# How do you find the antiderivative of # cosx^6#?

Let's examine each integral separately.

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To find the antiderivative of ( \cos^6(x) ), you can use the reduction formula for powers of cosine, which is derived from integration by parts.

[ \int \cos^n(x) , dx = \frac{\cos^{n-1}(x) \sin(x)}{n} + \frac{n-1}{n} \int \cos^{n-2}(x) , dx ]

Using this reduction formula, for ( \int \cos^6(x) , dx ), we can start by letting ( n = 6 ):

[ \int \cos^6(x) , dx = \frac{\cos^5(x) \sin(x)}{6} + \frac{5}{6} \int \cos^4(x) , dx ]

Now, for the remaining integral ( \int \cos^4(x) , dx ), we can again apply the reduction formula:

[ \int \cos^4(x) , dx = \frac{\cos^3(x) \sin(x)}{4} + \frac{3}{4} \int \cos^2(x) , dx ]

Since ( \cos^2(x) = \frac{1 + \cos(2x)}{2} ), we substitute this back into the equation:

[ \int \cos^4(x) , dx = \frac{\cos^3(x) \sin(x)}{4} + \frac{3}{8}x + \frac{3}{8}\sin(2x) + C ]

Now, substituting this back into the original integral, we get:

[ \int \cos^6(x) , dx = \frac{\cos^5(x) \sin(x)}{6} + \frac{5}{6} \left( \frac{\cos^3(x) \sin(x)}{4} + \frac{3}{8}x + \frac{3}{8}\sin(2x) \right) + C ]

This can be further simplified if needed.

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When evaluating a one-sided 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 one-sided 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 one-sided 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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