Question

What is the Integral of `sec^4 (x) tan^4 (x) dx`?

Answer 1

This is one of those problems where you just have to improvise to see what you can come up with.

Two ways I can think of to do this:

• Finding a way to let `du = secxtanxdx`. Thus, find a way for `u = secx` to be reasonably easy to do.
• Finding a way to let `du = sec^2dx`. Thus, find a way for `u = tanx` to be reasonably easy to do.

I already tried the first way and it didn't go well because both trig functions had the same exponent, so here is the second way.

`int sec^4xtan^4xdx`

Separate to achieve `sec^2x` as a product term.

`= int sec^2xsec^2x(tanx)^4dx`

Now transform `secx` terms that aren't the `sec^2x` you want to preserve to achieve `tanx` terms via the trig relationship `sec^2x = tan^2x + 1`.

`= int sec^2x (tan^2x + 1)(tanx)^4dx`

Now that we have `tanx` terms we can isolate as `u`... Let `u = tanx`. Then, `du = sec^2xdx`.

`=> int (u^2 + 1)(u^4)du`

`= int u^6 + u^4du`

`= u^7/7 + u^5/5`

Pretty much done now. Substitute `u = tanx` back in to get:

`= color(blue)(tan^7x/7 + tan^5x/5 + C)`

And just to check that this worked...

`d/(dx)[tan^7x/7 + tan^5x/5 + C]`

`= tan^6x*sec^2x + tan^4x*sec^2x` (chain rule)

`= sec^2x(tan^6x + tan^4x)` (factor)

`= sec^2x(tan^4x(tan^2x + 1))` (factor)

`= sec^2x(tan^4xsec^2x)` (trig identity)

`= color(green)(sec^4xtan^4x)` (associate/distribute)