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

`(\mathrm d)/(\mathrm d\alpha)\tan\alpha=\sec^2\alpha`

so we get the antiderivative

`\frac13\tan\alpha+C`

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

`\frac13\tan3\theta+C`