Can someone tell me how to evaluate the integral:

∫tan(2x)sec³(2x)dx

I don't know what to do when sec is cubed and which substitution to make, or if there even is a substitution, if I express sec³(2x) as sec²(2x)sec(2x) and change sec²(2x) to (tan²(2x)+1), then let u=tan(2x) du/dx=sec²(2x) but I only have a sec(2x) left to cancel, :(

\int tan(2x)sec^{3}(2x)dx

Make the sub:

u=2x, \;\ \frac{1}{2}du=dx

\frac{1}{2}\int tan(u)sec^{3}(u)du

Now, make the sub w=sec(u), \;\ dw=sec(u)tan(u)du

See? The dw takes up everything but sec^{2}(u), which we use w^2 on

So, we have:

\frac{1}{2}\int w^{2}dw

Integrate:

\frac{1}{6}w^{3}

Resub:

\frac{1}{6}sec^{3}(u)

resub:

\frac{1}{6}sec^{3}(2x)+C

The C is a hanger on. Add it if the professor is anal.