To solve the integral:

$\int \sec(x) \, dx$

we use a standard method involving a clever multiplication and division by $(\sec(x) + \tan(x))$. Here's the step-by-step process:

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Step 1: Multiply and divide by $(\sec(x) + \tan(x))$

$\int \sec(x) \, dx = \int \frac{\sec(x)(\sec(x) + \tan(x))}{\sec(x) + \tan(x)} \, dx$

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Step 2: Use substitution

Let: $u = \sec(x) + \tan(x) \quad \text{so that} \quad du = (\sec(x)\tan(x) + \sec^2(x)) \, dx$

The numerator $(\sec(x)\tan(x) + \sec^2(x))$ matches exactly with $du$. Hence, the integral becomes: $\int \sec(x) \, dx = \int \frac{1}{u} \, du$

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Step 3: Solve the resulting integral

$\int \frac{1}{u} \, du = \ln|u| + C$

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Step 4: Substitute back $u = \sec(x) + \tan(x)$

$\ln|u| + C = \ln|\sec(x) + \tan(x)| + C$

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Final Answer:

$\int \sec(x) \, dx = \ln|\sec(x) + \tan(x)| + C$

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Would you like to explore how this solution applies to specific bounds or see a detailed derivation of substitution?

Related Questions:

1. How is the substitution $u = \sec(x) + \tan(x)$ derived?
2. Why is the absolute value necessary in the logarithm?
3. Can this integral be expressed differently using trigonometric identities?
4. What happens if we compute the definite integral of $\sec(x)$ over a specific interval?
5. Can this solution be extended to integrals involving $\sec^n(x)$?

Tip:

For functions like $\sec(x)$ and $\csc(x)$, their integrals often involve logarithms and clever substitutions based on trigonometric identities.