∫√(1-x^2)dx 积分上限1 下限0 求定积分

令x=sina
则√(1-x²)=cosa
dx=cosada
x=1,a=π/2
x=0,a=0
原式=∫(0→π/2)cos²ada
=∫(0→π/2)(1+cos2a)/2da
=1/4∫(0→π/2)(1+cos2a)d2a
=1/4*(2a+sin2a)(0→π/2)
=1/4*(2*π/2+sinπ)-1/4*(2*0+sin0)
=π/4