Question

求∫1/(2+sinx)dx的不定积分

需要过程，谢谢

Answer 1

原式=∫1/(2+sinx)dx=∫1/(2+sinx)dx=∫1/(2+cos(x-π/2))dx由cos2t=2(cost)^2-1可得：=∫1/(2+2[cos(x/2-π/4)]^2-1)dx=∫1/[1+2cos(x/2-π/4)^2] dx=tan(x/2-π/4)+C

Answer 2

let tg(x/2)=two can got sinx=2t/(1+t^2) and x=2arctgt.
so,dx=[2/(1+t^2)]dtthen ∫1/(2+sinx)dx =∫[1/(1+t+t^2)]dt =∫1/[(t+1/2)^2+3/4]dt
=4/3∫1/{[(2t+1)/√3]^2+1}
=2/√3*arctg[(2t+1)/√3]+C