Question

Evaluate the following integrals. $$\int_{0}^{\ln 2} \frac{e^{3 x}-e^{-3 x}}{e^{3 x}+e^{-3 x}} d x$$

Short answer

Question: Evaluate the given integral: $$\int_{0}^{\ln 2}\frac{e^{3x}-e^{-3x}}{e^{3x}+e^{-3x}} dx$$ Answer: $$\frac{1}{6}(63-2\ln(65)+2\ln(2))$$

Step-by-step solution

Simplify the integrand

Begin by finding a common denominator for the terms in the numerator and denominator. Multiply the numerator and denominator by \(e^{3x}\) to obtain: $$\int_{0}^{\ln 2}\frac{(e^{3x}-e^{-3x})(e^{3x})}{(e^{3x}+e^{-3x})(e^{3x})} dx$$ This simplifies to: $$\int_{0}^{\ln 2}\frac{e^{6x}-1}{e^{6x}+1} dx$$

Integrate the simplified integrand

Now, we need to integrate this integrand. In order to solve this we perform a substitution. Set \(u=e^{6x}\), then \(\frac{du}{dx}=6e^{6x}\), which would lead us to: $$\int\frac{u-1}{u+1}\frac{1}{6}du$$ Now we have to change our limits of integration. When \(x=0\), \(u=e^{6(0)}=1\). When \(x=\ln 2\), \(u=e^{6(\ln 2)}=2^6=64\). Thus, our integral is: $$\frac{1}{6}\int_1^{64}\frac{u-1}{u+1} du$$

Performintegration and evaluate the definite integral

Integrating the quotient, we have \(1/6\) times the integral of the sum of two functions: $$\frac{1}{6}\int_1^{64}\left(1-\frac{2}{u+1}\right) du$$ The first part can be integrated, and we are left with: $$\frac{1}{6}\left[\int_1^{64} du -2 \int_1^{64}\frac{1}{u+1} du\right]$$ Integrating each part: $$\frac{1}{6}\left[(u-2\ln(u+1))\right]_1^{64}$$

Apply the limits of integration

Now plug in the bounds and simplify and calculate the integral: $$\frac{1}{6}\left[(64-2\ln(65))-(1-2\ln(2))\right]$$ So our final answer is: $$\frac{1}{6}(63-2\ln(65)+2\ln(2))$$

Key concepts

Substitution Method

The Substitution Method is a powerful technique in calculus, particularly useful in evaluating integrals. It allows us to transform an integral into an easier form. The idea is to substitute part of the integral with a new variable, simplifying the integrand. In this problem, the substitution involved setting \( u = e^{6x} \), which transformed a complex expression into a simpler one. This makes it easier to integrate. We also have to modify the differential from \( dx \) to \( du \). The derivative \( du = 6e^{6x}dx \) means we substitute \( dx \) with \( \frac{1}{6}du \). Remember, the key steps in using substitution are:

• Choose a substitution \( u \) that simplifies the integral.
• Express \( dx \) in terms of \( du \).
• Change the limits of integration to values corresponding to \( u \) if dealing with a definite integral.

Substitution not only simplifies the integrand but often converts a complicated definite integral into one with easier bounds.

Definite Integrals

Definite integrals represent the area under a curve within a specific interval \( [a, b] \). Unlike indefinite integrals, definite integrals produce a numerical value, not a function. The limits of integration are crucial. They define where the integration starts and ends. In our example, the original limits were \( x = 0 \) and \( x = \ln 2 \). But after substitution, these were transformed with \( u = e^{6x} \) to new limits \( u = 1 \) and \( u = 64 \). Calculating a definite integral involves evaluating the antiderivative at these two points and subtracting:

• Determine \( F(b) - F(a) \), where \( F(x) \) is the antiderivative of \( f(x) \).

Always pay attention to correctly substituting the limits if substitution is involved. This is vital to ensure the numerical result is accurate.

Exponential Functions

Exponential functions feature expressions where the variable is an exponent, like \( e^{3x} \) or \( e^{-3x} \). These functions grow or decay rapidly. In the integral problem, exponential functions in the numerator and denominator were combined. Simplifying such expressions is often necessary because exponential growth can make straightforward calculations cumbersome. Multiplying polynomials of exponentials, as done by \( e^{3x}(e^{3x} - e^{-3x}) \), transforms the expression into a polynomial, making it possible to use substitution. Whenever dealing with exponential functions:

• Consider their properties of growth and decay.
• Look for opportunities to simplify the expression by combining or factoring terms.
• Understand that these functions often require logarithmic or substitution techniques for integration.

Mastering exponential functions is essential in calculus as they frequently appear in real-world applications, from population models to financial calculations.