Definite Integral

👨🏻‍🏫 學術台數學

[匿名]

2012-01-20 4:24 am

Find ∫ dx/(1 + cos^2 x) from 0 to π.

更新1:

To : 自由自在 Please explain why : (1) the lower limit of the 2nd integral is -ve infinity. (2) arctan (infinity) = arctan (- infinity) = π/2. Can it also be 3π/2, 5π/2,......

更新2:

To : 自由自在 Thank you for your explanation that we have to split the integral into 2 parts at π/2 because tan x is not continuous at π/2. However, the function as a whole sec^2 x/(2 + tan^2 x) = 1/(1 + cos^2 x) is continuous at x = π/2. So what is the guideline for when we should split?

回答 (2)

用戶9112

2012-01-20 4:53 am

✔ 最佳答案

http://i1090.photobucket.com/albums/i376/Nelson_Yu/int-1939.jpg

圖片參考：http://i1090.photobucket.com/albums/i376/Nelson_Yu/int-1939.jpg

2012-01-20 14:04:15 補充：
If you draw the curve of y= tan x, you can see that x=pi/2 is a discontinuous point. When x is < pi/2, tan x is positive, when x is > pi/2, tan x is negative.
Remember when you do the substitute, the entire range must be continuous, this is the reason why I have splitted the integral into 2.

2012-01-20 14:06:45 補充：
As to your second question, it is usual to pick the principle value of arctan function from -pi/2 to pi/2.
For the sake of discussion, if you want to pick 3pi/2, then you have to pick the corresponding value for arctan(0). As far as the entire range is continuous, you will get the same result.

2012-01-21 11:26:10 補充：
You have to consider the continuity of the substitution u = tan x as well.

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hung

2012-01-20 6:06 am

arctan 既 range 淨係取 [-pi/2 , pi/2]

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