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Question Number 205590 by Lindemann last updated on 25/Mar/24

J=∫_0 ^1 (√(1−x^4 ))dx

$${J}=\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}−{x}^{\mathrm{4}} }{dx} \\ $$

Commented by lepuissantcedricjunior last updated on 25/Mar/24

J=∫_0 ^1 (√(1−x^4 ))dx   posons x^4 =t=>x=(t)^(1/4)   ⇔dx=(dt/(4t^(3/4) ))=  qd: { ((x→0)),((x→1)) :}=> { ((t→0)),((t→1)) :}  J=(1/4)∫_0 ^1 t^(−(3/4)) (1−t)^(1/2) dt=(1/4)∫_0 ^1 t^((1/4)−1) (1−t)^((3/2)−1) dt  J=(1/4)(((𝚪((1/4))×𝚪((3/2)))/(𝚪((7/4)))))=(1/4)((((1/2)Γ((1/2))𝚪((1/4)))/((3/4)𝚪((3/4)))))  =((Γ((1/4))(√𝛑))/(6𝚪((3/4))))  J=((Γ((1/4))(√𝛑))/(6Γ((3/4))))  ..............le puissant Dr...................

$$\boldsymbol{{J}}=\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}−\boldsymbol{{x}}^{\mathrm{4}} }\boldsymbol{{dx}}\:\:\:\boldsymbol{{posons}}\:\boldsymbol{{x}}^{\mathrm{4}} =\boldsymbol{{t}}=>\boldsymbol{{x}}=\sqrt[{\mathrm{4}}]{\boldsymbol{{t}}} \\ $$$$\Leftrightarrow\boldsymbol{{dx}}=\frac{\boldsymbol{{dt}}}{\mathrm{4}\boldsymbol{{t}}^{\frac{\mathrm{3}}{\mathrm{4}}} }= \\ $$$$\boldsymbol{{qd}}:\begin{cases}{\boldsymbol{{x}}\rightarrow\mathrm{0}}\\{\boldsymbol{{x}}\rightarrow\mathrm{1}}\end{cases}=>\begin{cases}{\boldsymbol{{t}}\rightarrow\mathrm{0}}\\{\boldsymbol{{t}}\rightarrow\mathrm{1}}\end{cases} \\ $$$$\boldsymbol{{J}}=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{1}} \boldsymbol{{t}}^{−\frac{\mathrm{3}}{\mathrm{4}}} \left(\mathrm{1}−\boldsymbol{{t}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \boldsymbol{{dt}}=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{1}} \boldsymbol{{t}}^{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} \left(\mathrm{1}−\boldsymbol{{t}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}−\mathrm{1}} \boldsymbol{{dt}} \\ $$$$\boldsymbol{{J}}=\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\boldsymbol{\Gamma}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)×\boldsymbol{\Gamma}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)}{\boldsymbol{\Gamma}\left(\frac{\mathrm{7}}{\mathrm{4}}\right)}\right)=\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\frac{\mathrm{1}}{\mathrm{2}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\boldsymbol{\Gamma}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)}{\frac{\mathrm{3}}{\mathrm{4}}\boldsymbol{\Gamma}\left(\frac{\mathrm{3}}{\mathrm{4}}\right)}\right) \\ $$$$=\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\sqrt{\boldsymbol{\pi}}}{\mathrm{6}\boldsymbol{\Gamma}\left(\frac{\mathrm{3}}{\mathrm{4}}\right)} \\ $$$$\boldsymbol{{J}}=\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\sqrt{\boldsymbol{\pi}}}{\mathrm{6}\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)} \\ $$$$..............{le}\:{puissant}\:\boldsymbol{{D}}{r}................... \\ $$

Commented by mr W last updated on 26/Mar/24

please post your answer to a question  as “answer”, not as “comment”! thanks!

$${please}\:{post}\:{your}\:{answer}\:{to}\:{a}\:{question} \\ $$$${as}\:``{answer}'',\:{not}\:{as}\:``{comment}''!\:{thanks}! \\ $$

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