积分号（2到-2）根号下4-x^2(sinx+1)dx
问题描述：
积分号（2到-2）根号下4-x^2(sinx+1)dx
答
原式=∫(-2,2) 根号(4-x^2)*sinx dx +∫(-2,2) 根号(4-x^2) dx
因为f(x)=根号(4-x^2)*sinx是奇函数 g(x)=根号(4-x^2)是偶函数
所以原式=2∫(0,2) 根号(4-x^2) dx
令x=2sint dx=2costdt
原式=4∫(0,π/2) 2(cost)^2 dt
=4∫(0,π/2) 1+cos2t dt
=(4t-2sin2t)|(0,π/2)
=2π
答
∫[2,-2]√(4-x^2)(sinx+1)dx=∫[2,-2]√(4-x^2)sinxdx+∫[2,-2]√(4-x^2)dx=0+x√(4-x^2)|[2,-2] +∫[2,-2]dx/√(4-x^2)=∫[2,-2]d(x/2)√(1-(x/2)^2)=arcsin(x/2)|[2,-2]=-π/2-π/2=-π ∫[2,-2]√(4-x^2)sinxdx ...