Question

How do you integrate `int cos^3(x/3)dx`?

Answer 1

`3sin(x/3)-sin^3(x/3)+C`

Explanation:

First let `t=x/3`. This implies that `dt=1/3dx`. We then see that:

`intcos^3(x/3)dx=3intcos^3(x/3)1/3dx=3intcos^3(t)dt`

To do this, split up `cos^3(t)` into `cos^2(t)cos(t)` and then rewrite `cos^2(t)` using the identity `sin^2(theta)+cos^2(theta)=1=>cos^2(theta)=1-sin^2(theta)`.

`3intcos^3(t)dt=3intcos^2(t)cos(t)dt=3int(1-sin^2(t))cos(t)dt`

Now let `s=sin(t)`, so `ds=cos(t)dt`. Luckily we already have this in the integrand!

`3int(1-sin^2(t))cos(t)dt=3int(1-s^2)ds`

Integrating term by term using `ints^nds=s^(n+1)/(n+1)` where `n!=-1`.

`3int(1-s^2)ds=3(s-s^3/3)=3s-s^3`

From `s=sin(t)` and `t=x/3` we see that `s=sin(x/3)`. Also add the constant of integration.

`3s-s^3=3sin(x/3)-sin^3(x/3)+C`