Problem 54 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

The value of the integral is \( \frac{2}{3 \ln(3)} \).

Step 1: Identify Parts for Integration by Substitution

Given the integral \( \int_{0}^{\pi / 4}\left(\frac{1}{3}\right)^{\tan t} \sec ^{2} t dt \), we notice that the function \( (1/3)^{\tan t} \) involves the derivative of \( \tan t \) in the form of \( \sec^2 t \). This suggests that we can use the substitution method where \( u = \tan t \).

Step 2: Substitution

Let \( u = \tan t \), then the derivative \( du = \sec^2 t \, dt \). This substitution transforms the integral with bounds \( t = 0 \) and \( t = \pi/4 \) into \( u = 0 \) and \( u = 1 \) as \( \tan 0 = 0 \) and \( \tan(\pi/4) = 1 \). Thus, the integral becomes \( \int_{0}^{1}\left(\frac{1}{3}\right)^u \, du \).

Step 3: Integrate the New Function

Now, evaluate the integral \( \int_{0}^{1} \left(\frac{1}{3}\right)^u \, du \). This is a basic exponential integral. The integral of \( a^u \) is given by \( \frac{a^u}{\ln a} \). Substitute \( a = \frac{1}{3} \).

Step 4: Evaluate the Definite Integral

Integrate the function: \( \int_{0}^{1} \left(\frac{1}{3}\right)^u \, du = \left[ \frac{\left(\frac{1}{3}\right)^u}{\ln(1/3)} \right]_{0}^{1} \). Plug in the bounds: \( \left[ \frac{\left(\frac{1}{3}\right)^1}{\ln(1/3)} \right] - \left[ \frac{\left(\frac{1}{3}\right)^0}{\ln(1/3)} \right] = \left[ \frac{1/3}{\ln(1/3)} \right] - \left[ \frac{1}{\ln(1/3)} \right] \).

Step 5: Simplify the Result

The expression simplifies to \((\frac{1}{3} - 1) / \ln(1/3) = -\frac{2}{3} / \ln(1/3)\). Remember that \( \ln(1/3) = -\ln(3) \), hence \(-\frac{2}{3} / -\ln(3) = \frac{2}{3 \ln(3)}\).

Key Concepts

Definite Integrals

Definite integrals are a fundamental concept within calculus that allow us to calculate the "net area" under a curve from one point to another on the x-axis. Unlike indefinite integrals, which provide a family of functions, definite integrals give us a specific numerical value.

When we evaluate a definite integral, we are essentially summing up infinitely small slices to find this area. These slices can be interpreted as the "accumulation" of all values the function takes between the two bounds.

• A definite integral has limits of integration, typically written as \( \int_a^b f(x) \, dx \), where \(a\) and \(b\) are the lower and upper bounds of integration.
• Notice the limits cause the "+ C" in indefinite integrations to be unnecessary, as any constants cancel out when calculating the definite value.
• The process involves first finding an antiderivative of the function and then evaluating it at the upper limit and subtracting the value at the lower limit.

In our solution, we begin by rewriting the original integral with appropriate substitution, turning it into a much more manageable form before leveraging properties of definite integrals to solve it.

Exponential Functions

Exponential functions describe quantities that grow or decay at a constant relative rate. They take the form \(f(x) = a^x\), where \(a\) is a positive constant. A key property of exponential functions is their rapid growth compared to polynomial functions.

In calculus, working with exponential functions often requires a careful approach because their rate of change is proportional to the function's current value, making them especially pivotal in contexts like population growth models and radioactive decay.

• In any exponential function \(a^x\), if \(0 < a < 1\), the function denotes exponential decay, while \(a > 1\) denotes exponential growth.
• The derivative of an exponential function \(f(x) = a^x\) includes a natural logarithm: \(f'(x) = a^x \ln(a)\).
• During integration, this derivative property helps us handle exponentials by dividing instead of multiplying by \(\ln(a)\).

In solving our integral, applying this property allows us to integrate the function involving \((1/3)^u\) smoothly. It emphasizes how exponential functions' unique growth characteristics influence integral calculus.

Calculus Techniques

Calculus provides several powerful techniques for solving complex integrals, such as integration by substitution, integration by parts, and partial fraction decomposition. The integration by substitution technique simplifies the process of finding integrals that match the direct form of integration.

Integration by substitution is analogous to making a change of variables, similar to using the chain rule in reverse. It's particularly effective for integrals where one part is the derivative of another; this helps to simplify the integral into a more straightforward form.

• The main goal is to substitute a new variable \(u\) for a selected part \(g(x)\) of the integral and directly identify its derivative \(du\) as another part of the integral.
• By transforming the integrand and limits of integration, the integral can be evaluated more easily.
• Subsequently, after solving the integral in terms of \(u\), you reverse the substitution, returning to the original variable.

In our exercise, we used substitution to let \(u = \tan t\), which simplified our calculations. This technique is invaluable, particularly when dealing with definite integrals, as it helps transform the bounds accordingly, preserving the integral's result accuracy.