∫e^x[(1-cosy)dx-(y-siny)dy],其中c為區(qū)域 0≤x≤π,0≤y≤sinx的邊界曲線取正向.求曲線積分
P(x,y)=e^x(1-cosy) -對(duì)y求偏導(dǎo)數(shù)=e^xsiny
Q(x,y)=e^x(siny-y) -->對(duì)x求偏導(dǎo)數(shù)=e^xsiny-ye^x
I=∫∫(e^xsiny-ye^x-e^xsiny)dxdy
=-∫∫(ye^x)dydy
=-[ ∫[0,π]e^xdx * ∫[0,sinx]ydy ]
=-(1/2)∫[0,π](sin^2x)e^xdx
因?yàn)閟in^2x =1/2(cos2x-1)
I=-(1/2)∫[0,π](1/2(cos2x-1))e^xdx
=1/4[ ∫[0,π]e^cos2x dx - ∫[0,π]e^x dx ]
又∫e^cos2x dx =（e^xcos2x+2e^xsin2x)/5
I=1/4(1-e^π)
正確答案是1/5(1-e^π).
請(qǐng)指出我哪里計(jì)算錯(cuò)了.給出正確步驟