Question

计算不定积分？

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Answer 1

(1)
let
tanx = (√3/2)tanu
dtanx =(√3/2)(secu)^2 du
∫ dx/[3+ (sinx)^2 ]
=∫ (secx)^2/[3(secx)^2+ (tanx)^2 ] dx
=∫ (secx)^2/[4(tanx)^2+3 ] dx
=∫ dtanx/[4(tanx)^2+3 ]
=∫ (√3/2)(secu)^2 du/[ 3(secu)^2]
=(√3/6)u + C
=(√3/6)arctan(2tanx/√3) + C
(2)
let
t= tan(x/2)
dt =(1/2)[sec(x/2)]^2 dx
dx = [2/(t^2+1) ] dt
∫ dx/(3+ cosx)
=∫ [2/(t^2+1) ] dt /[2+ 2/(1+t^2)]
=∫ dt/(2+t^2)
=(√2/2)arctan(t/√2) + C
=(√2/2)arctan[ tan(x/2)/√2] + C

Answer 2