∫e^xcosxdx=∫cosxd(e^x)
= cosx.e^x + ∫sinx .(e^x) dx
= cosx.e^x + ∫sinx .d(e^x)
= cosx .e^x + sinx.e^x - ∫ cosx(e^x) dx
2∫e^xcosxdx =cosx .e^x + sinx.e^x
∫e^xcosxdx =(1/2)[cosx .e^x + sinx.e^x] + C谢谢！那么极限lim(x—>0) (∫上限x 下限0e的t次方costdt)除以(sinx)怎么算啊？lim(x->0)∫(0->x)e^tcostdt / (sinx) (0/0)=lim(x->0)e^xcosx / cosx=lim(x->0)e^x=1