Question

How do you find the antiderivative of `int cos^3x dx`?

Answer 1

`sinx - 1/3sin^3x`

Explanation:

Change `cos^3x`:

`cos^3x = cos^2x*cosx = (1-sin^2)*cosx`

So we can write it as

`int cosx*(1-sin^2x) dx`

`int cosx dx - int cosxsin^2x dx`

`sinx - int cosxsin^2x dx`

Then we can solve using u substitution

`u = sinx` and `du = cosxdx`

`sinx - int u^2 du` (substituting u)

`sinx - 1/3u^3`

`sinx - 1/3sin^3x` (replug in x)

How to do these problems

I just learned how to solve these problems an hour ago at school! Basically, the gist is to use trig in order to "make" the derivative of one of the parts of the function. In this case, we only had cosine, so we had to make a sine component in order to use u substitution.