对e^xcos^2xdx求定积分,上限派=180'下限0

Question

如题

Answer 1

∫(0到π) (e^xcos�0�5x) dx=∫(0到π) [e^x*1/2*(1+cos2x)] dx=1/2*∫(0到π) e^x dx+1/2*∫(0到π) e^x*cos2x dx=1/2*(e^π-1)+1/2*∫(0到π) e^x*cos2x dx 独立求∫(0到π) e^x*cos2x dx=∫(0到π) e^x d(1/2*sin2x)=e^x*1/2*sin2x(0到π)-1/2*∫(0到π) e^x*sin2x dx,分部积分=-1/2*∫(0到π) e^x d(-1/2*cos2x)=-1/2*-1/2*cos2x*e^x(0到π)-(-1/2)(-1/2)∫(0到π) e^x*cos2x dx,分部积分=1/4*(e^π-1)-1/4*∫(0到π) e^x*cos2x dx(1+1/4)∫(0到π) e^x*cos2x dx=1/4*(e^π-1)∫(0到π) e^x*cos2x dx=1/5*(e^π-1) ∴原式=1/2*(e^π-1)+1/2*[1/5*(e^π-1)]=3/5*(e^π-1)

Answer 2

分部积分。先化成二分之一e∧x d（sin2x）。再套公式两次。