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Chapter 6: Problem 67

Miscellaneous integrals 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

Expert verified
Question: Evaluate the definite integral \(\int_{0 }^{\ln 2}\frac{e^{3 x}-e^{-3 x}}{e^{3 x}+e^{-3 x}} dx\). Answer: \(3\ln|\cosh(3\ln 2)|\)

Step by step solution

01

Find the antiderivative

To find the antiderivative of \(\frac{e^{3x}-e^{-3x}}{e^{3x}+e^{-3x}}\), firstly we observe that the function has the form of \(\frac{e^{ax}-e^{-ax}}{e^{ax}+e^{-ax}}\), where \(a=3\). This function can be rewritten as \(\frac{2\sinh(ax)}{2\cosh(ax)}\), which simplifies to \(\tanh(ax)\). Thus, we need to find the antiderivative of \(3\tanh(3x)\): \(\int 3\tanh(3x) dx\) Now, we know that the derivative of \(\cosh(x)\) is \(\sinh(x)\), and the derivative of \(\sinh(x)\) is \(\cosh(x)\). Therefore, the antiderivative of \(\tanh(x)\) is \(\ln| \cosh(x) | + C\), where C is the constant of integration. So, the antiderivative of \(3\tanh(3x)\) is \(3\ln|\cosh(3x)| + C\).
02

Evaluate the antiderivative at the upper and lower boundaries

Now, we will evaluate the antiderivative at the upper boundary \(\ln 2\) and subtract the antiderivative evaluated at the lower boundary 0: \(3\ln|\cosh(3\ln 2)| - 3\ln|\cosh(3 \cdot 0)|\) \(\cosh(3 \cdot 0)\) is equal to \(\cosh(0)\), and \(\cosh(0)\) equals 1. Therefore, the second expression is 3 times the natural logarithm of 1: \(3\ln|1| = 0\). So, the remaining expression after subtracting the evaluation at the lower boundary is just: \(3\ln|\cosh(3\ln 2)|\)
03

Simplify the result

Lastly, we can simplify the result: \(\int_{0}^{\ln 2} \frac{e^{3 x}-e^{-3 x}}{e^{3 x}+e^{-3 x}} d x = 3\ln|\cosh(3\ln 2)|\) This is the final answer for evaluating the definite integral.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integration Techniques
Understanding integration techniques is vital for computing integrals accurately. The example provided involves the use of substitution and recognition of hyperbolic function identities. To tackle integrals involving complex expressions like \(e^{ax} - e^{-ax}\) divided by \(e^{ax} + e^{-ax}\), spotting that the expression resembles a hyperbolic function is a key step.

Substitution can simplify the integral, for example, by setting \(u = ax\) and then integrating with respect to \(u\). This not only makes the integral more manageable but also allows the use of known antiderivatives for hyperbolic functions. Being familiar with integration techniques ensures versatility when faced with a variety of integral problems.
Hyperbolic Functions
The hyperbolic functions like \(\sinh(x)\), \(\cosh(x)\), and \(\tanh(x)\) are analogs of the trigonometric functions but for hyperbolas, as trigonometric functions are to the unit circle.

For instance, \(\sinh(x) = \frac{e^x - e^{-x}}{2}\) and \(\cosh(x) = \frac{e^x + e^{-x}}{2}\) are crucial to understand as they form the hyperbolic counterpart of sine and cosine functions.

Relationship to Exponential Functions

Notice how each hyperbolic function is expressed in terms of exponential functions, which links the concepts of hyperbolic and exponential functions. Recognizing these relationships makes it easier to integrate functions that seem complicated at first glance.
Antiderivative Calculations
Performing antiderivative calculations is a fundamental aspect of finding definite integrals. In the given example, identifying the antiderivative of \(3\tanh(3x)\) is central to solving the integral. It's important to draw on known derivatives of hyperbolic functions, as it's known that the antiderivative of \(\tanh(x)\) is \(\ln |\cosh(x)| + C\).

Once the antiderivative is found, evaluating it at the bounds of integration completes the process. Understanding the properties of hyperbolic functions, such as \(\cosh(0) = 1\), simplifies the evaluation step. Mastery of these calculations enhances a student's ability to tackle a wide array of problems involving definite integrals.

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