定积分(0到a) ∫x^2*(√[(a - x)⼀(a + x)] dx

令x = asinz，dx = acosz
x = 0 => z = 0
x = a => sinz = 1 => z = π/2
∫(0→a) x²√[(a - x)/(a + x)] dx
= ∫(0→a) x² * [√(a - x)√(a - x)]/[√(a + x)√(a - x)] dx
= ∫(0→a) x² * (a - x)/√(a² - x²) dx
= ∫(0→π/2) (asinz)²(a - asinz)/(acosz) * (acosz dz)
= ∫(0→π/2) (a³sin²z)(1 - sinz) dz
= a³∫(0→π/2) sin²z dz - a³∫(0→π/2) sin³z dz
= (a³/2)∫(0→π/2) (1 - cos2z) dz - a³∫(0→π/2) (cos²z - 1) d(cosz)
= (a³/2)[z - (1/2)sin2z] |(0→π/2) - a³[(1/3)cos³z - cosz] |(0→π/2)
= (a³/2)(π/2) - a³[((1/3)(- 1) - (- 1)) - ((1/3)(1) - 1)]
= a³(π - 4a)/4

利用换元法，令x=asiny，则0原式=(0到π) ∫(asiny)^2·acosy·d(asiny)
=(0到π)a^4·∫(siny)^2·(cosy)^2·dy
=(0到π)(a^4/4)·∫(sin2y)^2·dy
=(0到π)(a^4/8)·∫(1-cos4y)·dy
=(0到π)(a^4/32)·∫(1-cos4y)·d(4y)
=(a^4/32)·(4y-sin4y)|(0到π)
=a^4π/8