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)/4a³[(1/3)cos³z - cosz] |(0→π/2)a³[((1/3)(- 1) - (- 1)) - ((1/3)(1) - 1)]好像不對(duì)。cosπ/2是0果然算錯(cuò),本來(lái)只想給算個(gè)不定積分就罷了...(a^3/2)[z - (1/2)sin2z](0->Pi/2) - a^3[(1/3)cos^3z - cosz](0->Pi/2)= (a^3/2)[Pi/2] - a^3[- ((1/3)(- 1) - (1))]= (1/12)(3Pi - 16)a^3= (a^3/2)[Pi/2] - a^3[- ((1/3)(- 1) - (1))]= (1/12)(3Pi - 16)a^3好像不對(duì)（1/12)(3Pi - 8)a^3{(a³/2)[z - (1/2)sin(2z)] - a³[(1/3)cos³z - cosz]} |(0→π/2)= πa³/4 - 2a³/3= (π/4 - 2/3)a³如果你還不信的話(huà)，自己動(dòng)手算好了，我的小學(xué)加減乘除知識(shí)太差了。。。果然高深知識(shí)學(xué)多了就會(huì)忘掉基本知識(shí)的。。。