# How do you integrate #int dx/(x^2+25)# using trig substitutions?

Let

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Please see below.

A good way to sort out trig substitutions is to recall the pythagorean identities in trig.

Eventually (perhaps by going through the list of pythagorean identities) we recall that

So

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To integrate ( \int \frac{dx}{x^2 + 25} ) using trigonometric substitution, we let ( x = 5 \tan(\theta) ), which implies ( dx = 5 \sec^2(\theta) d\theta ). Substituting these expressions into the integral, we get:

[ \int \frac{5 \sec^2(\theta) d\theta}{25\tan^2(\theta) + 25} ]

Simplify the expression:

[ \int \frac{5 \sec^2(\theta) d\theta}{25(\tan^2(\theta) + 1)} ] [ = \int \frac{5 \sec^2(\theta) d\theta}{25\sec^2(\theta)} ] [ = \frac{1}{5} \int d\theta ]

Integrating ( d\theta ) gives:

[ \frac{1}{5} \theta + C ]

Finally, we substitute back for ( \theta ) using the original substitution ( x = 5 \tan(\theta) ). Since ( \tan(\theta) = \frac{x}{5} ), we can find ( \theta ) as ( \arctan(\frac{x}{5}) ).

Therefore, the final result is:

[ \frac{1}{5} \arctan\left(\frac{x}{5}\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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