∫（上限1 下线0）√（1-x的平方）dx怎么算?

令 x=sinθ,∵x∈[0,1],∴可令θ∈[0,π/2]
则 dx=d(sinθ)=cosθdθ
∫(0,1)(√（1-x²）)dx
=∫(0,π/2)cosθd(sinθ)
=∫(0,π/2)cos²θdθ
=∫(0,π/2)(1+cos2θ)/2 dθ
=1/2∫(0,π/2)dθ + 1/4∫(0,π)costdt
=1/2×(π/2-0) + 1/4(sinπ-sin0)
=π/4