求积分题目:被积函数1/(3+cosx)的原函数

Question

求积分题目:被积函数1/(3+cosx)的原函数

Answer 1

另u=tanx/2 则cosx=(1-u^2)/(1+u^2) dx=2/(1+u^2)du 带人原函数得1/(2+u^2) du =1/2^1/2(1/(1+(u/(1/2^1/2)))d(u/(1/2^1/2))=1/2^1/2arctan((tanx/2)/(1/2^1/2))