经过我和conlui的讨论和努力,
题目终于有了结果,在此我感谢他!
还有侯宇诗的
有:
sinx+sin2x+...+sinnx= - [cos(n+1)x+cosnx-cosx-1]/2sinx
证明:
左边=-2sinx[sinx+sin2x+...+sinnx]/(-2sinx)
=[cos2x-cos0+cos3x-cosx+...+cosnx-cos(n-2)x+cos(n+1)x-cos(n-1)x]/(-2sinx)
=- [cos(n+1)x+cosnx-cosx-1]/2sinx=右边
等式得证
所以
有:
(cos(n+1)x+cosnx)/sinx
=-2(sinx+sin2x+...+(sinnx)+cosx/2sinx+1/2sinx
所以
∫(cos(n+1)x+cosnx)dx/sinx
=-2积分:[(sinx+sin2x+...+(sinnx)+cosx/2sinx+1/2sinx]dx
=2[cosx+1/2cos2x+...+1/n*cosnx]+ln|2sinx|+ln|tanx/2|+C