高等数学 求定积分 这样做对面？后面怎么做啊

Question

高等数学 求定积分
这样做对面？后面怎么做啊

Answer 1

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

希望能帮到你, 望采纳. 祝学习进步

Answer 2

前面是正确的,接下来可利用公式(sin2t)^2=(1-cos4t)/2继续完成本回答被提问者采纳

Answer 3

对 根据 cos4t=1-2sin^2(2t) 得 sin^2(2t)=(1-cos4t)/2 代入即可

Answer 4

原式=a^4/4∫(0,π/2) (1-cos4t)/2 dt
=a^4/8 ×（t-1/4sin4t）|(0,π/2)
=a^4/8 ×π/2
=(πa^4)/16

Answer 5

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