Question Number 45670 by arvinddayama01@gmail.com last updated on 15/Oct/18
$$\int{cos}^{−\mathrm{1}} \left({sinx}\right){dx}=? \\ $$
Commented by maxmathsup by imad last updated on 15/Oct/18
$${let}\:{I}\:=\int\:{arccos}\left({sinx}\right){dx}\:\:{changement}\:{arcos}\left({sinx}\right)={t}\:\Rightarrow{sinx}={cost} \\ $$$$\Rightarrow{x}={arcsin}\left({cost}\right)\:\Rightarrow{dx}=−{sint}\:\frac{\mathrm{1}}{\sqrt{\mathrm{1}−{cos}^{\mathrm{2}} {t}}}{dt}=−{dt}\:\Rightarrow \\ $$$${I}\:=\int\:{t}\:\left(−{dt}\right)\:=−\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\:\:+{c}\:=−\frac{\mathrm{1}}{\mathrm{2}}\left({arcos}\left({sinx}\right)\right)^{\mathrm{2}} \:+{c}\:. \\ $$
Answered by ARVIND DAYAMA last updated on 15/Oct/18
$$\because{cos}^{−\mathrm{1}} \left({sinx}\right)={t} \\ $$$$−\frac{\mathrm{1}}{\sqrt{\mathrm{1}−{sin}^{\mathrm{2}} {x}}}.{cosxdx}={dt} \\ $$$${dx}=−{dt} \\ $$$$−\int{tdt} \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}}{t}^{\mathrm{2}} +{C} \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}}\left\{{cos}^{−\mathrm{1}} \left({sinx}\right)\right\}^{\mathrm{2}} +{C} \\ $$$$ \\ $$