Letu=1+sin(x)cos(x)=1+(1/2)sin(2x)
anddu=cos(2x)dx→dx=du/cos(2x)
So∫cos(2x)/(1+sin(x)cos(x))dx
=∫1/udu
=ln|u|+C
=ln|1+sin(x)cos(x)|+C
or=ln|sin(2x)+2|+C
anddu=cos(2x)dxwhy?1/2sin2xDX1/2怎么消失了呢?
du=d[1+(1/2)sin(2x)]=0+1/2*cos(2x)*(2x)'=1/2*cos(2x)*2=cos(2x)Doyouunderstand?Ifnot,plzaskmeagain.