Let be the function we are integrating. Let denote the region for which and that contains the value of where . Then we note that and the sines of the integral alternate, so we can just calculate the first one and everything else will be bounded (in particular by ). With a bit of Taylor approximation, we can bound the size of each by , and noting that is always positive for , we can replace the with and then bound by . This gives us the bound we claimed above and we are done.

## Solution (final part)

Let be the function we are integrating. Let denote the region for which and that contains the value of where . Then we note that and the sines of the integral alternate, so we can just calculate the first one and everything else will be bounded (in particular by ). With a bit of Taylor approximation, we can bound the size of each by , and noting that is always positive for , we can replace the with and then bound by . This gives us the bound we claimed above and we are done.

Jacob Steinhardt