Question

积分∫dx/3+cosx=?请尽可能详细,谢谢.

Answer 1

令u=tan（x/2)

cosx=(1-u^2)/(1+u^2) dx=2/(1+u^2)du

1/(3+cosx)=1/{2+[2/(1+u^2)]}

所以原始变为：∫[1/(3+cosx)]dx=∫1/(u^2+2)du=√2/2*arctan[√2/2*tan(x/2)]+c