Question

这题都是三角函数的不定积分怎么求？

Answer 1

半角代换。
令 u = tan(x/2), 则 sinx = 2u/(1+u^2),
cosx = (1-u^2)/(1+u^2), dx = 2du/(1+u^2)
I = ∫ [2u(1-u^2)/(1+u^2)^2]/(1+4u-u^2)/(1+u^2)]2du/(1+u^2),
= ∫(4u(1-u^2)du/[(1+4u-u^2)(1+u^2)^2] 再化为有理分式部分分式， 本题很麻烦。

Answer 2

∫dx/sin2x+2sinx
=∫dx/2sinx(cosx+1)
=∫dx/8sin(x/2)cos(x/2){cox(x/2)}^2
=1/4∫1/sin(x/2)cos(x/2)dtan(x/2)
=1/4∫(cos(x/2)/sin(x/2)+sin(x/2)/cos(x/2)dtan(x/2)
=1/4∫1/tan(x/2)dtan(x/2)+1/4∫tan(x/2)dtan(x/2)
=1/4ln绝对值tan(x/2)+1/8{tan(x/2)}^2+C