Let [tex]\alpha=3\theta[/tex], so that [tex]\mathrm d\alpha=3\,\mathrm d\theta[/tex].
[tex]\displaystyle\int\sec^23\theta\,\mathrm d\theta=\frac13\int\sec^2\alpha\,\mathrm d\alpha[/tex]
Recall that
[tex]\dfrac{\mathrm d}{\mathrm d\alpha}\tan\alpha=\sec^2\alpha[/tex]
so we get the antiderivative
[tex]\dfrac13\tan\alpha+C[/tex]
and back-substitute to get it in terms of [tex]\theta[/tex]:
[tex]\dfrac13\tan3\theta+C[/tex]
