Question

请问这一题怎么解，帮忙解答一下过程。

请问这一题怎么解，帮忙解答一下过程。请大家帮忙看一下这道题目的过程应该怎么解答，帮忙解答一下过程，谢谢大家。

Answer 1

前者定积分偶倍奇零
=2∫(0.π)xsin³xdx
=2∫x(cos²x-1)dcosx
=2/3∫xdcos³x-2xcosx+2∫cosxdx
=2xcos³x/3-2/3∫cos³xdx+2π+2sinx
=-2π/3+2π-2/3∫(1-sin²x)dsinx
=4π/3
后者换元u=-x整理
=(∫(-π.π)cos²x/(1+e^-x)dx+∫(π.-π)cos²(-u)/(1+e^u)d(-u))/2
=1/2∫(-π.π)cos²xdx
=∫(0.π)cos²xdx
=∫(-π/2.π/2)cos²xdx
=2∫(0.π/2)cos²xdx
=∫cos2x+1dx
=x+sin2x/2
=π/2
∴=4π/3+π/2=11π/6