# How do you integrate #int 1/sqrt(x^2+1)# by trigonometric substitution?

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To integrate (\int \frac{1}{\sqrt{x^2 + 1}}) by trigonometric substitution, let (x = \tan(\theta)), then (\sqrt{x^2 + 1} = \sqrt{\tan^2(\theta) + 1} = \sec(\theta)). Differentiate both sides to find (dx). Substituting these expressions into the integral and simplifying using trigonometric identities, you will get (\int \sec^2(\theta) d\theta), which is straightforward to integrate as it equals (\tan(\theta) + C). Finally, substitute back (x = \tan(\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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