Integral -- Beta function, Bessel function?

  • #1
LagrangeEuler
703
19
Integral
[tex]\int^{\pi}_0\sin^3xdx=\int^{\pi}_0\sin x \sin^2xdx=\int^{\pi}_0\sin x (1-\cos^2 x)dx=\frac{4 \pi}{3}[/tex]
Is it possible to write integral ##\int^{\pi}_0\sin^3xdx## in form of Beta function, or even Bessel function?
 

Answers and Replies

  • #2
pasmith
Homework Helper
2,427
1,033
Since [itex]B(\frac12,\frac12) = \pi[/itex] you have [tex]
\int_0^{\pi} \sin^3 x\,dx = \tfrac43B(\tfrac12,\tfrac12) = \tfrac43 \int_0^1 u^{-1/2}(1 - u)^{-1/2}\,du.[/tex]
 

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