Problem 37 - Calculus Early Transcendentals

Question

Evaluate each definite integral. $$\int_{0}^{1} \cosh ^{3} 3 x \sinh 3 x d x$$

Step-by-Step Solution

Answer

In this case, the technique used to solve the definite integral is integration by parts. This technique is useful for dealing with functions that have both hyperbolic sine (sinh) and cosine (cosh). The steps involve identifying the substitution for integration by parts, applying the integration by parts formula, simplifying the resulting integral, and evaluating the simplified integral at the bounds.

Step 1: Identify the substitution for integration by parts.

We need to choose appropriate parts of our function to integrate and differentiate in the integration by parts formula. Let $$ \begin{cases} u = \cosh^3(3x), & \;\mathrm{then}\; du = 9\cosh^2(3x)\sinh(3x) dx\\ dv = \sinh(3x) dx, & \;\mathrm{then}\; v = \int \sinh(3x) dx = \frac{1}{3} \cosh(3x) \end{cases} $$

Step 2: Apply the integration by parts formula.

Apply the integration by parts formula, which states that $$ \int u \, dv = uv - \int v \, du $$ We apply this formula to our problem: $$ \int_{0}^{1} \cosh ^{3} 3 x \sinh 3 x d x = \left[\frac{1}{3}\cosh^3(3x)\cosh(3x)\right]_{0}^{1} - \int_{0}^{1} \frac{1}{3}\cosh(3x)(9\cosh^2(3x)\sinh(3x)) dx $$

Step 3: Simplify the resulting integral.

Now, we simplify the expression inside the integral: $$ \int_{0}^{1} \cosh ^{3} 3 x \sinh 3 x d x = \left[\frac{1}{3}\cosh^3(3x)\cosh(3x)\right]_{0}^{1} - 3 \int_{0}^{1} \cosh^3(3x)\sinh(3x) dx $$ Notice that the integral on the right side is just the original integral multiplied by a constant. Let's represent the original integral by \(I\), so: $$ I = \left[\frac{1}{3}\cosh^3(3x)\cosh(3x)\right]_{0}^{1} - 3I $$ Now we can solve for \(I\): $$ 4I = \left[\frac{1}{3}\cosh^3(3x)\cosh(3x)\right]_{0}^{1} $$ $$ I = \frac{1}{12}\left[\cosh^3(3x)\cosh(3x)\right]_{0}^{1} $$

Step 4: Evaluate the simplified integral at the bounds.

Finally, we evaluate the simplified integral at the bounds \(0\) and \(1\) to find the value of the definite integral: $$ I = \frac{1}{12}\left(\cosh^3(3\cdot 1)\cosh(3\cdot 1) - \cosh^3(3\cdot 0)\cosh(3\cdot 0)\right) $$ $$ I = \frac{1}{12}\left(\cosh^3(3)\cosh(3) - \cosh^3(0)\cosh(0)\right) $$ $$ I = \frac{1}{12}\left(\cosh^3(3)\cosh(3) - 1\right) $$ This is the value of the given definite integral: $$\int_{0}^{1} \cosh ^{3} 3 x \sinh 3 x d x = \frac{1}{12}\left(\cosh^3(3)\cosh(3) - 1\right) $$

Key Concepts

Integration by Parts

Integration by parts is a technique for evaluating integrals, especially when the integrand is a product of two functions. This method is derived from the product rule for differentiation, making it a powerful tool for solving complex integrals. The integration by parts formula is:

• \( \int u \, dv = uv - \int v \, du \)

In this formula:

• \( u \) is a function chosen from the integrand that when differentiated, simplifies.
• \( dv \) is the remaining part of the integrand.
• The task is to differentiate \( u \) to get \( du \), and integrate \( dv \) to obtain \( v \).

In our definite integral example \( \int_{0}^{1} \cosh^3(3x) \sinh(3x) \, dx \), we used integration by parts to break it down into simpler components by letting:

• \( u = \cosh^3(3x) \)
• \( dv = \sinh(3x) \, dx \)

Through this setup, the original problem can be simplified, allowing us to manage the integration process better. By applying the above steps, integration by parts can help solve integrals involving products of algebraic, exponential, logarithmic, and trigonometric functions efficiently.

Hyperbolic Functions

Hyperbolic functions are analogues of trigonometric functions that arise naturally in the context of a hyperbola, similar to how trigonometric functions relate to a circle. The two primary hyperbolic functions are sine hyperbolic, \( \sinh(x) \), and cosine hyperbolic, \( \cosh(x) \). They are defined as:

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

These definitions make them very useful in calculus, especially for solving problems involving integrals, such as in our example. They have properties similar to trigonometric functions:

• Derivative of \( \sinh(x) \) is \( \cosh(x) \).
• Derivative of \( \cosh(x) \) is \( \sinh(x) \).

The integral of hyperbolic functions comes in handy while solving definite integrals involving hyperbolic terms, as they simplify many integrals that are otherwise complex. In the given exercise, both \( \cosh(3x) \) and \( \sinh(3x) \) are used, demonstrating their integration and differentiation properties to solve the integral effectively. Recognizing patterns and relationships between hyperbolic functions aids in the necessary manipulation required during integration.

Substitution Method

The substitution method, also known as u-substitution, is a technique used to simplify the process of integration by making the integrand easier to handle. It involves substituting part of the integrand with a single variable to reduce the complexity of the expression. The steps typically involve:

• Identifying a part of the integrand to substitute with a new variable \( u \).
• Calculating \( du \), the differential of \( u \).
• Rewriting the integrand in terms of \( u \) and \( du \).
• Solving the integral and substituting back the original variable.

In our example, while integration by parts was the primary method, substitution can also help in managing parts of the integral. It was utilized in defining components like \( u = \cosh^3(3x) \) and \( dv = \sinh(3x) \, dx \), effectively turning a complex integral into more manageable parts. This technique is crucial in handling non-linear and composite functions, transforming them into simpler forms that are more straightforward to integrate. It is an essential strategy in an integrator's toolkit and is often used in conjunction with other methods, such as integration by parts, to solve intricate integrals efficiently.