积分的题

这类题一般都使用万能代换，请看以下解法一。
解法一：设 tan(x/2)=t (-π<x<π),则dx=2dt/(1+t²)
∵sinx=2sin(x/2)cos(x/2)/[sin²(x/2)+cos²(x/2)]
=2tan(x/2)/[1+tan²(x/2)]
=2t/(1+t²)
cosx=[cos²(x/2)-sin²(x/2)]/[sin²(x/2)+cos²(x/2)]
=[1-tan²(x/2)]/[1+tan²(x/2)]
=(1-t²)/(1+t²)
∴ 1+sinx-cosx=1+2t/(1+t²)-(1-t²)/(1+t²)
=2+2t/(1+t²)-2/(1+t²)
∴原式=∫dx/(1+sinx-cosx)
=∫[2+2t/(1+t²)-2/(1+t²)]dt
=2∫dt+∫d(1+t²)/(1+t²)-2∫dt/(1+t²)
=2t+ln(1+t²)-2arctant+C (C是积分常数)
=2tan(x/2)+ln[1+tan²(x/2)]-x+C
解法二：原式=∫dx/(1+sinx-cosx)
=∫dx/(sinx+1-cosx)
=∫dx/[2sin(x/2)cos(x/2)+2sin²(x/2)]
=∫[1/sin²(x/2)]d(x/2)/[cot(x/2)+1]