For [tex]\sin^2x[/tex] and [tex]\cos^2x[/tex], it's helpful to remember the half-angle identities:
[tex]\sin^2x=\dfrac{1-\cos2x}2[/tex]
[tex]\cos^2x=\dfrac{1+\cos2x}2[/tex]
So
[tex]\displaystyle\int\sin^2x\,\mathrm dx=\frac12\int(1-\cos2x)\,\mathrm dx=\frac x2-\frac{\sin2x}4=C[/tex]
[tex]\displaystyle\int\cos^2x\,\mathrm dx=\frac12\int(1+\cos2x)\,\mathrm dx=\frac x2+\frac{\sin2x}4=C[/tex]
For [tex]\tan^2x[/tex], the Pythagorean identity suffices:
[tex]\sin^2x+\cos^2x=1\implies \tan^2x+1=\sec^2x[/tex]
which means
[tex]\displaystyle\int\tan^2x\,\mathrm dx=\int(\sec^2x-1)\,\mathrm dx=\tan x-x+C[/tex]
since [tex]\dfrac{\mathrm d}{\mathrm dx}\tan x=\sec^2x[/tex].
