Question

What is `int(cos(x))^4 dx`?

Answer 1

`int (cos(x))^4 dx = 1/32[12x + 8sin(2x) + sin(4x)]`

Explanation:

While initially appearing to be a really annoying integral, we can actually exploit trig identities to break this integral down into a series of simple integrals that we are more familiar with.

The identity we will be using is:
`cos^2(x) = (1 + cos(2x))/2`

This lets us manipulate our equation as such:
`int cos^4(x)dx = int (1 + cos(2x))/2 * (1 + cos(2x))/2dx`
`= 1/4 int (1 + cos(2x))(1 + cos(2x))dx`
`=1/4int (1+ 2cos(2x) + cos^2(2x))dx`

We can now apply our rule again to eliminate the cos^2(2x) inside the parenthetical:
`1/4int (1+ 2cos(2x) + cos^2(2x))dx`
`= 1/4int (1+ 2cos(2x) + (1 + cos(4x))/2)dx`
`= 1/8int (2+ 4cos(2x) + 1 + cos(4x))dx`
`= 1/8int (3+ 4cos(2x) + cos(4x))dx`

Now we actually have a fairly simple integration problem, we can distribute the integral into our parenthetical so that:
`= 1/8[int3dx + 4intcos(2x)dx + intcos(4x)dx]`

Each of these trig integrals is handled with the simple rule that `int cos(ax)dx = 1/a sin(ax)`.

Thus,
`= 1/8[3x + 2 sin(2x) + 1/4 sin(4x)]`
`= 1/32[12x + 8sin(2x) + sin(4x)]`