Question

How do you find the integral of `sin(2 pi t) dt`?

Answer 1

`(-cos(2pit))/(2pi)+C`

Explanation:

We have:

`intsin(2pit)dt`

Substitute:

`u=2pit" "=>" "du=2pidt`

In order to have `du` in our integral expression, we must multiply the inside by `2pi`. However, we also must balance this by multiplying the outside by `1"/"2pi`.

`=1/(2pi)intsin(2pit)*2pidt`

Substituting, we obtain:

`=1/(2pi)intsin(u)du`

This is a common integral:

`=-1/(2pi)cos(u)+C`

Back substituting for `u`:

`=(-cos(2pit))/(2pi)+C`