Question

sinx^3的不定积分

Answer 1

(sinx)^是个什么式子?
∫sinxdx=-cosx+c;
∫(sinx)^5dx=∫[(sinx)^4]*sinxdx=∫{[(sinx)^2]}^2d(-cosx)=∫[1-(-cosx)^2]^2d(-cosx)=∫[1-2(-cosx)^2+(-cosx)^4]d(-cosx)=-cosx+(2/3)(cosx)^x-(1/5)(cosx)^5+c
也可以用降幂公式先把sinx的幂降下来:
(sinx)^2=(1/2)(1-cos2x)
(sinx)^3=(1/4)(3sinx-sin3x)
(sinx)^4=(1/8)(3-4cos2x+cos4x)

Answer 2

∫ (sinx)^3 dx
=-∫ (sinx)^2 dcosx
=∫ [(cosx)^2-1] dcosx
=(1/3)(cosx)^3-cosx + C