Question Number 45970 by maxmathsup by imad last updated on 19/Oct/18
$${find}\:\int\:\:\frac{{arcsin}\left(\mathrm{2}{x}\right)}{\sqrt{\mathrm{1}−\mathrm{4}{x}^{\mathrm{2}} }}{dx} \\ $$
Commented by maxmathsup by imad last updated on 20/Oct/18
$${changement}\:{arcsin}\left(\mathrm{2}{x}\right)={t}\:\:{give}\:\:\mathrm{2}{x}={sint}\:\Rightarrow\mathrm{2}{dx}={cost}\:{dt}\:\Rightarrow \\ $$$$\int\:\frac{{arcsin}\left(\mathrm{2}{x}\right)}{\sqrt{\mathrm{1}−\mathrm{4}{x}^{\mathrm{2}} }}{dx}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int\:\:\frac{{t}}{{cost}}\:{cost}\:{dt}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\int\:{t}\:{dt}\:=\frac{{t}^{\mathrm{2}} }{\mathrm{4}}\:+{c}\: \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left({arcsin}\left(\mathrm{2}{x}\right)\right)^{\mathrm{2}} \:+{c}\:. \\ $$
Answered by last updated on 19/Oct/18
![∵sin^(−1) 2x=t (1/(√(1−4x^2 )))dx=(1/2)dt =(1/2)∫tdt =(1/4)[sin^(−1) 2x]^2 +c](Q45978.png)
$$\because{sin}^{−\mathrm{1}} \mathrm{2}{x}={t} \\ $$$$\frac{\mathrm{1}}{\sqrt{\mathrm{1}−\mathrm{4}{x}^{\mathrm{2}} }}{dx}=\frac{\mathrm{1}}{\mathrm{2}}{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int{tdt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left[{sin}^{−\mathrm{1}} \mathrm{2}{x}\right]^{\mathrm{2}} +{c} \\ $$