Question

求不定积分一题 ∫dx/(3+cosx)

Answer 1

简单计算一下即可，答案如图所示

Answer 2

1.利用万能代换公式化为有理函数的不定积分,即令t=tan(x/2),则x=2arctant,cosx=(1-t^2)/(1+t^2)
∫dx/(3+cosx)=∫d(2arctant)/[3+(1-t^2)/(1+t^2)]=2∫(1/1+t^2)/[3+(1-t^2)/(1+t^2)]dt=...有理函数一定可以积出来,最后只须把t回代即可
2.凑微分法：3+cosx=3[(sin(x/2))^2+(cos(x/2))^2]+[(cos(x/2))^2-sin(x/2))^2]=2(sin(x/2))^2+4(cos(x/2))^2
∫dx/(3+cosx)=∫dx/[2(sin(x/2))^2+4(cos(x/2))^2]=∫[1/(cos(x/2))^2]{1/[2(tan(x/2))^2+4]}dx=∫[2d(tan(x/2))^2]/[2(tan(x/2))^2+4]=∫du/(u^2+2)=(1/√2)arctan(u/√2)+c
u=（tan(x/2))^2
主要是数学符号不好表达,看上去混乱了点,其实很清晰简单的