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!

如题所述

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.
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