Re: [微積] 98桃園縣教甄Q19
※ 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
討論串 (同標題文章)