∫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 = sinadx = cosa dax= 0,a= 0x =1 ,a= π/2I =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