Problem 49 - Thomas Calculus

Question

Evaluate the integrals in Exercises \(41-60\) $$\int \frac{\operatorname{sech} \sqrt{t} \tanh \sqrt{t} d t}{\sqrt{t}}$$

Step-by-Step Solution

Answer

The integral evaluates to \(-2 \operatorname{sech} \sqrt{t} + C\).

Step 1: Identify the Substitution

To simplify the integral, we'll first perform a substitution. Notice the expressions \(\operatorname{sech} \sqrt{t}\) and \(\tanh \sqrt{t}\). Let's set \(u = \sqrt{t}\), then \(t = u^2\) and \(dt = 2u \, du\).

Step 2: Express dt in Terms of du

Substitute \(dt\) with \(2u \, du\) to transform the integral. The integral becomes \(\int \frac{\operatorname{sech} u \tanh u}{u} \cdot \frac{2u \, du}{\sqrt{t}} = \int \operatorname{sech} u \tanh u \cdot 2 \, du. \)

Step 3: Simplify the Integral

The integral simplifies to \(2 \int \operatorname{sech} u \tanh u \, du.\) We know from standard calculus identities that the derivative of \(\operatorname{sech} u\) is \(-\operatorname{sech} u \tanh u\). Therefore, the integral of \(\operatorname{sech} u \tanh u\) is \(-\operatorname{sech} u\).

Step 4: Evaluate the Simplified Integral

Given the standard result, \(2 \int \operatorname{sech} u \tanh u \, du = -2 \operatorname{sech} u + C\). Thus, integrating gives us \( -2 \operatorname{sech} u + C \).

Step 5: Substitute Back in Terms of t

Recall that \(u = \sqrt{t}\). Substitute back to express the result in terms of \(t\): \(-2 \operatorname{sech} \sqrt{t} + C\).

Key Concepts

Calculus Substitution

Substituting variables is a powerful technique in calculus. It's used to simplify complex integrals by transforming them into a more manageable form. In our original exercise, substitution helps in simplifying the given integral involving hyperbolic functions.
The process begins by identifying an expression within the integral that can be redefined in terms of a new variable, often termed as "u". For example, in our case, substituting \(u = \sqrt{t}\) makes the problem easier. This converts the original variable \(t\) into \(u^2\) and requires that \(dt\) is expressed in terms of \(du\).
Using calculus identities, when \(t = u^2\), the derivative \(dt\) becomes \(2u \, du\). This step is crucial because it enables the whole integral to transition smoothly from the variable \(t\) to the variable \(u\), making it a simpler problem to tackle.

Hyperbolic Functions

Hyperbolic functions like \(\operatorname{sech}\) and \(\tanh\) appear regularly in calculus, similar to trigonometric functions. These functions are particularly useful in many areas of math and engineering, especially when describing certain types of geometry and growth models.

• \(\operatorname{sech}(x)\) is the hyperbolic secant function, which is the reciprocal of the hyperbolic cosine:
• \(\operatorname{sech}(x) = \frac{1}{\cosh(x)}\)
• \(\tanh(x)\) is the hyperbolic tangent function, defined as the ratio of sinh and cosh:
• \(\tanh(x) = \frac{\sinh(x)}{\cosh(x)}\)

These functions can be handled through calculus using identities similar to trigonometric identities. Many calculus problems simplify when recognizing and applying these hyperbolic identities.

Integral Evaluation

Evaluating integrals is one of the big goals in calculus. It involves finding a function whose derivative is the integrand, essentially working backward. In our exercise, integral evaluation is the final step after substituting variables and identifying related derivative relationships.
Once the integral is simplified, as in our case to \(2 \int \operatorname{sech} u \tanh u \, du\), the goal is to evaluate it using known calculus truths. Here, we draw on recognition that the derivative of \(\operatorname{sech} u\) is \(-\operatorname{sech} u \tanh u\). This allows the integral of \(\operatorname{sech} u \tanh u\) to be quickly evaluated via an antiderivative which is \(-\operatorname{sech} u\), leading us to the final answer.

Calculus Identities

Calculus identities are essential tools which provide shortcuts to simplify complex expressions and evaluations. They freely connect derivatives and integrals of functions with similar properties. In this exercise, they were used to find the derivative and integral of hyperbolic functions.

• The derivative of \(\operatorname{sech} u\) is \(-\operatorname{sech} u \tanh u\).
• This can be deduced from fundamental differential relationships between hyperbolic functions.

Identifying these identities is key to resolving many calculus problems efficiently. Understanding calculus identities helps in transforming a daunting task into a more straightforward one.