Question

请问这个怎么算啊 高等数学三角函数定积分？

Answer 1

∫(1+cosx)^2(sinx)^3(1+2cosx)dx
= -∫(1+cosx)^2(sinx)^2(1+2cosx)dcosx
= -∫(1+cosx)^2[1-(cosx)^2](1+2cosx)dcosx 记 u = cosx
= -∫(1+2u+u^2)(1-u^2)(1+2u)du
= - ∫(1+4u+4u^2-2u^3-5u^4-2u^5)du
= -[u+2u^2+(4/3)u^3-(1/2)u^4-u^5-(1/3)u^6]
= -[cosx+2(cosx)^2+(4/3)(cosx)^3-(1/2)(cosx)^4-(cosx)^5-(1/3)(cosx)^6]
[0, π/2] 代入， 得 1+2+4/3-1/2-1-1/3 = 5/2, 64π·5/2 = 160π