MQ26 --- Indefinite Integral

圖片參考：http://i1191.photobucket.com/albums/z467/robert1973/Jun12/Crazyint2.jpg
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2012-06-26 08:43:29 補充：
You mean which line?

2012-06-26 09:31:31 補充：
Image updated. Or you can visit:
http://i1191.photobucket.com/albums/z467/robert1973/Jun12/Crazyint2.jpg

2012-06-26 09:32:39 補充：
Shd be this one:
http://i1191.photobucket.com/albums/z467/robert1973/Jun12/Crazyint3.jpg

2012-06-26 16:02:55 補充：
(ln x) √(1 - x²)

2012-06-26 17:23:10 補充：
∫cscθdθ
= ∫ cscθ (cscθ + cot θ)/(cscθ + cot θ) dθ
= ∫ (csc^2 θ + csc θ cot θ)/(cscθ + cot θ) dθ
= - ∫ d(cscθ + cot θ)/(cscθ + cot θ) since d(cscθ + cot θ)/dθ = - csc θ cot θ - csc^2 θ
= - ln |cscθ + cot θ| + C

2012-06-26 21:51:39 補充：
If you just input ∫cscx dx, it will give:
log [sin (x/2)] - log [cos (x/2)]
= log [tan (x/2)]
By t-formula:
sin x = 2t/(1 + t^2) and cos x = (1 - t^2)/(1 + t^2) where t = tan (x/2)
Hence
csc x + cot x = (1 + cos x)/sin x
With 1 + cos x = 2/(1 + t^2), csc x + cot x = 1/t

2012-06-26 21:52:18 補充：
- ln |csc x + cot x| = - ln |1/t|
= ln |t|
= ln |tan (x/2)|

which is the same.

To 001:
Your answer is wrong because in the 7th line last term it should be minus.