Problem 90 - Thomas Calculus in SI Units

Question

Evaluate the integrals in Exercises \(83-92.\) $$\int_{0}^{\pi / 4}\left(\frac{1}{3}\right)^{\tan t} \sec ^{2} t d t$$

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{2}{3|\ln(3)|} \).

Step 1: Recognize the function and its derivative

The integrand is \( \left(\frac{1}{3}\right)^{\tan t} \sec^2 t \). Recognize that if we let \( u = \tan t \), then the derivative \( \frac{du}{dt} = \sec^2 t \). This means the structure of the integral suggests using a substitution method since \( \sec^2 t dt = du \).

Step 2: Use substitution to change the variable

With the substitution \( u = \tan t \), the differential becomes \( du = \sec^2 t \, dt \). Substituting these into the integral changes it to \( \int \left(\frac{1}{3}\right)^{u} \, du \).

Step 3: Evaluate the new integral

The integral \( \int \left(\frac{1}{3}\right)^u \, du \) is equal to \( \int e^{u \ln(1/3)} \, du \), which evaluates to \( \frac{e^{u \ln(1/3)}}{\ln(1/3)} + C \).

Step 4: Change limits of integration

Convert the limits from \( t \) to \( u \). When \( t = 0 \), \( u = \tan 0 = 0 \). When \( t = \pi/4 \), \( u = \tan(\pi/4) = 1 \). Thus, the limits of integration in terms of \( u \) are from 0 to 1.

Step 5: Substitute back and evaluate the definite integral

Substitute back to get \( \frac{(1/3)^u}{\ln(1/3)} \) and evaluate from 0 to 1:\[ \left[ \frac{(1/3)^u}{\ln(1/3)} \right]_0^1 = \frac{(1/3)^1 - (1/3)^0}{\ln(1/3)} = \frac{1/3 - 1}{\ln(1/3)} = \frac{-2/3}{\ln(1/3)}. \]

Step 6: Compute the expression

Simplifying, we find: \( \frac{-2/3}{\ln(1/3)} \), which can be rewritten using the properties of logarithms as \( \frac{2}{3 |\ln(3)|} \) because \( \ln(1/3) = -\ln(3) \).

Key Concepts

Substitution Method

The substitution method is a powerful tool in calculus for solving integrals. It's particularly useful when you encounter integrals with a function and its derivative present. The idea is to simplify the integral by substituting a part of it with a new variable. This method often makes a complex integral easier to evaluate.

In the exercise provided, we noticed that the integrand is \( \left(\frac{1}{3}\right)^{\tan t} \sec^2 t \). The presence of \( \tan t \) and its derivative \( \sec^2 t \) suggests substitution could be beneficial.

• To apply the substitution, set a new variable such as \( u = \tan t \).
• Differentiate it to find \( du = \sec^2 t \, dt \).

When substituted into the integral, it simplifies the variable and integrates in terms of \( u \), turning the understanding of a trigonometric function into a more straightforward exponential function problem in this particular case.

Definite Integrals

Definite integrals are used to compute the net area under a curve within given boundaries. They are defined with limits of integration, and the result is a specific numerical value, unlike indefinite integrals where we have a general function plus a constant.

For the integral in consideration, the bounds are from \( t = 0 \) to \( t = \pi/4 \). Upon substitution, these limits change in terms of our new variable \( u \), as follows:

• When \( t = 0 \), \( u = \tan(0) = 0 \).
• When \( t = \pi/4 \), \( u = \tan(\pi/4) = 1 \).

This transforms the integral into evaluating the area from 0 to 1 in the new variable. The process not only clarifies computations but also ensures we're working within the appropriate framework for solving the given problem. The final numerical result represents the accumulated change or area from the start to end within the specified range.

Trigonometric Functions

Trigonometric functions are fundamental in mathematics, often appearing in integrals and derivatives. They describe relationships between angles and sides of triangles and are periodic, meaning they repeat their values in regular intervals.

In the provided exercise, the function \( \tan t \) and its derivative \( \sec^2 t \) play a crucial role.

• \( \tan t \) is the ratio of the opposite side to the adjacent side in a right triangle.
• \( \sec^2 t \), the derivative of \( \tan t \), represents the rate of change of the tangent function, which is pivotal for substitution.

The integral's set-up relies on the interplay between \( \tan t \) and \( \sec^2 t \), using their mathematical properties to transform and solve the integral effectively. Understanding these functions can illuminate how different integrals can be approached and solved.