Question

(cosx)^3不定积分

∫(cosx)^3dx
=∫(cosx)^2dsinx
=∫[1-(sinx)^2]dsinx

=∫dsinx-∫(sinx)^2dsinx //为什么∫dsinx=sinx了?

=sinx-(1/3)(sinx)^3+C

Answer 1

∫ ƒ'(x) dx = ∫ [dƒ(x)/dx] dx = ∫ d[ƒ(x)] = ƒ(x) + C嘛
这里ƒ'(x) = cosx，ƒ(x) = sinx
理所当然∫ d(sinx) = ∫ (sinx)' dx = sinx + C，积分与微分抵消