Question

求(1—sinx三次方)的定积分,积分上限是π,积分下限是0

Answer 1

3)(cosx)^3
代入上下限;3
嗯，你的答案是正确的，得到定积分为π-4/，我写程序计算了一下∫1-(sinx)^3dx
=∫1+sinx-(sinx)^3-sinxdx
=∫1+sinx[1-(sinx)^2]-sinxdx
=∫1+sinx(cosx)^2-sinxdx
=∫1-sinxdx+∫sinx(cosx)^2dx
=∫1-sinxdx-∫(cosx)^2dcosx
=x+cosx-(1/

Answer 2

∫(0→π) (1 - sin³x) dx
= ∫(0→π) dx - ∫(0→π) sin³x dx
= [ x ] |(0→π) + ∫(0→π) (1 - cos²x) d(cosx)
= π + [ (cosx - 1/3 * cos³x) ] |(0→π)
= π + [(- 1) - (1/3)(- 1)] - [(1) - (1/3)(1)]
= π - 4/3