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