Integration Questions

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 byMathspace 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 byTawa11 last updated on 11/Jul/22

$$\mathrm{Great}\:\mathrm{sir} \\$$