Question

高数定积分？

详细点

Answer 1

结果是0。注意积分上限是t，不是x。t的函数对x求导，如果t与x无关，导数为0.

Answer 2

∫(-π,π)[eˣcos²x/(1+eˣ)]dx
=∫(-π,π)½[eˣ(cos2x+1)/(1+eˣ)]dx
=∫(-π,π)½[eˣcos2x/(1+eˣ)]dx+∫(-π,π)½[eˣ/(1+eˣ)]dx
=∫(-π,π)½[cos2x·ln(1+eˣ)]+∫(-π,π)½[1/(1+eˣ)]d(1+eˣ)
=½{cos2x·ln(1+eˣ)|(-π,π)-∫(-π,π)[ln(1+eˣ)·dcos2x]}+½ln(1+eˣ)|(-π,π)
=ln[(1+e^π)/(1+e^-π)]+∫(-π,π)[ln(1+eˣ)·sin2x]dx
=π+∫(-π,π)[ln(1+eˣ)·sin2x]dx
∫(-π,π)[ln(1+eˣ)·sin2x]dx
=2∫(-π,π)[ln(1+eˣ)·sinx]dsinx
=2ln(1+eˣ)·sin²x|(-π,π)-2∫(-π,π)sinxd[ln(1+eˣ)·sinx]
=-2∫(-π,π)sinx[ln(1+eˣ)·cosx+eˣsin²x/(1+eˣ)]dx
=-∫(-π,π)sin2x·ln(1+eˣ)dx-2∫(-π,π)eˣsin²x/(1+eˣ)]dx
∴∫(-π,π)[ln(1+eˣ)·sin2x]dx=-∫(-π,π)eˣsin²x/(1+eˣ)]dx
∴I=∫(-π,π)[eˣ(1-sin²x/(1+eˣ)]dx=π-∫(-π,π)eˣsin²x/(1+eˣ)]dx
即：∫(-π,π)eˣsin²x/(1+eˣ)]dx=½∫(-π,π)[eˣ/(1+eˣ)]=½ln[(1+eˣ)](-π,π)=½π
∴I=π-½π=½π