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。
1个回答
展开全部
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
本回答被提问者和网友采纳
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询