看板 Math 關於我們 聯絡資訊
※ 引述《mk426375 (時雨)》之銘言: : ∫(e^2x) (cosx) dx : = e^2x*sinx - ∫2(e^2x)(sinx) dx : = e^2x*sinx - 2[e^2x(-cosx) - ∫2(e^2x)(-cosx) dx] : = e^2x*sinx + 2e^2x*cosx - 4∫e^2x(cosx) dx : = (1/5)*[e^2x*sinx+2e^2x*cosx] + C Let me do it using differentiation rather than integration (e^{2x} cos x)' = 2e^{2x} cos x - e^{2x} sin x ...........(1) (e^{2x} sin x)' = 2e^{2x} sin x + e^{2x} cos x ...........(2) (1)*2+(2) (2e^{2x} cos x + e^{2x} sin x)' = 5e^{2x} cos x -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 112.104.91.81