Question Number 6519 by benny last updated on 30/Jun/16
$$\:\underset{\:\mathrm{0}} {\overset{{a}} {\int}}\:\:\:\frac{\mathrm{1}}{{x}+\sqrt{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} }}\:{dx}\:= \\ $$
Commented by Tawakalitu. last updated on 30/Jun/16
$$\frac{\Pi}{\mathrm{4}} \\ $$
Commented by Tawakalitu. last updated on 01/Jul/16
$${x}\:=\:{a}\:{sint}\: \\ $$$${dx}\:=\:{a}\:{cost}\:{dt}\: \\ $$$$\therefore\:\int\frac{{dx}}{{x}\:+\:\sqrt{{a}^{\mathrm{2}} \:−\:{x}^{\mathrm{2}} }}\: \\ $$$$=\:\int\frac{{a}\:{cost}\:{dt}}{{a}\:{sint}\:+\:\sqrt{{a}^{\mathrm{2}} \:−\:{a}^{\mathrm{2}} \:{sin}^{\mathrm{2}} {t}}}\:\:=\:\int\frac{{a}\:{cost}\:{dt}}{{a}\:{sint}\:+\:\sqrt{{a}^{\mathrm{2}} \:−\:{a}^{\mathrm{2}} \:\left(\mathrm{1}\:−\:{cos}^{\mathrm{2}} {t}\right)}}\:=\:\int\frac{{a}\:{cost}\:{dt}}{{a}\:{sint}\:+\:\sqrt{{a}^{\mathrm{2}} \:−\:{a}^{\mathrm{2}} \:+\:{a}^{\mathrm{2}} {cos}^{\mathrm{2}} {t}}} \\ $$$$=\:\int\frac{{a}\:{cost}\:{dt}}{{a}\:{sint}\:+\:\sqrt{{a}^{\mathrm{2}} \:{cos}^{\mathrm{2}} {t}}} \\ $$$$=\:\int\frac{{a}\:{cost}\:{dt}}{{a}\:{sint}\:+\:{a}\:{cost}}\:\:=\:\:\int\frac{{a}\:{cost}\:{dt}}{{a}\left({sint}\:+\:{cost}\right)} \\ $$$$=\:\int\frac{{cost}\:{dt}}{{sint}\:+\:{cost}} \\ $$$$=\:\int\frac{\frac{\mathrm{1}}{\mathrm{2}}\:×\:\mathrm{2}\:{cost}\:{dt}}{{sint}\:+\:{cost}} \\ $$$$=\:\int\frac{\frac{\mathrm{1}}{\mathrm{2}}\:×\:\left({cost}\:+\:{cost}\right)\:{dt}}{{sint}\:+\:{cost}} \\ $$$$=\:\int\frac{\frac{\mathrm{1}}{\mathrm{2}}\:×\:\left({cost}\:+\:{sint}\:+\:{cost}\:−\:{sint}\right)\:{dt}}{{sint}\:+\:{cost}}\: \\ $$$$=\:\int\frac{\frac{\mathrm{1}}{\mathrm{2}}\left({cost}\:+\:{sint}\right){dt}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\left({cost}\:−\:{sint}\right){dt}}{{cost}\:+\:{sint}} \\ $$$$=\:\int\frac{\mathrm{1}}{\mathrm{2}}\:×\:\frac{{cost}\:+\:{sint}}{{cost}\:+\:{sint}}\:{dt}\:+\:\int\frac{\mathrm{1}}{\mathrm{2}}\:×\:\frac{{cost}\:−\:{sint}}{{cost}\:+\:{sint}}\:{dt} \\ $$$$=\:\int\frac{\mathrm{1}}{\mathrm{2}}\:{dt}\:+\:\int\frac{\mathrm{1}}{\mathrm{2}}\:×\:\frac{{cost}\:−\:{sint}}{{cost}\:+\:{sint}}\:{dt} \\ $$$${let}\:{u}\:=\:{cost}\:+\:{sint},\:\:\:\frac{{du}}{{dt}}\:=\:−\:{sint}\:+\:{cost}\:,\:{du}\:=\:\left({cost}\:−\:{sint}\right)\:{dt} \\ $$$$=\:\int\frac{\mathrm{1}}{\mathrm{2}}\:{dt}\:+\:\int\frac{\mathrm{1}}{\mathrm{2}}\:×\:\frac{{du}}{{u}} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}{t}\:+\:{ln}\left({u}\right)\:+\:{C} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}{t}\:+\:{ln}\left({cost}\:+\:{sint}\right)\:+\:{C} \\ $$$${But},\:{x}\:=\:{a}\:{sint}\:\:\:….\:\:\:{x}\:=\:\mathrm{0},\:{a} \\ $$$$\mathrm{0}\:=\:{a}\:{sint}\:\Rightarrow\:{sint}\:=\:\mathrm{0}\:\Rightarrow\:{t}\:=\:{sin}^{−\mathrm{1}} \mathrm{0}\:\Rightarrow\:{t}\:=\:\mathrm{0} \\ $$$${Again}, \\ $$$${a}\:=\:{a}\:{sint}\:\Rightarrow\:{sint}\:=\:\mathrm{1}\:\Rightarrow\:{t}\:=\:{sin}^{−\mathrm{1}} \left(\mathrm{1}\right)\:\Rightarrow\:{t}\:=\:\mathrm{90}\:\Rightarrow\:{t}\:=\:\frac{\Pi}{\mathrm{2}} \\ $$$$\left[\frac{\mathrm{1}}{\mathrm{2}}{t}\:+\:{ln}\left({sint}\:+\:{cost}\right)\:+\:{C}\right]_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\Pi}{\mathrm{2}}\right)+{ln}\left({sin}\frac{\Pi}{\mathrm{2}}+{cos}\frac{\Pi}{\mathrm{2}}\right)+{C}\:−\:\left[\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{0}\right)+{ln}\left({sin}\mathrm{0}+{cos}\mathrm{0}\right)\:+\:{C}\right] \\ $$$$=\:\frac{\Pi}{\mathrm{4}}\:+\:{ln}\left(\mathrm{1}+\mathrm{0}\right)+{C}−\mathrm{0}−{ln}\left(\mathrm{0}+\mathrm{1}\right)−{C} \\ $$$$=\:\frac{\Pi}{\mathrm{4}}\:+\:{ln}\left(\mathrm{1}\right)\:+\:{C}\:−\:\mathrm{0}\:−\:{ln}\left(\mathrm{1}\right)\:−\:{C} \\ $$$$=\:\frac{\Pi}{\mathrm{4}}\: \\ $$$$ \\ $$$${DONE}\:! \\ $$$${Force}\:×\:{Distance}\:=\:{ISE} \\ $$$$ \\ $$
Commented by sandy_suhendra last updated on 01/Jul/16
$${It}'{s}\:{a}\:{great}\:{answer}.\: \\ $$$${But}\:{I}\:{think}\:{there}\:{is}\:{only}\:{a}\:{little}\:{mistake}\:{in}\:{typing} \\ $$$${sin}\:{t}\:=\:\mathrm{1}\:\Rightarrow\:{t}\:=\:\mathrm{90}\:\left({not}\:\mathrm{45}\right)\:=\:\frac{\pi}{\mathrm{2}} \\ $$
Commented by Tawakalitu. last updated on 01/Jul/16
$${corrected}.\:\mathrm{10}{Q} \\ $$