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

Question

Emily Ammerman · Accepted Answer

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

The best I can do is pick one or two and explain using those of the numerous methods and notations that may or may not have been introduced to students at the time this question was posed:

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

The best course of action is to think like this:

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)#.

As a result of the Mean Value Theorem, any function with the same derivative deviates from this by only a constant.

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 will likely be introduced to the standard "u-substitution" mechanics; however, that is merely the mechanics behind the previously stated reasoning.)

Elijah Abram · Answer

undefined The antiderivative of sin(4t) is -(1/4)cos(4t) + C, where C is the constant of integration.