Question

求1/（1+cosx）^2的原函数

要有具体过程

Answer 1

求原函数即是求此函数的不定积分:∫1/（1+cosx）^2dx.
解答如下：
由半角公式可得：cos(x)=[1-tan(x/2)^2]/[1+tan(x/2)^2].
设u=tan(x/2)(-π<x<π),则x=2arctan(u)
所以dx=(2/1+u^2)du
所以
∫1/（1+cosx）^2dx=∫1/[1+(1-u^2)/(1+u^2)]^2*(2/1+u^2)du
=∫1/[2/(1+u^2)]^2*(2/1+u^2)du
=(1/2)∫(1+u^2)du
=(1/2)(1/2u+1/3u^3)+c
=1/4u+1/6u^3+c
=1/4tan(x/2)+1/6(tan(x/2))^3+c

Answer 2

∫1/（1+cosx）^2dx
=∫1/[2(cos(x/2))^2]^2dx
=∫1/[4(cos(x/2))^4]dx
=∫(sec(x/2))^4/4dx
=1/4*∫[1+(tan(x/2))^2](sec(x/2))^2dx
=1/4*∫[1+(tan(x/2))^2]d[2(tan(x/2))^2]
=1/2*(tan(x/2))+1/2*1/3*(tan(x/2)^3)+c
=(tan(x/2)^3)/6+(tan(x/2))/2+c

Answer 3

y=arccos[(1/根号x)-1]

Answer 4

用积分慢慢求就行，我求了求太麻烦，没有化到底！