∫sin^3xdx =- 1/3 * cos3x +c [注：(- 1/3* cos3x)'= sin3x
∫sin^3xdx=∫(sinx)^2*sinxdx=-∫(1-(cosx)^2)dcosx=-∫dcosx+∫(cosx)^2dcosx=-cosx+(1/3)*(cosx)^3+C
sin^3(x)=sinx(1-cos^2(x))∫sin^3xdx=∫sinxdx-∫cos^2(x)sin(x)dx=-cosx+(1/3)cos^3(x)+C
∫sin^3xdx=∫(sinx)^2*sinxdx=-∫(1-(cosx)^2)dcosx=-∫dcosx+∫(cosx)^2dcosx=-cosx+(1/3)*(cosx)^3+C
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