就是求不定積分
∫sinx^5dx
=-∫sinx^4dcosx
=-∫[(1-cosx^2]^2dcosx
=-∫(1-2cosx^2+cosx^4)dcosx
=-∫1dcosx+∫2cosx^2dcosx-∫cosx^4dcosx
=-cosx+2/3cosx^3-1/5cox^5
驗(yàn)算結(jié)果
對-cosx+2/3cosx^3-1/5cox^5 求導(dǎo)
=(-cosx)'+(2/3cosx^3)'-(1/5cox^5)'
=sinx+2/3*3*cosx^2(cosx)'-1/5*5cosx^4(cosx)'
=sinx-2cosx^2sinx+cosx^4sinx
=sinx(1-2cosx^2+cosx^4)
=sinx(1-cosx^2)^2
=sinx(sinx^2)^2
=sinx^5