Question

How do you find the integral of `cos^5(3x)dx`?

Answer 1

`=1/3(sin(3x)-2(sin(3x))^3/3+(sin(3x))^5/5)+C`

Explanation:

First let `u = 3x, du = 3dx` to simplify the integral. Then we have:

`1/3` `intcos^5(u)du`

Since the power of cos(u) is odd, we can use the procedure of saving one `cos(u)` and using the identity `cos(u)^2=(1-sin^2(u))`

`=>` `1/3` `int(1-sin^2(u))^2cos(u)du`

Remember we made it `(1-sin^2(u))^2` all squared because our original function is `cos(u)^4`

After this we need to use another substitution. For this time: let `z = sin(u)`, `dz = cos(u)du`

`=>` `1/3` `int(1-z^2)^2dz`

Now it's a simple polynomial integral we can evaluate it straightforwardly

`=` `1/3` `int(1-2z^2+z^4)dz`

`=1/3(z-2z^3/3+z^5/5)+C`

We started with `x` we substitute back to `x`

`=1/3(sin(u)-2(sin(u))^3/3+(sin(u))^5/5)+C`
`=1/3(sin(3x)-2(sin(3x))^3/3+(sin(3x))^5/5)+C`