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1-1-1-t-4-dt-




Question Number 208335 by Shrodinger last updated on 12/Jun/24
∫_(−1) ^1 (√(1−t^4 ))dt
$$\int_{−\mathrm{1}} ^{\mathrm{1}} \sqrt{\mathrm{1}−{t}^{\mathrm{4}} }{dt} \\ $$
Answered by Frix last updated on 12/Jun/24
∫_(−1) ^1 (√(1−t^4 ))dt=2∫_0 ^1 (√(1−t^4 ))dt =^(u=sin^(−1)  t^2 )   =∫_0 ^(π/2) (sin u)^(−(1/2)) (cos u)^2 du=  =(1/2)B ((1/4), (3/2)) =((Γ ((1/4)) Γ ((3/2)))/(2Γ ((7/4))))=(((√π) Γ ((5/4)))/(Γ ((7/4))))=  =(1/(3(√(2π))))(Γ ((1/4)))^2 ≈1.74803837
$$\underset{−\mathrm{1}} {\overset{\mathrm{1}} {\int}}\sqrt{\mathrm{1}−{t}^{\mathrm{4}} }{dt}=\mathrm{2}\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\sqrt{\mathrm{1}−{t}^{\mathrm{4}} }{dt}\:\overset{{u}=\mathrm{sin}^{−\mathrm{1}} \:{t}^{\mathrm{2}} } {=} \\ $$$$=\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\left(\mathrm{sin}\:{u}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{cos}\:{u}\right)^{\mathrm{2}} {du}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{B}\:\left(\frac{\mathrm{1}}{\mathrm{4}},\:\frac{\mathrm{3}}{\mathrm{2}}\right)\:=\frac{\Gamma\:\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\:\Gamma\:\left(\frac{\mathrm{3}}{\mathrm{2}}\right)}{\mathrm{2}\Gamma\:\left(\frac{\mathrm{7}}{\mathrm{4}}\right)}=\frac{\sqrt{\pi}\:\Gamma\:\left(\frac{\mathrm{5}}{\mathrm{4}}\right)}{\Gamma\:\left(\frac{\mathrm{7}}{\mathrm{4}}\right)}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}\sqrt{\mathrm{2}\pi}}\left(\Gamma\:\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\right)^{\mathrm{2}} \approx\mathrm{1}.\mathrm{74803837} \\ $$

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