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Question Number 153860 by SANOGO last updated on 11/Sep/21

Answered by puissant last updated on 11/Sep/21

$$Q={\int }_0^{\pi }{sin}^{\sqrt{3}}xdx$$$$=2{\int }_0^{\frac{\pi }{2}}{sin}^{\sqrt{3}}xdx$$$$\to I=2{\int }_0^{\frac{\pi }{2}}{sin}^{\sqrt{3}}xdx$$$$posonsf\left(\alpha \right)={\int }_0^{\frac{\pi }{2}}{sin}^{\alpha -1}xdx$$$$\Rightarrow f\left(\alpha \right)=\frac{1}{2}\beta (\frac{\alpha }{2};\frac{1}{2})=\frac{\sqrt{\pi }\mathrm{\Gamma }\left(\frac{\alpha }{2}\right)}{2\mathrm{\Gamma }\left(\frac{\alpha +1}{2}\right)}$$$$pour\alpha =\sqrt{3}+1,ona:$$$$Q=2\times \frac{\sqrt{\pi }\mathrm{\Gamma }(\frac{\sqrt{3}}{2}+\frac{1}{2})}{2\mathrm{\Gamma }(\frac{\sqrt{3}}{2}+1)}..$$$$\therefore \because Q=\frac{\sqrt{\pi }\mathrm{\Gamma }(\frac{\sqrt{3}}{2}+\frac{1}{2})}{\mathrm{\Gamma }(\frac{\sqrt{3}}{2}+1)}....$$$$$$$$..........\mathcal{L}epuissant..........$$

Commented by SANOGO last updated on 11/Sep/21

$$toujourlecerveaudescerveaumerci$$

Commented by SANOGO last updated on 11/Sep/21

$$propremonsuperieur$$

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