# What is #int csc^4(x) cot^6(x) dx #?

As a result, the integral becomes:

Adding term by term, this provides us with

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To integrate (\int \csc^4(x) \cot^6(x) , dx), use the trigonometric identity (\csc^2(x) = \cot^2(x) + 1).

First, rewrite the integral using this identity: [ \int \csc^4(x) \cot^6(x) , dx = \int \csc^2(x) (\cot^2(x) \cot^4(x)) , dx ]

Substitute (\csc^2(x) = \cot^2(x) + 1): [ \int (\cot^2(x) + 1) (\cot^2(x) \cot^4(x)) , dx = \int (\cot^6(x) + \cot^2(x) \cot^4(x)) , dx ]

Now, let's integrate each term separately:

- (\int \cot^6(x) , dx):

Rewrite (\cot^6(x)) as (\cot^2(x) \cdot \cot^4(x)): [ \int \cot^6(x) , dx = \int \cot^2(x) \cdot \cot^4(x) , dx ]

Let (u = \cot(x)) and (du = -\csc^2(x) , dx). The integral becomes: [ -\int u^4 , du = -\frac{1}{5}u^5 + C = -\frac{1}{5}\cot^5(x) + C ]

- (\int \cot^2(x) \cot^4(x) , dx):

Let (w = \cot(x)) and (dw = -\csc^2(x) , dx). The integral becomes: [ -\int w^3 , dw = -\frac{1}{4}w^4 + C = -\frac{1}{4}\cot^4(x) + C ]

Combine the results: [ \int \csc^4(x) \cot^6(x) , dx = -\frac{1}{5}\cot^5(x) - \frac{1}{4}\cot^4(x) + C ]

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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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