∫ 1/(e^x+e^(2-x))dx
=∫e^x/(e^2x+e^2)dx
=∫ 1/(e^2x+e^2)de^x
不妨令t=e^x,则有
∫ 1/(e^x+e^2)de^x
=∫1/(t^2+e^2)dt
=1/e∫ 1/[(t/e)^2+1]d(t/e)
==1/e*arctan(t/e)+C
∫cos^2xsin^3xdx
= ∫ cos^2x * sin²x * sinx dx
= ∫ cos^2x * (1-cos²x) d(-cosx)
= ∫ (cos^4x - cos^2x) d(cosx)
= (1/5)cos^7x - (1/3)cos^5x + C
希望能帮到你, 望采纳. 祝学习进步
追问19题最后一刚为什么积出来是7次方和5次方?
追答打快了, 选错了数字键
本回答被提问者采纳