定义域为1-x^2≥0所以,x∈[-1,1]令x=sint（t∈[-π/2,π/2]）则原式y=(sint)^2+2sintcsot=sin^2 t+sin2t=[(1-cos2t)/2]+sin2t=sin2t-(1/2)cos2t+(1/2)=(√5/2)*sin(2t-φ)+(1/2)（其中φ=arcsin(√5/5)）已知t∈[-π...