Question

积分号（2到-2）根号下4-x^2(sinx+1)dx

Answer 1

∫[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 ...