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Question Number 187453 by norboyev last updated on 17/Feb/23

$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\sqrt{\mathrm{sin}{x}}\mathrm{d}{x}=? \\$$

Answered by Frix last updated on 17/Feb/23

$$\int\sqrt{\mathrm{sin}\:{x}}\:{dx}=\:\left[\mathrm{Elliptic}\:\mathrm{Integral}\right] \\$$$$=−\mathrm{2}{E}\:\left(\frac{\pi}{\mathrm{4}}−\frac{{x}}{\mathrm{2}}\mid\mathrm{2}\right)\:+{C} \\$$$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\sqrt{\mathrm{sin}\:{x}}\:{dx}\approx.\mathrm{642977634658} \\$$

Answered by CElcedricjunior last updated on 21/Feb/23

$$\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\boldsymbol{{sinx}}}\boldsymbol{{dx}}=\boldsymbol{{k}} \\$$$$\boldsymbol{{posons}}\:\boldsymbol{{sinx}}=\boldsymbol{{t}}^{\mathrm{2}} =>\boldsymbol{{dx}}=\frac{\mathrm{2}\boldsymbol{{t}}}{\:\sqrt{\mathrm{1}−\boldsymbol{{t}}^{\mathrm{2}} }}\boldsymbol{{dt}} \\$$$$\boldsymbol{{qd}}:\begin{cases}{\boldsymbol{{x}}−>\mathrm{0}\:}\\{\boldsymbol{{x}}−>\mathrm{1}}\end{cases}=>\begin{cases}{\boldsymbol{{t}}−>\mathrm{0}}\\{\boldsymbol{{t}}−>\boldsymbol{{sin}}\left(\mathrm{1}\right)}\end{cases} \\$$$$\boldsymbol{{k}}=\int_{\mathrm{0}} ^{\boldsymbol{{sin}}\left(\mathrm{1}\right)} \frac{\mathrm{2}\boldsymbol{{t}}^{\mathrm{3}} }{\:\sqrt{\mathrm{1}−\boldsymbol{{t}}^{\mathrm{2}} }}\boldsymbol{{dt}} \\$$$$\begin{cases}{\boldsymbol{{u}}=\boldsymbol{{t}}^{\mathrm{2}} }\\{\boldsymbol{{v}}'=\frac{\boldsymbol{{t}}}{\:\sqrt{\mathrm{1}−\boldsymbol{{t}}^{\mathrm{2}} }}}\end{cases}=>\begin{cases}{\boldsymbol{{u}}'=\mathrm{2}\boldsymbol{{t}}}\\{\boldsymbol{{v}}=−\sqrt{\mathrm{1}−\boldsymbol{{t}}^{\mathrm{2}} }}\end{cases}\bigstar{Moivre} \\$$$$\boldsymbol{{k}}=−\mathrm{2}\boldsymbol{{si}}\overset{\mathrm{2}} {\boldsymbol{{n}}}\left(\mathrm{1}\right)\mid\boldsymbol{{cos}}\left(\mathrm{1}\right)\mid+\mathrm{4}\int_{\mathrm{0}} ^{\boldsymbol{{sin}}\left(\mathrm{1}\right)} \boldsymbol{{t}}\sqrt{\mathrm{1}−\boldsymbol{{t}}^{\mathrm{2}} }\boldsymbol{{dt}} \\$$$$\boldsymbol{{k}}=−\boldsymbol{{sin}}\left(\mathrm{1}\right)\boldsymbol{{sin}}\left(\mathrm{2}\right)+\mathrm{4}\left[−\frac{\mathrm{2}}{\mathrm{3}}\sqrt{\left(\mathrm{1}−\boldsymbol{{t}}^{\mathrm{2}} \right)}\right]_{\mathrm{0}} ^{\boldsymbol{{sin}}\left(\mathrm{1}\right)} \:\blacksquare\mathscr{C}{edric}\:{junior} \\$$$$\boldsymbol{{k}}=\boldsymbol{{sin}}\left(\mathrm{1}\right)\boldsymbol{{sin}}\left(\mathrm{2}\right)−\frac{\mathrm{8}}{\mathrm{3}}\boldsymbol{{co}}\overset{\frac{\mathrm{3}}{\mathrm{2}}} {\boldsymbol{{s}}}\left(\mathrm{1}\right)+\frac{\mathrm{8}}{\mathrm{3}} \\$$$$\\$$$$\\$$$$\\$$