∫sinx/(1-sinx)dx

1个回答

  • 参考以下

    ∫sinx/(1+sinx)dx

    =∫(1+sinx-1)/(1+sinx)dx

    =∫[1-1/(1+sinx)]dx

    =∫dx-∫dx/(1+sinx)

    =x-∫dx/[sin²(x/2)+cos²(x/2)+2sin(x/2)cos(x/2)]

    =x-∫dx/[sin(x/2)+cos(x/2)]²

    =x-∫sec²(x/2)/[tan(x/2)+1]²dx

    =x-∫d[tan(x/2)]/[tan(x/2)+1]²dx

    =x-∫d[tan(x/2)+1]/[tan(x/2)+1]²dx

    =x+1/[tan(x/2)+1]+C (C是积分常数)