题目

问一下∫1/cos（2x）dx如何积分
如题
能不能给我解法？我这里∫ 1/cos（nx）dx答案是(1/n)(ln|cos(nx/2)-sin(nx/2)|-ln|sin(nx/2)+cos(nx/2)|)+c，这怎么解出来的？

提问时间：2020-11-20

答案

1/cos(nx) = 1/(cos²(nx/2) - sin²(nx/2))
= (1/2)((cos(nx/2) - sin(nx/2))/(cos(nx/2) + sin(nx/2)) - (-cos(nx/2) - sin(nx/2))/(cos(nx/2) - sin(nx/2)))
∫(cos(nx/2) - sin(nx/2))/(cos(nx/2) + sin(nx/2))dx
=(2/n)∫(1/t)dt （令(cos(nx/2) + sin(nx/2)=t）
=(2/n)ln|t|
=(2/n)ln|cos(nx/2) + sin(nx/2)| + C1
同理
∫(-cos(nx/2) - sin(nx/2))/(cos(nx/2) - sin(nx/2))dx
=(2/n)ln|cos(nx/2) - sin(nx/2)| + C2
∫1/cos(nx)dx = ∫1/(cos²(nx/2) - sin²(nx/2))dx
= ∫(1/2)((cos(nx/2) - sin(nx/2))/(cos(nx/2) + sin(nx/2)) - (-cos(nx/2) - sin(nx/2))/(cos(nx/2) - sin(nx/2)))dx
=(1/2)((2/n)ln|cos(nx/2) + sin(nx/2)| - (2/n)ln|cos(nx/2) - sin(nx/2)|) + C
=(1/n)(ln|cos(nx/2) + sin(nx/2)| - ln|cos(nx/2) - sin(nx/2)|) + C

问题补充：能不能给我解法?我这里∫ 1/cos（nx）dx答案是(1/n)(ln|cos(nx/2)-sin(nx/2)|-ln|sin(nx/2)+cos(nx/2)|)+c,这怎么解出来的?
验证一下不就知道了,
(1/n)(ln|cos(nx/2)-sin(nx/2)|-ln|sin(nx/2)+cos(nx/2)|)+c
求导 = (1/n)(n/2)
((-sin(nx/2)-cos(nx/2))/(cos(nx/2)-sin(nx/2))
- (cos(nx/2)-sin(nx/2))/(sin(nx/2)+cos(nx/2)))
=(1/2)(-2)/(cos²(nx/2) - sin²(nx/2))
= -1/cos(nx)
所以答案错了