let
t= tan(x/2)
dt = (1/2)[ sec(x/2)]^2 dx
dx = 2dt/(1+t^2)
sinx = 2sin(x/2).cos(x/2) =2t/(1+t^2)
∫ dx/(3+sinx)
=∫ [2dt/(1+t^2)]/[3 +2t/(1+t^2) ]
=2∫ dt/ (3t^2+2t+3)
=(2/3) ∫ dt/[ t^2+(2/3)t+1]
=(2/3)(3√2/4)arctan[(3t+1)/(2√2)] +C
=(√2/2)arctan[(3t+1)/(2√2)] +C
=(√2/2)arctan{ [3tan(x/2)+1]/(2√2) } +C
consider
t^2+(2/3)t+1 = (t+ 1/3)^2 + 8/9
let
t+1/3 = (2√2/3) tanu
dt =(2√2/3) (secu)^2 du
∫ dt/[ t^2+(2/3)t+1]
=∫ (2√2/3) (secu)^2 du/[ (8/9) (secu)^2 ]
= (3√2/4)u +C
= (3√2/4)arctan[(3t+1)/(2√2)] +C
追问为什么sinx=2t/1+t^2
追答sin(x/2) = t/√(1+t^2)
cos(x/2)= 1/√(1+t^2)
sinx = 2sin(x/2) .cos(x/2) = 2t/(1+t^2)