How do you find the antiderivative of #1−cos(4x)#?
Let's say that, as an illustration unrelated to the issue:
This would imply that:
And in the end, that
due to the chain rule.
If so, the following steps should be taken when resolving the issue:
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To find the antiderivative of (1  \cos(4x)), we can use the following steps:

Recognize that (\int 1  \cos(4x) , dx) involves two separate integrals: one for (1) and one for (\cos(4x)).

The antiderivative of (1) with respect to (x) is simply (x).

To find the antiderivative of (\cos(4x)), we use the substitution method. Let (u = 4x), then (du = 4 , dx), or equivalently, (dx = \frac{1}{4} , du).

Now, we rewrite the integral in terms of (u): (\int \cos(u) \cdot \frac{1}{4} , du).

The antiderivative of (\cos(u)) is (\sin(u)). So, the antiderivative of (\cos(4x)) with respect to (u) is (\sin(u)).

Finally, we substitute (u = 4x) back into the expression, giving us (\frac{1}{4} \sin(4x)).

Combining the antiderivatives of (1) and (\cos(4x)), we get the antiderivative of (1  \cos(4x)) as (x + \frac{1}{4} \sin(4x) + C), where (C) is the constant of integration.
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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