Problem 60 - Thomas Calculus in SI Units

Question

Evaluate the integrals. $$\int_{0}^{\ln 10} 4 \sinh ^{2}\left(\frac{x}{2}\right) d x$$

Step-by-Step Solution

Answer

\(\frac{99}{10} - 2 \ln 10\)

Step 1: Apply the Double Angle Identity

Recall the hyperbolic double angle identity: \(\sinh^2(u) = \frac{1}{2}(\cosh(2u) - 1)\). Substitute \(u = \frac{x}{2}\), so \(\sinh^2\left(\frac{x}{2}\right) = \frac{1}{2}(\cosh(x) - 1)\).

Step 2: Simplify the Integral

Substitute the identity into the integral to get: \(\int_{0}^{\ln 10} 4 \cdot \frac{1}{2}(\cosh(x) - 1) dx = 2 \int_{0}^{\ln 10} (\cosh(x) - 1) dx\).

Step 3: Separate and Evaluate the Integrals

Split the integral into two separate integrals: \(2 \int_{0}^{\ln 10} \cosh(x) dx - 2 \int_{0}^{\ln 10} 1 dx\). Evaluate each integral separately. Recall that \(\int \cosh(x) dx = \sinh(x)\).

Step 4: Calculate \(\int \cosh(x) dx\)

Evaluate \( 2 \int_{0}^{\ln 10} \cosh(x) dx = 2 \left[ \sinh(x) \right]_{0}^{\ln 10} = 2 (\sinh(\ln 10) - \sinh(0)) = 2(\frac{10 - 1/10}{2} - 0) = 2 \times \frac{99}{20} = \frac{99}{10}.\)

Step 5: Calculate \(\int 1 dx\)

Evaluate \(- 2 \int_{0}^{\ln 10} 1 dx = - 2 (\ln 10 - 0) = -2 \ln 10.\)

Step 6: Combine Both Parts of Step 3

Add up results from Steps 4 and 5: \(\frac{99}{10} - 2 \ln 10.\)

Step 7: Final Answer

Thus, the evaluated integral \(\int_{0}^{\ln 10} 4 \sinh^2\left(\frac{x}{2}\right) dx\) equals \(\frac{99}{10} - 2 \ln 10\).

Key Concepts

Hyperbolic Functions

Hyperbolic functions are mathematical functions similar to trigonometric functions but are based on hyperbolas instead of circles. There are two primary hyperbolic functions:

• Hyperbolic Sine: \(\sinh(x) = \frac{e^x - e^{-x}}{2}\)
• Hyperbolic Cosine: \(\cosh(x) = \frac{e^x + e^{-x}}{2}\)

These functions share several properties with their trigonometric counterparts, such as involving identities that help simplify expressions. In calculus, they often arise when solving integrals and differential equations.For example, in the exercise, we transformed \(\sinh^2\) using the double angle identity. Doing this standardizes calculations, making it easier to solve or simplify integrals involving hyperbolic functions. Knowing how to apply these properties effectively is crucial when handling integrals involving hyperbolic functions.

Double Angle Identity

The double angle identity is a powerful tool in both trigonometric and hyperbolic mathematics. For hyperbolic functions, the identity allows us to rewrite expressions into simpler forms. This identity for hyperbolic functions is \(\sinh^2(u) = \frac{1}{2}(\cosh(2u) - 1)\) By using this identity, we can break down complex expressions into simpler components that are easier to integrate.

Application in the Exercise

In the exercise, the given integral involves \(\sinh^2\left(\frac{x}{2}\right)\). By substituting \(u = \frac{x}{2}\) into the identity, we convert \(\sinh^2\left(\frac{x}{2}\right)\) into \(\frac{1}{2}(\cosh(x) - 1)\). This substitution simplifies the problem, turning the integral into a simpler one with \(\cosh(x) - 1\), making it more straightforward to solve. Understanding and using identities like this one is essential when simplifying and evaluating integrals in calculus.

Definite Integrals

Definite integrals are a fundamental concept in calculus used to find the area under a curve over a particular interval. They provide a numerical value that represents this area precisely.In the exercise, we evaluated the integral \(\int_{0}^{\ln 10} 4 \sinh^2\left(\frac{x}{2}\right) dx\) which was transformed and simplified to two separate integrals. Each definite integral provides a specific evaluation:

• The integral of \( \cosh(x) \) over \([0, \ln(10)]\) gives \( \sinh(x) \)
• Direct calculation allows us to evaluate straightforward constants, such as \( \int 1 \, dx \)

Finally, by combining the results from individual integrals, we reach the evaluated answer.

The Importance of Limits

The limits \(0\) to \(\ln(10)\) define the interval over which we're calculating the integral. They are critical to evaluating definite integrals because they specify the region of interest on the curve. Knowing how to set up and evaluate definite integrals accurately is an important part of calculus, allowing for applications ranging from computing areas to determining accumulated values over time.