# How do you integrate #int 1/sqrt(x^2-6) # using trigonometric substitution?

So substituting the above values for their respective functions, we get the integral of the said main function.

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To integrate (\int \frac{1}{\sqrt{x^2 - 6}}) using trigonometric substitution, we follow these steps:

- Recognize that the expression under the square root resembles a trigonometric identity, particularly (\sec^2(\theta) - 1 = \tan^2(\theta)).
- Substitute (x = \sqrt{6}\sec(\theta)), which implies (dx = \sqrt{6}\sec(\theta)\tan(\theta)d\theta).
- Rewrite the integral in terms of (\theta) using the substitution.
- Solve the integral in terms of (\theta).
- Substitute back (x) in terms of (\theta).
- Simplify the result.

After substitution, the integral becomes:

[ \int \frac{1}{\sqrt{6\sec^2(\theta) - 6}} \cdot \sqrt{6}\sec(\theta)\tan(\theta)d\theta ]

Simplify this expression to:

[ \int \frac{1}{\sqrt{6\tan^2(\theta)}} \cdot \sqrt{6}\sec(\theta)\tan(\theta)d\theta ]

This simplifies further to:

[ \int \frac{1}{\sqrt{6}\tan(\theta)} \cdot \sqrt{6}\sec(\theta)\tan(\theta)d\theta ]

The (\sqrt{6}\tan(\theta)) terms cancel out, leaving:

[ \int d\theta ]

This integral evaluates to (\theta + C). Finally, substitute back (x = \sqrt{6}\sec(\theta)), so (\theta = \arccos\left(\frac{x}{\sqrt{6}}\right)).

Thus, the result is (\arccos\left(\frac{x}{\sqrt{6}}\right) + 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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