Question

求∫1/(2+sinx)dx的不定积分

Answer 1

令t=tan(x/2),则x=2arctant,所以dx=2/(1+t^2)dt

由万能公式:sinx=2tan(x/2)/(1+(tan(x/2))^2)=2t/(1+t^2),

则原式=(1/2)∫d(t+1/2)/[(t+1/2)^2+(根号3/2)^2]

=(1/根号3)arctan[2(t+1/2)/根号3]+C

=(1/根号3)arctan[2(arctan(x/2)+1/2)/根号3]+C