∫−π3π3(π+4x3)dx2−cos(|x|+π3)=

I1=∫−π3π3πdx2−cos(x+π3), dropping the odd term         =2π∫0π3dx2−cos(x+π3),x+π3=t         =2π∫π32π3dt2−cost,u=tant2         =4π∫133du2(1+u2)−(1−u2)          =4π∫133du1+3u2=4π3tan−13u|133            =4π3(tan−13−tan−11)=4π3tan−1(12)