selene84
幼苗
共回答了22个问题采纳率:90.9% 举报
∫e^xsin^2xdx = ∫e^x(1-cos(2x))/2dx = 1/2 e^x - 1/2 ∫e^x cos(2x)dx
而 ∫e^x cos(2x)dx = 1/2 ∫e^x d[sin(2x)]
= 1/2 [ e^x [sin(2x)] - ∫e^x [sin(2x)] dx ]
= 1/2 [ e^x [sin(2x)] + 1/2 [ ∫e^x d[cos(2x)] ]
= 1/2 [ e^x [sin(2x)] + 1/2 [ e^x [cos(2x)] - ∫e^x [cos(2x)]dx ]
= 1/2 e^x [sin(2x)] + 1/4[ e^x [cos(2x)] - 1/4∫e^x [cos(2x)]dx
合并两边的 ∫e^x cos(2x)dx ,得
5/4 ∫e^x cos(2x)dx = 1/2 e^x [sin(2x)] + 1/4[ e^x [cos(2x)]
即 ∫e^x cos(2x)dx = 2/5 e^x [sin(2x)] + 1/5[ e^x [cos(2x)]
代回第一行的式子,即求出结果.
1年前
9