Question Number 196788 by ERLY last updated on 31/Aug/23
$${calculer}\:\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$$<{erly}\:{rolvinst}> \\ $$$$ \\ $$
Answered by MrGHK last updated on 31/Aug/23
$$\:{let}\:{x}^{\mathrm{2}} ={u}\:{x}={u}^{\frac{\mathrm{1}}{\mathrm{2}}} \:{dx}=\frac{\mathrm{1}}{\mathrm{2}}{u}^{−\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} {u}^{−\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{1}−{u}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} {du}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} {u}^{\left(−\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{1}\right)−\mathrm{1}} \left(\mathrm{1}−{u}\right)^{\left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{1}\right)−\mathrm{1}} {du} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\beta\left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{3}}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)}{\Gamma\left(\mathrm{2}\right)}=\frac{\mathrm{1}}{\mathrm{2}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{\pi}\left(\frac{\sqrt{\pi}}{\mathrm{2}}\right)\right)=\frac{\pi}{\mathrm{4}} \\ $$
Answered by MrGHK last updated on 31/Aug/23
![let x=sin(u) u=arcsin(x) dx=cos(u)du ∫(√(1−sin^2 (u)))cos(u)du=∫(√(cos^2 (u)))cos(u)du=∫cos^2 (u)du=(u/2)+(1/4)sin(2u) ((arcsin(x))/2)+(1/4)sin(2arcsin(x))∣_0 ^1 =[((arcsin(1))/2)+(1/4)sin(2arcsin(1))]−[((arcsin(0))/2)+(1/4)(2arcsin(0))] (π/4)+0=(π/4)](https://www.tinkutara.com/question/Q196803.png)
$${let}\:{x}={sin}\left({u}\right)\:\:{u}={arcsin}\left({x}\right)\:{dx}={cos}\left({u}\right){du} \\ $$$$\int\sqrt{\mathrm{1}−{sin}^{\mathrm{2}} \left({u}\right)}{cos}\left({u}\right){du}=\int\sqrt{{cos}^{\mathrm{2}} \left({u}\right)}{cos}\left({u}\right){du}=\int{cos}^{\mathrm{2}} \left({u}\right){du}=\frac{{u}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}{sin}\left(\mathrm{2}{u}\right) \\ $$$$\frac{{arcsin}\left({x}\right)}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}{sin}\left(\mathrm{2}{arcsin}\left({x}\right)\right)\mid_{\mathrm{0}} ^{\mathrm{1}} =\left[\frac{{arcsin}\left(\mathrm{1}\right)}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}{sin}\left(\mathrm{2}{arcsin}\left(\mathrm{1}\right)\right)\right]−\left[\frac{{arcsin}\left(\mathrm{0}\right)}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{2}{arcsin}\left(\mathrm{0}\right)\right)\right] \\ $$$$\frac{\pi}{\mathrm{4}}+\mathrm{0}=\frac{\pi}{\mathrm{4}} \\ $$