# How do you integrate #int 1/(xsqrt(25-x^2))# by trigonometric substitution?

Rearrange. Substitute. Voila!

Then we set:

and hence

Our integral with the variable t now reads:

Whence we rearrange and get:

which is:

And this integral is solved in the following way:

We make a small but cunning rearrangement, namely:

So that our integral now reads :

Which is just :

The one half comes about from the partial fraction decomposition. This is just :

However this is a solution expressed through the variable u, but back substitution is done trivially if needed.

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To integrate ( \int \frac{1}{x\sqrt{25-x^2}} ) by trigonometric substitution, use the substitution ( x = 5\sin(\theta) ). Then, ( dx = 5\cos(\theta) d\theta ). Substitute these into the integral, simplify, and integrate using trigonometric identities.

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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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