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Ch. 4 - Applications of Derivatives
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Capitolo 4, Problema 4.7.56

Finding Indefinite Integrals


In Exercises 17–56, find the most general antiderivative or indefinite integral. You may need to try a solution and then adjust your guess. Check your answers by differentiation.


∫csc θ/(csc θ − sin θ) dθ

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Start by rewriting the integrand to express everything in terms of sine and cosine functions. Recall that \(\csc \theta = \frac{1}{\sin \theta}\). So rewrite the integral as \(\int \frac{\frac{1}{\sin \theta}}{\frac{1}{\sin \theta} - \sin \theta} \, d\theta\).
Simplify the denominator by finding a common denominator inside it: \(\frac{1}{\sin \theta} - \sin \theta = \frac{1 - \sin^2 \theta}{\sin \theta}\). Use the Pythagorean identity \(1 - \sin^2 \theta = \cos^2 \theta\) to rewrite the denominator as \(\frac{\cos^2 \theta}{\sin \theta}\).
Now the integrand becomes \(\frac{\frac{1}{\sin \theta}}{\frac{\cos^2 \theta}{\sin \theta}} = \frac{1}{\sin \theta} \times \frac{\sin \theta}{\cos^2 \theta} = \frac{1}{\cos^2 \theta} = \sec^2 \theta\).
Recognize that the integral simplifies to \(\int \sec^2 \theta \, d\theta\). Recall the antiderivative of \(\sec^2 \theta\) is \(\tan \theta + C\).
Write the most general antiderivative as \(\tan \theta + C\), where \(C\) is the constant of integration. To verify, differentiate \(\tan \theta + C\) and confirm you get back the original integrand.

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