Question

定积分(0到a) ∫x^2*(√[(a - x)/(a + x)] dx

Answer 1

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

追问

a³[(1/3)cos³z - cosz] |(0→π/2)

a³[((1/3)(- 1) - (- 1)) - ((1/3)(1) - 1)]好像不对。

cosπ/2是0

追答

果然算错,本来只想给算个不定积分就罢了...
(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

追问

= (a^3/2)[Pi/2] - a^3[- ((1/3)(- 1) - (1))]
= (1/12)(3Pi - 16)a^3好像不对

（1/12)(3Pi - 8)a^3

追答

{(a³/2)[z - (1/2)sin(2z)] - a³[(1/3)cos³z - cosz]} |(0→π/2)
= πa³/4 - 2a³/3
= (π/4 - 2/3)a³
如果你还不信的话，自己动手算好了，我的小学加减乘除知识太差了。。。
果然高深知识学多了就会忘掉基本知识的。。。

Answer 2

利用换元法，令x=asiny，则0<x<a时0<y<π
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