Problem 39 - Thomas Calculus

Question

Evaluate the integrals in Exercises \(13-48\) . $$ \int \frac{1}{t^{2}} \cos \left(\frac{1}{t}-1\right) d t $$

Step-by-Step Solution

Answer

The integral is \(-\sin\left(\frac{1}{t} - 1\right) + C\).

Step 1: Substitution

Let's perform a substitution to simplify the integral. Choose the substitution: \( u = \frac{1}{t} - 1 \). Then, differentiate both sides: \( \frac{du}{dt} = -\frac{1}{t^2} \). Therefore, \( du = -\frac{1}{t^2} \, dt \) or \( dt = -t^2 \, du \). Substitute \( u \) and \( dt \) in the integral: \[ \int \frac{1}{t^2} \cos \left(\frac{1}{t} - 1\right) dt = \int \cos(u) (-du) = -\int \cos(u) \, du \].

Step 2: Integrate with Respect to u

Now, we solve the integral \( -\int \cos(u) \, du \). The integral of \( \cos(u) \) with respect to \( u \) is \( \sin(u) \). Thus, after integrating, we have: \( -\int \cos(u) \, du = -\sin(u) + C \), where \( C \) is the integration constant.

Step 3: Substitute Back in Terms of t

Now, substitute back \( u = \frac{1}{t} - 1 \). Therefore, the solution becomes: \( -\sin\left(\frac{1}{t} - 1\right) + C \).

Step 4: Final Expression

The evaluated integral \( \int \frac{1}{t^{2}} \cos \left(\frac{1}{t}-1\right) d t \) simplifies to \( -\sin\left(\frac{1}{t} - 1\right) + C \). This is the final answer.

Key Concepts

Substitution Method

The substitution method is a powerful tool in integral calculus that helps simplify complex integrals. By transforming the variable of integration, the integral becomes easier to evaluate. In this solution, we use substitution to make the integration process more manageable.

• First, we identify a part of the integrand that we can simplify. Here, it is beneficial to let \( u = \frac{1}{t} - 1 \). This choice simplifies the expression inside the cosine function of the integrand.
• After defining \( u \), we find the derivative \( \frac{du}{dt} = -\frac{1}{t^2} \) and rearrange to express \( dt \) in terms of \( du \).
• Substituting \( u \) and \( dt \) in the original integral changes it into a form involving \( \int \cos(u) (-du) \), which is much easier to solve.

The substitution method essentially reduces the complexity by changing the function's perspective, enabling us to solve integrals that would otherwise seem daunting.

Trigonometric Integration

Trigonometric integration involves solving integrals that contain trigonometric functions such as sine, cosine, tangent, etc. In this problem, the function involves a cosine term.

• The integral \( \int \cos(u) \ du \) is a standard trigonometric integral, which solves to \( \sin(u) + C \).
• These standard results are essential to know, as they enable us to quickly perform the integration once the substitution has simplified the integrand.
• Integral calculus often involves combining techniques, and many problems can be broken down into manageable parts using substitutions, followed by straightforward trigonometric integrations.

Recognizing the trigonometric pattern allows us to apply known integral results, speeding up the problem-solving process and ensuring accuracy.

Definite Integral

While the original exercise doesn't specifically ask for a definite integral, it's useful to understand this concept.

• A definite integral calculates the accumulation of quantities, giving us an exact numerical value over an interval, typically from \( a \) to \( b \).
• Definite integrals differ from indefinite integrals because they represent the total area under a curve bounded by the x-axis and vertical lines \( x = a \) and \( x = b \).
• When evaluating a definite integral, the process involves finding the antiderivative, substituting the upper and lower limits, and computing the difference between these two values.

Even though the given problem is an indefinite integral, grasping how definite integrals would work assists in understanding the broader applications of integral calculus.

Antiderivative

The concept of an antiderivative is central to solving indefinite integrals. An antiderivative is essentially a function whose derivative equals the original function we started with.

• The process of finding an antiderivative is the reverse process of differentiation.
• In the example, the antiderivative of \( \cos(u) \) is found to be \( \sin(u) \), which means the derivative of \( \sin(u) \) returns us back to \( \cos(u) \).
• Each family of antiderivatives includes an arbitrary constant \( C \), representing the vertical shift of the function. This constant becomes particularly important in solving initial value problems.

Ultimately, understanding antiderivatives provides the foundation for integrating functions and finding solutions that would otherwise remain hidden.