[tex]\displaystyle13\tan^2x\sec^2x\,\mathrm dx[/tex]
Let [tex]y=\tan x[/tex], so that [tex]\mathrm dy=\sec^2x\,\mathrm dx[/tex]. Then you have
[tex]\displaystyle13\int y^2\,\mathrm dy=\dfrac{13}3y^3+C[/tex]
and so
[tex]\displaystyle\int13\tan^2x\sec^2x\,\mathrm dx=\dfrac{13}3\tan^3x+C[/tex]
