Question

How do you find the antiderivative of `int cos(pit)cos(sin(pit))dt`?

Answer 1

`1/pisin(sin(pit))+C`

Explanation:

Let's try to simplify the trig function within the trig function by letting `u` be the inside function, that is, `u=sin(pit)`. Differentiating this shows that `du=picos(pit)dt`. (Recall to use the chain rule.)

We currently have `cos(pit)dt` in the integral, so we need the factor of `pi`.

`intcos(pit)cos(sin(pit))dt=1/piintcos(sin(pit))*picos(pit)dt`

Substituting in:

`=1/piintcos(u)du`

This is a common integral:

`=1/pisin(u)+C`

Substituting back in `u=sin(pit)`:

`=1/pisin(sin(pit))+C`