Using the additivity of the integral:
∫π3−π3(|x|+tanx)2dx=∫0−π3(|x|+tanx)2dx+∫π30(|x|+tanx)2dx
In the first integral substitute t=−x:
∫0−π3(|x|+tanx)2dx=∫0π3(|−t|+tan(−t))2d(−t)
∫0−π3(|x|+tanx)2dx=∫π30(|t|−tant)2dt
Change the name of the integration variable to x again for clarity:
∫π3−π3(|x|+tanx)2dx=∫π30(|x|+tanx)2dx+∫π30(|x|−tanx)2dx
as in the interval of integration x is positive, then |x|=x:
∫π3−π3(|x|+tanx)2dx=∫π30(x+tanx)2dx+∫π30(x−tanx)2dx
using the linearity of the integral:
∫π3−π3(|x|+tanx)2dx=∫π30((x+tanx)2+(x−tanx)2)dx
∫π3−π3(|x|+tanx)2dx=∫π30(x2+2xtanx+tan2x+x2−2xtanx+tan2x)dx
∫π3−π3(|x|+tanx)2dx=2∫π30(x2+tan2x)dx
∫π3−π3(|x|+tanx)2dx=2∫π30x2dx+2∫π30tan2xdx
Using the trigonometric identity:
tan2x=sec2x−1
∫π3−π3(|x|+tanx)2dx=2∫π30x2dx+2∫π30(sec2x−1)dx
∫π3−π3(|x|+tanx)2dx=2∫π30x2dx+2∫π30sec2xdx−2∫π30dx
∫π3−π3(|x|+tanx)2dx=2[x33+tanx−x]π30
∫π3−π3(|x|+tanx)2dx=2π381−π3+√3