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Ch. 8 - Techniques of Integration
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Capitolo 8, Problema 8.5.74a

Evaluate ∫ sec θ dθ by:
a. Multiplying by (sec θ + tan θ) / (sec θ + tan θ) and then using a u-substitution.

Guida verificata passo dopo passo
1
Start with the integral: \(\int \sec \theta \, d\theta\).
Multiply the integrand by \(\frac{\sec \theta + \tan \theta}{\sec \theta + \tan \theta}\), which is equivalent to multiplying by 1, so the integral becomes \(\int \sec \theta \cdot \frac{\sec \theta + \tan \theta}{\sec \theta + \tan \theta} \, d\theta\).
Rewrite the numerator as \(\sec \theta (\sec \theta + \tan \theta) = \sec^2 \theta + \sec \theta \tan \theta\), so the integral is now \(\int \frac{\sec^2 \theta + \sec \theta \tan \theta}{\sec \theta + \tan \theta} \, d\theta\).
Let \(u = \sec \theta + \tan \theta\). Then compute \(\frac{du}{d\theta}\) by differentiating \(u\) with respect to \(\theta\): \(\frac{du}{d\theta} = \sec \theta \tan \theta + \sec^2 \theta\).
Notice that the numerator of the integrand matches \(\frac{du}{d\theta}\), so rewrite the integral as \(\int \frac{du}{u}\), which can be integrated using the natural logarithm function.

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Multiplying by a Conjugate Expression

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