Soit-I-1-0-t-t-1-t-dt-Comment-calculer-I-

Question Number 197891 by Erico last updated on 02/Oct/23

Soit I = {\int_{0}}^{1} \sqrt{t \sqrt{t (1 - t)}} dt

Comment calculer I

Answered by Mathspace last updated on 03/Oct/23

I = \int_{0}^{1} t^{\frac{1}{2}} . t^{\frac{1}{4}} . {(1 - t)}^{\frac{1}{4}} d t

= \int_{0}^{1} t^{\frac{3}{4}} {(1 - t)}^{\frac{1}{4}} d t

= B (\frac{3}{4} + 1, \frac{1}{4} + 1)

= B (\frac{7}{4}, \frac{3}{4}) = \frac{Γ (\frac{7}{4}) . Γ (\frac{3}{4})}{Γ (\frac{5}{2})}

Γ (\frac{7}{4}) = Γ (\frac{3}{4} + 1) = \frac{3}{4} Γ (\frac{3}{4})

I = \frac{3}{4} Γ^{2} (\frac{3}{4}) . \frac{1}{Γ (\frac{5}{2})} = \frac{3 Γ^{2} (\frac{3}{4})}{4 Γ (\frac{5}{2})}

Answered by Frix last updated on 02/Oct/23

u = \frac{\sqrt[4]{1 - t}}{\sqrt[4]{t}} \Rightarrow

I = 4 \underset{0}{\int^{\infty}} \frac{u^{4}}{{(u^{4} + 1)}^{3}} d u = \frac{3 \sqrt{2}}{32} π

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