Question

integrate sec^6(2x) (dont use reduction formula)

Answer 1

\[\int\limits \sec ^6 (2x) ~dx\]

Answer 2

First, simple \(u\)-sub \[\frac{1}{2} \int \sec^6u\,\,du\] Then
\[\frac{1}{2} \int (\sec^3u)^2\,\,du\]\[\frac{1}{2} \int ((1 + \tan^2u)\sec{u})^2\,\,du\]\[\frac{1}{2} \int (\sec{u} + \sec{u}\tan^2u)^2\,\,du\]\[\frac{1}{2} \int \sec^2u+2sec^2u\tan^2u + \sec^2u\tan^4u \,\,du\]

Answer 3

The first term is trivial.
\[\int \sec^2u\,\,du = \tan{u}\] Then, factor a \(\sec^2u\) out of the remaining two terms and use \(u\)-substitution again to get \[\int 2v^2 + v^4 \,\,dv\] which is also trivial.