# How do you integrate #int dx/sqrt(16+x^2)^2# by trigonometric substitution?

The square root and the exponent cancel, leaving just:

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To integrate ( \int \frac{dx}{\sqrt{16+x^2}^2} ) using trigonometric substitution, we can let ( x = 4\tan(\theta) ), where ( \theta ) is the trigonometric angle.

Then we find ( dx ) by differentiating ( x = 4\tan(\theta) ) with respect to ( \theta ).

Next, we express ( \sqrt{16+x^2} ) in terms of ( \tan(\theta) ) using the trigonometric identity ( \sec^2(\theta) = 1 + \tan^2(\theta) ).

After that, we substitute ( x ) and ( dx ) in the integral with expressions involving ( \theta ).

We simplify the integrand using the substitution and then integrate with respect to ( \theta ).

Finally, we substitute back ( x ) in terms of ( \theta ) to express the result in terms of ( x ).

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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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