Question

How do you integrate `int sec^4(5x)`?

Answer 1

`tan(5x)/5+tan^3(5x)/15+C`

Explanation:

Start first with a simple substitution to simplify the argument of the secant functions: let `u=5x` which implies that `du=5dx`. Then:

`intsec^4(5x)dx=1/5intsec^4(5x)(5dx)=1/5intsec^4(u)du`

When working with secant, it's important to keep the identities `d/dxsecx=secxtanx`, `d/dxtanx=sec^2x`, and `1+tan^2x=sec^2x` in mind.

In this case, we see that the integral

`=1/5intsec^2(u)sec^2(u)du=1/5int(1+tan^2(u))sec^2(u)du`

The point of doing this is that we have a function of tangent—`1+tan^2(u)`—paired with the derivative of tangent—`sec^2(u)`.

So, let `v=tan(u)` so that `dv=sec^2(u)du`. This gives us:

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

We can now integrate term by term:

`=1/5(v+v^3/3)=v/5+v^3/15`

Returning to our original variable `x` using `v=tan(u)` and `u=5x` then adding the constant of integration:

`=tan(u)/5+tan^3(u)/15=tan(5x)/5+tan^3(5x)/15+C`