一道不定积分题!求不定积分∫1/(sinx＋cosx)dx 要求：用2种方法求解!

解法一：万能代换!
令u＝tanx/2,则sinx＝2u/(1＋u²),cosx＝(1－u²)/(1＋u²),dx＝2du/(1＋u²),于是得
∫1/(sinx＋cosx)＝∫2/(1＋2u－u²)du
＝√2/2∫［1/(u－(1－√2))－1/(u－(1＋√2))］du
＝√2/2ln|(u－(1－√2))/(u－(1＋√2))|＋C
＝√2/2ln|(tanx/2－1＋√2)/(tanx/2－1－√2)＋C.
解法二：
∫dx/(sinx＋cosx)＝√2/2∫dx/(√2/2sinx＋√2/2cosx)＝√2/2∫dx/cos(x－π/4)
＝√2/2∫sec(x－π/4)d(x－π/4)
＝√2/2ln|sec(x－π/4)＋tan(x－π/4)|＋C.