不定积分一题求∫(1+sinx)/(1-sinx)dx希望过程详细点,谢谢~答案貌似是化简到带secx和tgx的.我化不到他的那步~

Question

不定积分一题
求∫(1+sinx)/(1-sinx)dx
希望过程详细点,谢谢~
答案貌似是化简到带secx和tgx的.我化不到他的那步~

Answer 1

∫(1+sinx)/(1-sinx)dx
=∫(sinx-1+2)/(1-sinx)dx
=∫(-1 + 2/(1-sinx))dx
=∫-1dx + 2∫1/((sin(x/2))^2+
(cos(x/2))^2-2sin(x/2)cos(x/2))dx
=-x + 2∫(1/(cos(x/2))^2)/
((tan(x/2))^2-2tan(x/2)+1)dx
=-x + 4∫1/(tan(x/2)-1)^2d(tan(x/2))
=-x + 4(-1/((tan(x/2)-1) + c
=-x - 4/(tan (x/2) -1) + c
汗,其实都是一样的啊...不过是转换一下而已
-4/(tan (x/2) -1)
=-4cos(x/2)/(sin(x/2) - cos(x/2))
=-4cos(x/2)*(sin(x/2)+cos(x/2))/
(sin(x/2))^2 - (cos(x/2))^2)
=-(2sinx + 2cosx + 2)/(-cosx)
=2tanx + 2secx + 2
所以答案是
-x + 2tanx + 2secx + c
写哪个都是对的