Question

What is the antiderivative of `sin(4t)`?

Answer 1

It is: `-1/4cos(4t) + C`

Explanation:

There are many method and notations that may or may not have been introduced to students when this question is asked. So the best I can do is to choose one or two and explain usung those:

The antiderivative of `sin(4t)` is, of course, a function whose derivative is `sin(4t)`

On eway to proceed is to reason as follow:

I know that the derivative with respect to `t` of `cosu` can be found using the chain rule:

`d/dt(cosu) = -sinu (du)/(dt)`

With `u = 4t` we would have:

`d/dt(cos(4t)) = -4sin(4t)` which is not quite what we want.

But a constant multiple just stays out front when we differentiate, so if we multiplied by `-1/4` we would get:

`d/dt(-1/4cos(4t)) = -1/4 [-4sin(4t)] = sin(4t)`

So `-1/4cos(4t)` is an antiderivative of `sin(4t)`.

Since any function with the same derivative differs from this by only a constant (an important consequence of the Mean Value Theorem),

we conclude that every antiderivative of `sin(4t)`.can be written in the form:

`-1/4cos(4t)+C` for some constant `C`.

(If you haven't already, you probably will be introduced to the standard mechanics of "u-substitution", but that's just mechanics for the reasoning used above.)