1/(2(cosx^2)+sinxcosx+sinx^2=secx^2/(2+tanx+tanx^2)   LET tanx=t   hence secx^2 dx=dt                                          1/(2+t+t^2)=1/(t+2)^2+2-1/4=1/(t+2)^2+(sqroot3/4)    applying formula for 1/(t^2+a^2)=1/atan^-1(t/a)    hence =(4/3)^1/2tan^-1((t+2)/(3/4)^1/2))   replace t  to get answer.   PLEASE RATE IF SATISFIED.