求定积分-1⼀2到1⼀2 根号(1-x^2)dx的值

求定积分-1/2到1/2 根号(1-x^2)dx的值
2024-11-08 02:44:56
有2个网友回答
网友(1):

令x=sinΘ
dx=cosΘdΘ
x=1/2,Θ=π/6
x=-1/2,Θ=-π/6
原式=∫(-π/6,π/6)cosΘ*cosΘdΘ
=∫(-π/6,π/6)(1+cos2Θ)/2*1/2d(2Θ)
=1/4*(sin2Θ+2Θ)|(-π/6,π/6)
=√3/4+π/6

网友(2):

Sqrt[3]/4 + \[Pi]/6