∫1(上标)-1(下标)√(1-x²)dx=?详细的过程.

∫1(上标)-1(下标)√(1-x²)dx=?详细的过程.
∫1(上标)-1(下标)√(1-x²)dx=?详细的过程.

∫1(上标)-1(下标)√(1-x²)dx=?详细的过程.
I=∫(-1,1)√(1-x²)dx
= 2∫(0,1)√(1-x²)dx
let x = sina
dx = cosa da
x= 0,a= 0
x =1 ,a= π/2
I =2∫(0,π/2) (cosa)^2 da
=2∫(0,π/2) ( cos2a+1)/2 da
= 2[sin(2a)/4 + a/2] (0,π/2)
= π/2