求x^2*(根号下a^2-x^2)dx的定积分 上限a,下限0 (a>0) 答案为 十六分之a的四次派，求过程。谢谢。
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x = asinz，dx = acosz dz
∫(0→a) x²√[a² - x²] dx
= ∫(0→π/2) [a²sin²z][acosz][acosz] dz
= a⁴∫(0→π/2) sin²zcos²z dz
= a⁴∫(0→π/2) [(1/2)sin2z]² dz
= [a⁴/4]∫(0→π/2) sin²2z dz
= [a⁴/4]∫(0→π/2) [(1 - cos4z)/2] dz
= [a⁴/8][z - (1/4)sin4z] |(0→π/2)
= [a⁴/8][π/2 - (1/4)(0)]
= a⁴π/16
∫(0→a) x²√[a² - x²] dx
= ∫(0→π/2) [a²sin²z][acosz][acosz] dz
= a⁴∫(0→π/2) sin²zcos²z dz
= a⁴∫(0→π/2) [(1/2)sin2z]² dz
= [a⁴/4]∫(0→π/2) sin²2z dz
= [a⁴/4]∫(0→π/2) [(1 - cos4z)/2] dz
= [a⁴/8][z - (1/4)sin4z] |(0→π/2)
= [a⁴/8][π/2 - (1/4)(0)]
= a⁴π/16
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