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Question Number 173148 by JordanRoddy last updated on 07/Jul/22

$$$$$$$$$${\int }_0^1{}^{\mathrm{n}}\sqrt{\mathrm{x}}\left(\arcsin \mathrm{x}\right)\mathrm{dx}$$$$$$$$$$$$$$$$$$$$$$

Answered by Mathspace last updated on 07/Jul/22

$$I_n={\int }_0^1\left({}^n\sqrt{x}\right)arcsinxdxwedo$$$$thechangementarcsinx=t\Rightarrow$$$$x=sintand$$$$I_n={\int }_0^{\frac{\pi }{2}}{\left(sint\right)}^{\frac{1}{n}}tcostdt$$$$={\int }_0^{\frac{\pi }{2}}t\left(cost{\left(sint\right)}^{\frac{1}{n}}\right)dt$$$$={\left[\frac{t}{1+\frac{1}{n}}{\left(sint\right)}^{1+\frac{1}{n}}\right]}_0^{\frac{\pi }{2}}$$$$-{\int }_0^{\frac{\pi }{2}}\frac{n}{n+1}{\left(sint\right)}^{1+\frac{1}{n}}dt$$$$\frac{\pi n}{2(n+1)}-\frac{n}{n+1}{\int }_0^{\frac{\pi }{2}}{\left(sint\right)}^{1+\frac{1}{n}}dt$$$$wehave{\int }_0^{\frac{\pi }{2}}{\left(cost\right)}^{2p-1}.{\left(sint\right)}^{2q-1}dt$$$$=B(\rho ,q)=\frac{\mathrm{\Gamma }\left(\rho \right).\mathrm{\Gamma }\left(q\right)}{\mathrm{\Gamma }(\rho +q)}\Rightarrow$$$$2\rho -1=0and2q-1=1+\frac{1}{n}$$$$\Rightarrow \rho =\frac{1}{2}andq=1+\frac{1}{2n}$$$${\int }_0^{\frac{\pi }{2}}{\left(sint\right)}^{1+\frac{1}{n}}dt=\frac{\mathrm{\Gamma }\left(\frac{1}{2}\right)\mathrm{\Gamma }(1+\frac{1}{2n})}{2\mathrm{\Gamma }(\frac{1}{2}+1+\frac{1}{2n})}$$$$=\frac{\sqrt{\pi }}{2}\times \frac{\mathrm{\Gamma }(1+\frac{1}{2n})}{\mathrm{\Gamma }(\frac{3}{2}+\frac{1}{2n})}\Rightarrow$$$$I_n=\frac{n\pi }{2(n+1)}-\frac{n}{n+1}.\frac{\sqrt{\pi }}{2}\times \frac{\mathrm{\Gamma }(1+\frac{1}{2n})}{\mathrm{\Gamma }(\frac{3}{2}+\frac{1}{2n})}$$

Commented by Mathspace last updated on 07/Jul/22

$$sorry..{\int }_0^{\frac{\pi }{2}}{\left(cost\right)}^{2\rho -1}{\left(sint\right)}^{2q-1}dt$$$$=\frac{1}{2}B(p,q)=\frac{\mathrm{\Gamma }\left(\rho \right).\mathrm{\Gamma }\left(q\right)}{2\mathrm{\Gamma }(\rho +q)}$$

Commented by Tawa11 last updated on 11/Jul/22

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

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