close
標題:
integration 4
發問:
How to integrate: e^x cos2x dx Please explain it step by step.
最佳解答:
Integration By Parts： ∫udv = uv - ∫vdu Method 1： ∫e^x cos 2x dx ... (1)=∫e^x d[(sin 2x)/2] Let u=e^x, v=(sin 2x)/2 ?du=e^x dx, dv=cos 2x dx ∫e^x d[(sin 2x)/2] =∫udv =uv - ∫vdu =e^x (sin 2x)/2 - ∫(sin 2x)/2 d(e^x) =e^x (sin 2x)/2 - ∫(sin 2x)/2 (e^x) dx =e^x (sin 2x)/2 - (1/2)∫e^x sin 2x dx =e^x (sin 2x)/2 - (1/2)∫e^x d[(-cos 2x)/2] Let u=e^x, v=(-cos 2x)/2 ?du=e^x dx, dv=sin 2x dx e^x (sin 2x)/2 - (1/2)∫e^x d[(-cos 2x)/2] =e^x (sin 2x)/2 - (1/2)∫udv =e^x (sin 2x)/2 - [uv/2 - (1/2)∫vdu] =e^x (sin 2x)/2 - uv/2 + (1/2)∫vdu =e^x (sin 2x)/2 - e^x (-cos 2x)/4 + (1/2)∫(-cos 2x)/2 d(e^x) =e^x (sin 2x)/2 + e^x (cos 2x)/4 - (1/4)∫(cos 2x) (e^x) dx =e^x (sin 2x)/2 + e^x (-cos 2x)/4 - (1/4)∫e^x cos 2x dx ... (2) Since (1)=(2), ∫e^x cos 2x dx=e^x (sin 2x)/2 + e^x (-cos 2x)/4 - (1/4)∫e^x cos 2x dx ∫e^x cos 2x dx + (1/4)∫e^x cos 2x dx=e^x (sin 2x)/2 + e^x (-cos 2x)/4 + C1 (5/4)∫e^x cos 2x dx=e^x (sin 2x)/2 + e^x (-cos 2x)/4 + C1 ∫e^x cos 2x dx=2 e^x (sin 2x)/5 + e^x (cos 2x)/5 + C, where C=C1/5 Method 2： ∫e^x cos 2x dx ... (1)=∫cos 2x d(e^x) =e^x (cos 2x) - ∫e^x d(cos 2x) =e^x (cos 2x) - ∫e^x (-2sin 2x) dx =e^x (cos 2x) + 2∫e^x sin 2x dx =e^x (cos 2x) + 2∫sin 2x d(e^x) =e^x (cos 2x) + 2 e^x sin 2x - 2∫e^x d(sin 2x) =e^x (cos 2x) + 2 e^x sin 2x - 2∫e^x (cos 2x) dx =e^x (cos 2x) + 2 e^x sin 2x - 2∫e^x (2cos 2x) dx =e^x (cos 2x) + 2 e^x sin 2x - 4∫e^x cos 2x dx ... (2) Since (1)=(2), ∫e^x cos 2x dx=e^x (cos 2x) + 2 e^x sin 2x - 4∫e^x cos 2x dx ∫e^x cos 2x dx + 4∫e^x cos 2x dx=e^x (cos 2x) + 2 e^x sin 2x + C1 5∫e^x cos 2x dx=e^x (cos 2x) + 2 e^x (sin 2x) + C1 ∫e^x cos 2x dx=2 e^x (sin 2x)/5 + e^x (cos 2x)/5 + C, where C=C1/5 2015-07-03 14:47:48 補充： Method 1 有 Let 哥到就開始用 by parts 而呢題就用左2次 by Parts 彧且你可以睇下 M1 同 M2 邊個會適合你多D M1會寫多D,但自己睇哥時會清楚D M2就快少少 2015-07-03 15:13:38 補充： ? . ? 2015-07-03 15:40:46 補充： 其實 M1 同 M2 係講緊 Method 1 同 Method 2 話說本身諗住寫 Method 1 同 Method 2 但按補充哥時佢話字數過多.... 所以最後就將 Integration By parts 同 Method 1 及 Method 2 呢D字將佢簡寫 0.0 最後寫左做 M1 同 M2
其他解答:
integration 4
發問:
How to integrate: e^x cos2x dx Please explain it step by step.
