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

Question: Evaluate the definite integral \(\int_{0}^{1} \cosh ^{3} 3 x \sinh 3 x d x\). Answer: \(\frac{1}{36}\left[\left( (\cosh 3)^4 - 1 \right)\right]\).

Step 1: Use integration by parts formula

Recall the integration by parts formula, which is given by: \(\int u dv = uv - \int v du\). In this case, let \(u = \cosh^3 3x\) and \(dv = \sinh 3x dx\). Now we will find \(du\) and \(v\).

Step 2: Find the derivatives and integrals

Differentiate \(u\) with respect to \(x\) to get \(du\): \(du = \frac{d}{dx}(\cosh^3 3x) = 9\cosh^2 3x \sinh 3x dx\). Integrate \(dv\) to find \(v\): \(v = \int \sinh 3x dx = \frac{1}{3}\cosh 3x + C\).

Step 3: Apply the integration by parts formula

Now we have all the components needed to apply the integration by parts formula: \(\int_{0}^{1} \cosh^3 3x \sinh 3x dx = \left[\cosh^3 3x \cdot \frac{1}{3}\cosh 3x\right]_0^1 - \int_{0}^{1}(\frac{1}{3}\cosh 3x)(9\cosh^2 3x \sinh 3x dx)\).

Step 4: Perform a substitution

Let's make a substitution to simplify the integral. Let \(t = \cosh 3x\). Then, \(dt = 3\sinh 3x dx\). Now we have: \(\int_{0}^{1}\cosh^{3} 3x \sinh 3x dx = \frac{1}{9}\int_{1}^{\cosh 3} t^3 dt\).

Step 5: Integrate w.r.t. t

Now, we integrate the simplified expression with respect to \(t\): \(\frac{1}{9}\int_{1}^{\cosh 3} t^3 dt = \frac{1}{36}\left[t^4\right]_{1}^{\cosh 3}\).

Step 6: Evaluate the definite integral

Now, we evaluate the definite integral by plugging the limits of integration into the expression: \(\frac{1}{36}\left[\left(\cosh^4 3\right) - 1\right] = \frac{1}{36}\left[\left( (\cosh 3)^4 - 1 \right)\right]\). So, the value of the definite integral is \(\int_{0}^{1} \cosh ^{3} 3 x \sinh 3 x d x = \frac{1}{36}\left[\left( (\cosh 3)^4 - 1 \right)\right]\).

Key Concepts

Integration by Parts

Integration by parts is a powerful technique used in calculus to transform the integration of a product of functions into a possibly easier form. This technique is especially useful when dealing with integrals that are difficult to evaluate directly. The formula for integration by parts is like a "reverse product rule" for integration: \[ \int u \, dv = uv - \int v \, du \]Here's how it works in detail:

• Identify parts of the integrand to assign as \(u\) and \(dv\). Usually, \(u\) is chosen as the part that becomes simpler when differentiated, and \(dv\) is what's left.
• Differentiate \(u\) to find \(du\), and integrate \(dv\) to determine \(v\).
• Substitute these into the integration by parts formula and simplify.

In our example, we set \(u = \cosh^3 3x\) and \(dv = \sinh 3x\, dx\). This choice simplifies the integral and demonstrates the usefulness of the integration by parts technique for dealing with hyperbolic functions.

Hyperbolic Functions

Hyperbolic functions are analogs of trigonometric functions, encountered often in calculus and complex analysis. They are essential in solving integrals involving exponentials, like those seen in our exercise.The most common hyperbolic functions are \(\sinh\) and \(\cosh\), defined as:

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

These functions exhibit identities similar to trigonometric functions, such as \(\cosh^2 x - \sinh^2 x = 1\). In the original exercise, recognizing the properties of hyperbolic functions allowed substitutions and simplifications that made the integral manageable. Hyperbolic functions often appear when dealing with hyperbolas, in equations of motion, and in certain types of differential equations, making them versatile and invaluable tools in calculus.

Integration Techniques

Integration techniques encompass a variety of methods used to solve definite and indefinite integrals. These techniques come into play when straightforward integration isn't feasible and include:

• Substitution: This involves changing variables to simplify the integral.
• Integration by Parts: Useful when the integrand is a product of two non-trivial functions.
• Trigonometric Identities & Hyperbolic Functions: Utilizing known identities to transform the integral into a simpler form.

In our exercise, both integration by parts and substitution were pivotal. After applying integration by parts, a substitution with \(t = \cosh 3x\) simplified the problem further. Being proficient in these techniques allows one to tackle a wide array of problems, each needing different approaches. Understanding when and where to use each method can lead to success in more complex calculus challenges.