令a=tanx
则原式=∫da/(2+a²)
=1/2*∫da/(1+a²/2)
=1/2*∫da/[1+(a/√2)²]
=√2/2*∫d(a/√2)/[1+(a/√2)²]
=(1/√2)*aectan(a/√2)+C
=(1/√2)*aectan(tanx/√2)+C