∫1(上标)-1(下标)√(1-x²)dx=?

∫1(上标)-1(下标)√(1-x²)dx=?
∫1(上标)-1(下标)√(1-x²)dx=?

∫1(上标)-1(下标)√(1-x²)dx=?
设x=sin⁡t,因此（-π/2≤t≤π/2）
则原式变为∫_(-π/2)^(π/2)▒cos⁡t d_sin⁡t =∫_(-π/2)^(π/2)▒〖（cos⁡t ）^2 〗 d_t
=∫_(π/(-2))^(π/2)▒(1+cos⁡2t)/2 d_t
=(t/2+sin⁡2t/4)｜_((-π)/2)^(π/2)
=π/2