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Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Capitolo 7, Problema 7.1.59

29–62. Integrals Evaluate the following integrals. Include absolute values only when needed.


∫₀ˡⁿ ² (e^{3x} − e^{−3x}) / (e^{3x} + e^{−3x}) dx

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1
Recognize that the integral involves the expression \( \frac{e^{3x} - e^{-3x}}{e^{3x} + e^{-3x}} \), which resembles the hyperbolic tangent function \( \tanh(u) = \frac{e^{u} - e^{-u}}{e^{u} + e^{-u}} \). Here, identify \( u = 3x \).
Rewrite the integral using the substitution: \( \frac{e^{3x} - e^{-3x}}{e^{3x} + e^{-3x}} = \tanh(3x) \). So the integral becomes \( \int_0^{\ln 2} \tanh(3x) \, dx \).
Use substitution to simplify the integral: let \( t = 3x \), then \( dt = 3 \, dx \) or \( dx = \frac{dt}{3} \). Change the limits accordingly: when \( x=0 \), \( t=0 \); when \( x=\ln 2 \), \( t=3 \ln 2 \).
Rewrite the integral in terms of \( t \): \( \int_0^{3 \ln 2} \tanh(t) \cdot \frac{1}{3} \, dt = \frac{1}{3} \int_0^{3 \ln 2} \tanh(t) \, dt \).
Recall the integral formula for hyperbolic tangent: \( \int \tanh(t) \, dt = \ln|\cosh(t)| + C \). Use this to write the antiderivative and then apply the limits to express the definite integral.

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