Question

这一题怎么解

高一数学

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

Answer 2

只解了（1），供参考。有事，（2）自己解决吧。