興趣 integral 之 6

As follows:

圖片參考：http://i707.photobucket.com/albums/ww74/stevieg90/0IIT1.gif


圖片參考：http://i707.photobucket.com/albums/ww74/stevieg90/0IIT2.gif


2009-04-08 18:32:26 補充：
Please check the update.

2009-04-09 13:41:15 補充：
哦,感謝chowchihong2004的指點,但我還是覺得分CASE比較清晰吧~^^

STEVIE-G™ :

我發現無須分case都行,無論遇到什麼情況 A=B , A= -B 也好
這個積分也可以用第三個來表示,
2/sqrt[(A-B)(A+B)] * arctan{ sqrt[(A-B)/(A+B)] tan(x/2) ] }
當遇到 A=B , A=-B 的時候, 只要取極限便可
lim x->0 arctanax / x = a

IN [1/(A+Bcosx) dx]

IN [udv] = uv - IN [vdu]

u = 1/(A+Bcosx) v = x

IN [1/(A+Bcosx) dx]

= x / (a+bcosx) - IN [x d(A+Bcosx)]

= x / (a+bcosx) - B * IN [x d(cosx)]

= x / (a+bcosx) - B (xcosx - IN [ cosx dx])

= x / (a+bcosx) - B (xcosx - sinx + C)

= x / (a+bcosx) - Bxcosx + Bsinx + C

Let I be the given integral .

If A = B, I = (1/A)tan(θ/2) + C
If A > B, I = [2/√(A^2-B^2)] arctan { √[(A-B)/(A+B)]tan(θ/2)} + C
If A < B, I = [1/√(B^2-A^2)] ln │ {tan(θ/2) + √[(B+A)/(B-A)]} / {tan(θ/2) - √[(B+A)/(B-A)]} │+ C

in each case where C is a constant.

2009-04-08 18:30:01 補充：
If A = -B, I = 1/[B tan(θ/2)] + C

If A <> -B and A > B, ...
If A <> -B and A < B, ...