Domanda del libro di testo
7–28. Derivatives Evaluate the following derivatives.
d/dx (ln (cos² x))
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Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243
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TOC.tableOfContents
- Ch. 1 - Functions210
- Ch. 2 - Limits371
- Ch. 3 - Derivatives633
- Ch. 4 - Applications of the Derivative412
- Ch. 5 - Integration328
- Ch. 6 - Applications of Integration331
- Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions177
- Ch. 8 - Integration Techniques394
- Ch. 9 - Differential Equations179
- Ch. 10 - Sequences and Infinite Series339
- Ch. 11 - Power Series218
- Ch.12 - Parametric and Polar Curves215
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
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Briggs 3rd EditionCh. 7 - Logarithmic, Exponential Functions, and Hyperbolic FunctionsProblema 7.1.33
Briggs 3rd EditionCh. 7 - Logarithmic, Exponential Functions, and Hyperbolic FunctionsProblema 7.1.33Capitolo 7, Problema 7.1.33
29–62. Integrals Evaluate the following integrals. Include absolute values only when needed.
∫ e^{2x} / (4 + e^{2x}) dx
Guida verificata passo dopo passo1
Identify the integral to solve: \(\int \frac{e^{2x}}{4 + e^{2x}} \, dx\).
Use substitution to simplify the integral. Let \(u = e^{2x}\). Then, compute \(du\): since \(u = e^{2x}\), differentiate to get $du = 2 e^{2x} dx$, or equivalently, \(dx = \frac{du}{2u}\).
Rewrite the integral in terms of \(u\): substitute \(e^{2x} = u\) and \(dx = \frac{du}{2u}\) into the integral to get \(\int \frac{u}{4 + u} \cdot \frac{du}{2u} = \int \frac{1}{4 + u} \cdot \frac{1}{2} du = \frac{1}{2} \int \frac{1}{4 + u} du\).
Integrate \(\frac{1}{4 + u}\) with respect to \(u\): recall that \(\int \frac{1}{a + u} du = \ln|a + u| + C\), so here the integral becomes \(\frac{1}{2} \ln|4 + u| + C\).
Substitute back \(u = e^{2x}\) to express the answer in terms of \(x\): the integral is \(\frac{1}{2} \ln|4 + e^{2x}| + C\).
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Integration by Substitution
Integration by substitution is a method used to simplify integrals by changing variables. It involves identifying a part of the integrand as a new variable, which transforms the integral into a simpler form. This technique is especially useful when the integral contains a composite function.
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Substitution With an Extra Variable
Exponential Functions
Exponential functions have the form e^{kx}, where e is Euler's number and k is a constant. Their derivatives and integrals often involve the same exponential function, making them straightforward to handle in calculus. Recognizing these functions helps in applying substitution effectively.
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6:13Exponential Functions
Handling Absolute Values in Integrals
Absolute values appear in integrals when dealing with logarithmic functions resulting from integration. They ensure the argument of the logarithm is positive, maintaining the function's domain. Including absolute values only when necessary avoids incorrect expressions and ensures the integral's correctness.
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Percorso guidato
05:59Initial Value Problems Example 2
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Domanda del libro di testo
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