Question Number 162398 by qaz last updated on 29/Dec/21
$$\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{arctan}\:\left(\mathrm{t}+\mathrm{sin}\:\mathrm{x}\right)−\mathrm{arctan}\:\mathrm{t}\right)\mathrm{dt}}{\mathrm{arctan}\:\mathrm{x}}=? \\ $$
Answered by aleks041103 last updated on 29/Dec/21
$${both}\:{go}\:{to}\:\mathrm{0} \\ $$$${use}\:{l}'{hopital} \\ $$$${ans}=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\frac{{d}}{{dx}}\int_{\mathrm{0}} ^{\mathrm{1}} {arctan}\left({t}+{sinx}\right)\:{dt}}{\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }} \\ $$$$\frac{{d}}{{dx}}\int_{\mathrm{0}} ^{\mathrm{1}} {arctan}\left({t}+{sinx}\right)\:{dt}= \\ $$$$=\frac{{d}}{{dx}}\int_{{sinx}} ^{\mathrm{1}+{sinx}} {arctan}\:{t}\:{dt}= \\ $$$$={arctan}\left(\mathrm{1}+{sinx}\right).\frac{{d}}{{dx}}\left(\mathrm{1}+{sinx}\right)\:−\:{arctan}\left({sinx}\right)\frac{{d}}{{dx}}{sin}\left({x}\right)= \\ $$$$={cosx}\left({arctan}\left(\mathrm{1}+{sinx}\right)−{arctan}\left({sin}\left({x}\right)\right)\right) \\ $$$$\Rightarrow{ans}.=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{{cosx}\left({arctan}\left(\mathrm{1}+{sinx}\right)−{arctan}\left({sin}\left({x}\right)\right)\right)}{\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }}= \\ $$$$=\frac{\mathrm{1}\left({arctan}\left(\mathrm{1}\right)−{arctan}\left(\mathrm{0}\right)\right)}{\frac{\mathrm{1}}{\mathrm{1}+\mathrm{0}}}={arctan}\left(\mathrm{1}\right)=\frac{\pi}{\mathrm{4}} \\ $$$$\Rightarrow{ans}.=\frac{\pi}{\mathrm{4}} \\ $$