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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Capitolo 7, Problema 7.7.73a

Evaluate the integrals in Exercises 67–74 in terms of
a. inverse hyperbolic functions.
73. ∫(from 0 to π)cos(x)dx/√(1+sin²x)

Guida verificata passo dopo passo
1
Recognize that the integral is \( \int_0^{\pi} \frac{\cos(x)}{\sqrt{1 + \sin^2(x)}} \, dx \). Notice the presence of \( \cos(x) \) and \( \sin(x) \) inside the integral, which suggests a substitution involving \( \sin(x) \).
Use the substitution \( t = \sin(x) \). Then, \( dt = \cos(x) \, dx \). This substitution will simplify the integral because \( \cos(x) \, dx \) can be replaced by \( dt \), and the limits of integration will change accordingly.
Change the limits of integration from \( x \) to \( t \): when \( x = 0 \), \( t = \sin(0) = 0 \); when \( x = \pi \), \( t = \sin(\pi) = 0 \). So the integral becomes \( \int_0^0 \frac{1}{\sqrt{1 + t^2}} \, dt \).
Notice that the new integral has the same upper and lower limits, which means the integral evaluates to zero. However, to understand the integral in terms of inverse hyperbolic functions, consider the indefinite integral \( \int \frac{1}{\sqrt{1 + t^2}} \, dt \).
Recall that \( \int \frac{1}{\sqrt{1 + t^2}} \, dt = \sinh^{-1}(t) + C \), where \( \sinh^{-1}(t) \) is the inverse hyperbolic sine function. This connects the integral to inverse hyperbolic functions as requested.

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