integration 4

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

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~~~