm{1}\right)\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }{y}\left({x}\right) \\ $$$$=\frac{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(−\mathrm{2}{x}+\mathrm{1}\right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{y}\left({x}\right)\:+\frac{\left(\mathrm{1}−\mathrm{2}{x}\right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{y}\left({x}\right) \\ $$$$=\mathrm{0}\:\:\Rightarrow\left(\mathrm{1}+{x}^{\mathrm{2}} \right){y}^{''} −\left(\mathrm{2}{x}−\mathrm{1}\right){y}^{'} =\mathrm{0}\:{so}\:{there}\:{is}\:{a}\:{error}\:{at}?{the}\: \\ $$$${question}! \\ $$
Commented by Abdo msup. last updated on 24/Nov/18
$${let}\:{I}\:=\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:{arcosx}\:{dx}\:\:{by}\:{parts} \\ $$$${I}\:=\:\left[{x}\:{arvosx}\right]_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\:−\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\frac{−{x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx} \\ $$$$=\frac{\pi}{\mathrm{6}}\:+\:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\:\frac{{xdx}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:\:{chsngement}\:{x}={sin}\theta{give} \\ $$$$\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\:\frac{{xdx}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \:\:\:\frac{{sin}\theta}{{cos}\theta}\:{cos}\theta\:{d}\theta\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \:{sin}\theta{d}\theta \\ $$$$=\left[−{cos}\theta\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \:=\mathrm{1}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\Rightarrow\:{I}\:=\frac{\pi}{\mathrm{6}}\:+\mathrm{1}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$