求不定积分∫1/(1+sinx)dx
1个回答

令 x=2t,则 dx = 2dt,sinx = sin(2t) = 2 sint cost

∫ dx / (1 + sinx)

= ∫ 2dt / (sin²t + cos²t + 2 sint cost)

= ∫ 2dt / (sin t + cos t)²

= ∫ 2dt / [√2 sin (t + π/4) ]²

= ∫ csc² (t + π/4) dt

= - ctg (t + π/4) + C

= - ctg (x/2 + π/4) + C

结果的其他表达方式:

tan(x/2 - π/4) + C

2 / [1 + ctg(x/2)] + C