Question

Integral of sec^2 3θ dθ

Answer 1

Let $\alpha=3\theta$, so that $\mathrm d\alpha=3\,\mathrm d\theta$.

$\displaystyle\int\sec^23\theta\,\mathrm d\theta=\frac13\int\sec^2\alpha\,\mathrm d\alpha$

Recall that

$\dfrac{\mathrm d}{\mathrm d\alpha}\tan\alpha=\sec^2\alpha$

so we get the antiderivative

$\dfrac13\tan\alpha+C$

and back-substitute to get it in terms of $\theta$:

$\dfrac13\tan3\theta+C$