integration?
回答 (2)
2016-02-27 12:21 pm
✔ 最佳答案
∫ (1-cos x) / (1+cos x) dx from x=0 to π/2= ∫ ( (1-cos x) / 2 ) / ( (1+cos x) / 2) dx from x=0 to π/2
= ∫ (sin^2 x/2) / (cos^2 x/2) dx from x=0 to π/2
= ∫ tan^2 x/2 dx from x=0 to π/2
= ∫ ( sec^2 x/2 - 1 ) dx from x=0 to π/2
= [ 2 tan x/2 - x ] from x=0 to π/2
= 2 - π/2
2016-02-27 12:03 pm
Mind (1-cos x)/(1+cos x)=((1-cos x)/2)/((1+cos x)/2)=(sin^2 x/2)/(cos^2 x/2)=tan^2 x/2=sec^2 x/2 - 1.
Hence ∫ (1-cos x)/(1+cos x) = ∫ (sec^2 x/2 - 1) dx = 2 tan x/2 - x + C.
And then plug in 0, pi/2, you can get:
(2 - pi/2) - 0 = 2 - pi/2.
Hence ∫ (1-cos x)/(1+cos x) = ∫ (sec^2 x/2 - 1) dx = 2 tan x/2 - x + C.
And then plug in 0, pi/2, you can get:
(2 - pi/2) - 0 = 2 - pi/2.
收錄日期: 2021-04-20 16:16:58
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