Recover an integral from a derivative

$\dfrac d{dx}(\tan^3x)=3\tan^2x\sec^2x$. Using $\tan^2x=\sec^2x-1$ gives $3(\sec^2x-1)\sec^2x=3\sec^4x-3\sec^2x$, as required.

Integrate the displayed derivative: $3\int\sec^4x\,dx-3\int\sec^2x\,dx=\tan^3x+C^{\prime\prime}$. Thus $3\int\sec^4x\,dx-3\tan x=\tan^3x+C^\prime$. Rearrange and absorb the factor into the arbitrary constant: $\int\sec^4x\,dx=\dfrac13\tan^3x+\tan x+C$.