∫x/sinx^2dx=∫2x/(1-cos2x)dx设tanx=u,x=arctanu,dx=1/(1+u²)du,cos2x=(1-u²)/(1+u²)原式=2∫arctanu/(2u²/(1+u²))*1/(1+u²)du=∫arctanu/u²du=-∫arctanud(1/u)=-arctanu*1/u...