Calculus - Integration

1. 性質:
(1)cos(2u)= 2tan(u)/[1+tan^2(u)]= (1-t^2)/ (1+ t^2) , u=arctan(t), t=tan(u)=tan(x/2)
(2)∫dx/(x^2 + k^2) = 1/k arctan(x/k) + C (k≠0)
(3)∫ 2 dx/ (x^2 - k^2) = 1/k ln | (x-k)/(x+k) | + C (k≠0)
2. 符號設定
設 a= α^2, A^2= (1-a)/(1+a) , B^2= (a-1)/(a+1)
3. 開始積分
(1)α^2 = 1時
原積分=∫dx/(1-cosx) =∫ (1+cosx)/(sinx)^2 dx
=∫[(csc x)^2 + csc x cot x ] dx= - cot x - csc x + C
(2) α^2 < 1時
令 x= 2u = 2 arctan t => cosx= cos(2u)= (1-t^2)/(1+t^2) (性質1)
dx= 2/(1+t^2) dt, 1-α^2 cos x= 1- a*(1-t^2)/(1+t^2)
原積分=∫ 2/[(1+a)t^2+(1- a)] dt ( a=α^2) ----(P)
= 2/(1+a)∫ dt/(t^2 + A^2) (A^2= (1-a)/(1+a) )
= 2/(1+a) * (1/A) arctan(t/A) + C (性質2)
= 2/√(1- α^4) arctan(t/A) + C ( A=√[(1- α^2)/(1+α^2)] )
Note: t=tan(u)= tan(x/2)
(3) α^2 > 1時
原式= (P)式= 1/(1+a)∫ 2/(t^2 - B^2) dt (B^2= (a-1)/(a+1) , a=α^2)
= 1/(1+a) * 1/B * ln | (t-B)/(t+B) | + C (性質3)
= 1/ √(α^4 - 1) * ln | (t-B)/(t+B) | + C ( t=tan(x/2))