Question

How do you integrate `int sec^6(3x)`?

Answer 1

`tan(3x)+2/3tan^3(3x)+1/5tan^5(3x)+C`

Explanation:

First let `u=3x` so that `du=3dx`. Then we have:

`I=intsec^6(3x)dx=1/3intsec^6(3x)3dx=1/3intsec^6(u)du`

We will now make use of the identity `tan^2(theta)+1=sec^2(theta)`.

`I=1/3intsec^4(u)sec^2(u)du=1/3int(sec^2(u))^2sec^2(u)du`

`I=1/3int(1+tan^2(u))^2sec^2(u)du`

Now we should perform the substitution `v=tan(u)`. Note that `dv=sec^2(u)du`, which is why we left a `sec^2(u)` floating around in the integrand.

`I=int(1+v^2)^2dv`

Expand `(1+v^2)^2=(1+v^2)(1+v^2)` and then integrate term by term:

`I=int(1+2v^2+v^4)dv=v+2/3v^3+1/5v^5`

From `v=tan(u)`:

`I=tan(u)+2/3tan^3(u)+1/5tan^5(u)`

From `u=3x`:

`I=tan(3x)+2/3tan^3(3x)+1/5tan^5(3x)+C`