# How do you integrate #int x/sqrt(x^2-25)# by trigonometric substitution?

You do not need a trigonometric substitution to solve this integral.

Substituting:

you have:

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To integrate (\frac{x}{\sqrt{x^2 - 25}}) by trigonometric substitution, you can use the substitution (x = 5\sec(\theta)), where (\theta) is the angle in a right triangle with adjacent side (5) and hypotenuse (x).

With this substitution, you can express (dx) as (5\sec(\theta)\tan(\theta)d\theta) and (\sqrt{x^2 - 25}) as (5\tan(\theta)). Substituting these expressions into the integral, you get:

[ \int \frac{x}{\sqrt{x^2 - 25}} dx = \int \frac{5\sec(\theta)\tan(\theta)}{5\tan(\theta)} \times 5\sec(\theta)\tan(\theta) d\theta ]

[ = \int 5\sec^2(\theta) d\theta ]

Now, integrate (\sec^2(\theta)) with respect to (\theta) to get:

[ = 5\tan(\theta) + C ]

Finally, revert back to the variable (x) using the relation (x = 5\sec(\theta)) and (\tan(\theta) = \frac{\sqrt{x^2 - 25}}{5}):

[ = 5\tan(\theta) + C = 5\left(\frac{\sqrt{x^2 - 25}}{5}\right) + C = \sqrt{x^2 - 25} + C ]

So, the integral of (\frac{x}{\sqrt{x^2 - 25}}) by trigonometric substitution is (\sqrt{x^2 - 25} + C), where (C) is the constant of integration.

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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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