Question Number 173148 by JordanRoddy last updated on 07/Jul/22 | ||
$$ \\ $$$$ \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:^{\mathrm{n}} \sqrt{\mathrm{x}}\:\left(\mathrm{arcsin}\:\mathrm{x}\right)\:\mathrm{dx} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$ | ||
Answered by Mathspace last updated on 07/Jul/22 | ||
$${I}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \left(^{{n}} \sqrt{{x}}\right){arcsinx}\:{dx}\:{we}\:{do} \\ $$$${the}\:{changement}\:{arcsinx}={t}\:\Rightarrow \\ $$$${x}={sint}\:{and} \\ $$$${I}_{{n}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left({sint}\right)^{\frac{\mathrm{1}}{{n}}} {t}\:{cost}\:{dt} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {t}\left({cost}\:\left({sint}\right)^{\frac{\mathrm{1}}{{n}}} \right){dt} \\ $$$$=\left[\frac{{t}}{\mathrm{1}+\frac{\mathrm{1}}{{n}}}\left({sint}\right)^{\mathrm{1}+\frac{\mathrm{1}}{{n}}} \right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{n}}{{n}+\mathrm{1}}\left({sint}\right)^{\mathrm{1}+\frac{\mathrm{1}}{{n}}} {dt} \\ $$$$\frac{\pi{n}}{\mathrm{2}\left({n}+\mathrm{1}\right)}−\frac{{n}}{{n}+\mathrm{1}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left({sint}\right)^{\mathrm{1}+\frac{\mathrm{1}}{{n}}} {dt} \\ $$$${we}\:{have}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left({cost}\right)^{\mathrm{2}{p}−\mathrm{1}} .\left({sint}\right)^{\mathrm{2}{q}−\mathrm{1}} {dt} \\ $$$$={B}\left(\rho,{q}\right)=\frac{\Gamma\left(\rho\right).\Gamma\left({q}\right)}{\Gamma\left(\rho+{q}\right)}\:\Rightarrow \\ $$$$\mathrm{2}\rho−\mathrm{1}=\mathrm{0}\:{and}\:\mathrm{2}{q}−\mathrm{1}=\mathrm{1}+\frac{\mathrm{1}}{{n}} \\ $$$$\Rightarrow\rho=\frac{\mathrm{1}}{\mathrm{2}}\:{and}\:{q}=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{n}} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left({sint}\right)^{\mathrm{1}+\frac{\mathrm{1}}{{n}}} {dt}=\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{n}}\right)}{\mathrm{2}\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{n}}\right)} \\ $$$$=\frac{\sqrt{\pi}}{\mathrm{2}}×\frac{\Gamma\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{n}}\right)}{\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}{n}}\right)}\:\Rightarrow \\ $$$${I}_{{n}} =\frac{{n}\pi}{\mathrm{2}\left({n}+\mathrm{1}\right)}−\frac{{n}}{{n}+\mathrm{1}}.\frac{\sqrt{\pi}}{\mathrm{2}}×\frac{\Gamma\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{n}}\right)}{\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}{n}}\right)} \\ $$ | ||
Commented by Mathspace last updated on 07/Jul/22 | ||
$${sorry}..\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left({cost}\right)^{\mathrm{2}\rho−\mathrm{1}} \left({sint}\right)^{\mathrm{2}{q}−\mathrm{1}} {dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{B}\left({p},{q}\right)=\frac{\Gamma\left(\rho\right).\Gamma\left({q}\right)}{\mathrm{2}\Gamma\left(\rho+{q}\right)} \\ $$ | ||
Commented by Tawa11 last updated on 11/Jul/22 | ||
$$\mathrm{Great}\:\mathrm{sir} \\ $$ | ||