求定积分

Question

x^2*(a^2-x^2)^1/2在0，-a的积分，-a是下限
我只要答案就可以了

Answer 1

是-a³/3

过程:
∫<-a,0> x²âˆš(a²-x²) dx
Let x=asinβ,dx=acosβdβ
When x=-a,β=-π/2
When x=0,β=0
=∫<-π/2,0> (asinβ)²âˆš(a²-a²sin²Î²) dβ
=a²âˆ«<-π /2,0> sin²Î²*√(a²cos²Î²) dβ
=a²âˆ«<-π/2,0> sin²Î²*acosβ dβ
=a³âˆ«<-π/2,0> sin²Î²cosβ dβ
=a³âˆ«<-π/2,0> sin²Î² d(sinβ)
=a³[sin³Î²/3]<-π/2,0>
=a³[(-1-0)/3]
=-a³/3

Answer 2

∫[(sinx)^7-(sinx)^9]^1/2dx =∫_[0,∏/2](sinx)^(7/2)cosxdx-∫_[∏/2,∏](sinx)^(7/2)cosxdx =∫_[0,∏/2](sinx)^(7/2)dsinx-∫_[∏/2,∏](sinx)^(7/2)dsinx =∫_[0,1]t^(7/2)dt-∫_[1,0]t^(7/2)dt =2∫_[0,1

Answer 3

Answer 4