∫e^x[(1-cosy)dx-(y-siny)dy],其中c为区域 0≤x≤π,0≤y≤sinx的边界曲线取正向.求曲线积分P(x,y)=e^x(1-cosy) -对y求偏导数=e^xsinyQ(x,y)=e^x(siny-y) -->对x求偏导数=e^xsiny-ye^xI=∫∫(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因为sin^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)/5I=1/4(1-e^π)正确答案是1/5(1-e^π).请指出我哪里计算错了.给出正确步骤

∫e^x[(1-cosy)dx-(y-siny)dy],其中c为区域 0≤x≤π,0≤y≤sinx的边界曲线取正向.求曲线积分
P(x,y)=e^x(1-cosy) -对y求偏导数=e^xsiny
Q(x,y)=e^x(siny-y) -->对x求偏导数=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
因为sin^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)
正确答案是1/5(1-e^π).
请指出我哪里计算错了.给出正确步骤