∫1(上標)-1(下標)√(1-x²)dx=?詳細的過程.
∫1(上標)-1(下標)√(1-x²)dx=?詳細的過程.
數(shù)學(xué)人氣：300 ℃時間：2019-10-17 06:50:23
優(yōu)質(zhì)解答
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
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