# How do you integrate #int 1/sqrt(e^(2x) -9)dx# using trigonometric substitution?

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To integrate (\int \frac{1}{\sqrt{e^{2x} - 9}} , dx) using trigonometric substitution, you can let (e^x = 3\sec(\theta)). Then, (dx = 3\sec(\theta)\tan(\theta) , d\theta). After substitution, simplify and integrate, the integral becomes: (\int \frac{3\sec(\theta)\tan(\theta)}{\sqrt{9\sec^2(\theta) - 9}} , d\theta). Simplify the expression inside the square root to obtain (\int \frac{3\sec(\theta)\tan(\theta)}{3\tan(\theta)} , d\theta). This simplifies to (\int \sec(\theta) , d\theta). Integrating (\sec(\theta)) yields (\ln|\sec(\theta) + \tan(\theta)| + C). Finally, substitute back for (\theta) using (e^x = 3\sec(\theta)) to obtain the final answer: (\ln|e^x + \sqrt{e^{2x} - 9}| + 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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