最佳解答:
Integration By Parts： ∫udv = uv - ∫vdu Method 1： ∫e^x cos 2x dx ... (1)=∫e^x d[(sin 2x)/2] Let u=e^x, v=(sin 2x)/2 ?du=e^x dx, dv=cos 2x dx ∫e^x d[(sin 2x)/2] =∫udv =uv - ∫vdu =e^x (sin 2x)/2 - ∫(sin 2x)/2 d(e^x) =e^x (sin 2x)/2 - ∫(sin 2x)/2 (e^x) dx =e^x (sin 2x)/2 - (1/2)∫e^x sin 2x dx =e^x (sin 2x)/2 - (1/2)∫e^x d[(-cos 2x)/2] Let u=e^x, v=(-cos 2x)/2 ?du=e^x dx, dv=sin 2x dx e^x (sin 2x)/2 - (1/2)∫e^x d[(-cos 2x)/2] =e^x (sin 2x)/2 - (1/2)∫udv =e^x (sin 2x)/2 - [uv/2 - (1/2)∫vdu] =e^x (sin 2x)/2 - uv/2 + (1/2)∫vdu =e^x (sin 2x)/2 - e^x (-cos 2x)/4 + (1/2)∫(-cos 2x)/2 d(e^x) =e^x (sin 2x)/2 + e^x (cos 2x)/4 - (1/4)∫(cos 2x) (e^x) dx =e^x (sin 2x)/2 + e^x (-cos 2x)/4 - (1/4)∫e^x cos 2x dx ... (2) Since (1)=(2), ∫e^x cos 2x dx=e^x (sin 2x)/2 + e^x (-cos 2x)/4 - (1/4)∫e^x cos 2x dx ∫e^x cos 2x dx + (1/4)∫e^x cos 2x dx=e^x (sin 2x)/2 + e^x (-cos 2x)/4 + C1 (5/4)∫e^x cos 2x dx=e^x (sin 2x)/2 + e^x (-cos 2x)/4 + C1 ∫e^x cos 2x dx=2 e^x (sin 2x)/5 + e^x (cos 2x)/5 + C, where C=C1/5 Method 2： ∫e^x cos 2x dx ... (1)=∫cos 2x d(e^x) =e^x (cos 2x) - ∫e^x d(cos 2x) =e^x (cos 2x) - ∫e^x (-2sin 2x) dx =e^x (cos 2x) + 2∫e^x sin 2x dx =e^x (cos 2x) + 2∫sin 2x d(e^x) =e^x (cos 2x) + 2 e^x sin 2x - 2∫e^x d(sin 2x) =e^x (cos 2x) + 2 e^x sin 2x - 2∫e^x (cos 2x) dx =e^x (cos 2x) + 2 e^x sin 2x - 2∫e^x (2cos 2x) dx =e^x (cos 2x) + 2 e^x sin 2x - 4∫e^x cos 2x dx ... (2) Since (1)=(2), ∫e^x cos 2x dx=e^x (cos 2x) + 2 e^x sin 2x - 4∫e^x cos 2x dx ∫e^x cos 2x dx + 4∫e^x cos 2x dx=e^x (cos 2x) + 2 e^x sin 2x + C1 5∫e^x cos 2x dx=e^x (cos 2x) + 2 e^x (sin 2x) + C1 ∫e^x cos 2x dx=2 e^x (sin 2x)/5 + e^x (cos 2x)/5 + C, where C=C1/5 2015-07-03 14:47:48 補充： Method 1 有 Let 哥到就開始用 by parts 而呢題就用左2次 by Parts 彧且你可以睇下 M1 同 M2 邊個會適合你多D M1會寫多D,但自己睇哥時會清楚D M2就快少少 2015-07-03 15:13:38 補充： ? . ? 2015-07-03 15:40:46 補充： 其實 M1 同 M2 係講緊 Method 1 同 Method 2 話說本身諗住寫 Method 1 同 Method 2 但按補充哥時佢話字數過多.... 所以最後就將 Integration By parts 同 Method 1 及 Method 2 呢D字將佢簡寫 0.0 最後寫左做 M1 同 M2
其他解答:
此文章來自奇摩知識+如有不便請留言告知
由於剛才一連撞題兩次，所以我暫時不作答，讓其他網友有更多機會發揮一下～ 以下的連結可以幫你檢查答案： http://www.wolframalpha.com/input/?i=int%20e%5Ex%20cos2x%20dx&t=crmtb01 2015-07-03 14:58:50 補充： 佢未必一定是讀香港的 curriculum。 2015-07-03 17:05:52 補充： 哈哈哈~ 我誤會了~~~
文章標籤
全站熱搜
留言列表
