一个不定积分……急!有理函数积分这一节的内容∫dx/(1+sinx+cosx)

Question

一个不定积分……急!有理函数积分这一节的内容
∫dx/(1+sinx+cosx)

Answer 1

设t=tan(x/2),则x=2arctant,sinx=2t/(1+t²),cosx=(1-t²)/(1+t²),dx=2dt/(1+t²)
故 ∫dx/(1+sinx+cosx)=∫[2dt/(1+t²)]/[1+2t/(1+t²)+(1-t²)/(1+t²)]
=∫[2dt/(1+t²)]/[2(1+t)/(1+t²)]
=∫dt/(1+t)
=ln│1+t│+C (C是积分常数)
=ln│1+tan(x/2)│+C.