Problem 59 - Thomas Calculus

Question

Evaluate the integrals in Exercises \(51-60 .\) $$ \int_{-\ln 2}^{0} \cosh ^{2}\left(\frac{x}{2}\right) d x $$

Step-by-Step Solution

Answer

\( \frac{1}{2} \ln 2 + \frac{3}{8} \)

Step 1: Understand the Integral

The given integral is \( \int_{-\ln 2}^{0} \cosh^{2}\left(\frac{x}{2}\right) dx \). The limits of integration run from \(-\ln 2\) to \(0\). Our goal is to find the antiderivative and evaluate it between these limits.

Step 2: Use the Hyperbolic Identity

Recall that \( \cosh^{2}(t) = \frac{1 + \cosh(2t)}{2} \). Thus, we rewrite the integrand as \( \cosh^{2}\left(\frac{x}{2}\right) = \frac{1 + \cosh(x)}{2} \). This simplifies our integral to \( \int_{-\ln 2}^{0} \frac{1 + \cosh(x)}{2} \, dx \).

Step 3: Split the Integral

Use the linearity of integrals to separate: \[ \int_{-\ln 2}^{0} \frac{1}{2} \, dx + \int_{-\ln 2}^{0} \frac{\cosh(x)}{2} \, dx. \] This simplification will allow us to solve each part individually.

Step 4: Evaluate the First Integral

The first integral is \( \int_{-\ln 2}^{0} \frac{1}{2} \, dx = \frac{1}{2}\left[ x \right]_{-\ln 2}^{0} = \frac{1}{2}(0 - (-\ln 2)) = \frac{1}{2}\ln 2 \).

Step 5: Evaluate the Second Integral

The second integral is \( \int_{-\ln 2}^{0} \frac{\cosh(x)}{2} \, dx = \frac{1}{2} \left[ \sinh(x) \right]_{-\ln 2}^{0} = \frac{1}{2}(\sinh(0) - \sinh(-\ln 2)) = \frac{1}{2}(0 + \sinh(\ln 2)) \).

Step 6: Simplify \( \sinh(\ln 2) \)

Using the identity \( \sinh(\ln a) = \frac{a^2 - 1}{2a} \), for \( a = 2 \), we have \( \sinh(\ln 2) = \frac{4 - 1}{4} = \frac{3}{4} \).

Step 7: Combine the Results

Add the results of the two integrals: \( \frac{1}{2} \ln 2 + \frac{1}{2} \times \frac{3}{4} = \frac{1}{2} \ln 2 + \frac{3}{8} \). This sum gives us the final evaluated value of the integral.

Key Concepts

Hyperbolic Functions

Hyperbolic functions are analogs of the trigonometric functions, but they are based on hyperbolas rather than circles. They frequently occur in various areas of calculus, including integration. The hyperbolic cosine function, denoted as \( \cosh(x) \), is defined using an exponential function formula: \( \cosh(x) = \frac{e^x + e^{-x}}{2} \).
\( \sinh(x) \), the hyperbolic sine function, is another such function, defined as \( \sinh(x) = \frac{e^x - e^{-x}}{2} \). These functions exhibit several useful relationships and identities similar to those of trigonometric functions. One key identity is \( \cosh^2(x) - \sinh^2(x) = 1 \), akin to the Pythagorean identity in trigonometry.
In our exercise, we used the identity \( \cosh^2(t) = \frac{1 + \cosh(2t)}{2} \) to simplify the integrand. This transformation makes integrating more manageable, as we've transformed a complex expression into one more conducive for integration techniques.

Antiderivative

The antiderivative, also known as the indefinite integral, is a function whose derivative is the given function. Finding the antiderivative involves determining a function \( F(x) \) such that \( F'(x) = f(x) \).
In the case of definite integrals, like our original exercise, we first find the antiderivative \( F(x) \) and then compute it over the interval \([a, b]\) to find the area under the curve between \( x = a \) and \( x = b \).

• Example: For the constant function \( f(x) = c \), the antiderivative is \( F(x) = cx + C \), where \( C \) is the constant of integration.
• For \( \cosh(x) \), the antiderivative is \( \sinh(x) \), since \( \frac{d}{dx}[\sinh(x)] = \cosh(x) \).

Finding antiderivatives allows us to simplify the computation of definite integrals, as seen in the exercise where splitting the integral led to the use of known antiderivatives.

Definite Integrals

Definite integrals represent the accumulation of quantities, such as areas under curves. They are expressed as \( \int_{a}^{b} f(x) \, dx \) where \( a \) and \( b \) are the limits of integration. They yield a specific numerical value rather than a function.
To evaluate a definite integral, we find an antiderivative of the function, \( F(x) \), and apply the Fundamental Theorem of Calculus:

• Calculate \( F(b) - F(a) \), where \( a \) is the lower limit and \( b \) is the upper limit.

In our exercise, the definite integral \( \int_{-\ln 2}^{0} \cosh^2(\frac{x}{2}) \, dx \) was evaluated using a simplified form split into two parts. Each component was integrated separately and then combined to find the total accumulation over the interval.

Integration Techniques

Integration techniques are strategies that simplify the calculation of integrals. They are crucial for making complex integrals tractable. Common techniques include:

• Substitution: Used when an integral contains a composition of functions. It simplifies the integrand to a basic form by changing the variable of integration.
• Integration by Parts: Useful for products of functions. It applies the formula \( \int u \, dv = uv - \int v \, du \).
• Trig and Hyperbolic Identities: These identities can simplify the integrand to a more integrable form.

In our exercise, we applied the hyperbolic identity \( \cosh^2(\frac{x}{2}) = \frac{1 + \cosh(x)}{2} \) to turn a complex expression into simpler, solvable integrals. This technique reduced the problem to evaluating two basic integrals, which were then computed individually.