Problem 94 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

The integral evaluates to \(-\frac{2}{3 \ln(1/3)}\)."

Step 1: Identify Substitution

To evaluate the integral \( \int_{0}^{\pi / 4}\left(\frac{1}{3}\right)^{\tan t} \sec ^{2} t \ dt \), recognize that substituting \( u = \tan t \) simplifies this problem, as the derivative \( du = \sec^2 t \ dt \) matches part of our integrand.

Step 2: Change Limits of Integration

Change the limits of the integral according to our substitution \( u = \tan t \). When \( t = 0 \), \( u = \tan 0 = 0 \). When \( t = \frac{\pi}{4} \), \( u = \tan \frac{\pi}{4} = 1 \). The limits for \( u \) are from 0 to 1.

Step 3: Substitute and Integrate

After substitution, the integral \( \int_{0}^{1} \left( \frac{1}{3} \right)^u \ du \) needs to be evaluated. The integral of \( \left( \frac{1}{3} \right)^u \) is \( \frac{\left( \frac{1}{3} \right)^u}{\ln\left( \frac{1}{3} \right)} \).

Step 4: Evaluate the Integral

Evaluate the integral from 0 to 1: \[ \left. \frac{(\frac{1}{3})^u}{\ln(\frac{1}{3})} \right|_0^1 = \frac{(\frac{1}{3})^1}{\ln(\frac{1}{3})} - \frac{(\frac{1}{3})^0}{\ln(\frac{1}{3})} \].

Step 5: Simplify Result

The final result simplifies to \( \frac{1}{3 \ln(\frac{1}{3})} - \frac{1}{\ln(\frac{1}{3})} = \frac{1 - 3}{3 \ln(\frac{1}{3})} = -\frac{2}{3 \ln(\frac{1}{3})} \).

Key Concepts

Definite Integrals

Definite integrals allow us to find the accumulated value of a function over a specific interval. Specifically, they measure the net area between the graph of the function and the x-axis across a given range. In this exercise, we tackled the definite integral from 0 to \( \frac{\pi}{4} \). The process of integration involves calculating the antiderivative of the function, and then evaluating it at the upper and lower bounds of the interval.
For our problem, using the substitution \( u = \tan t \) transformed the given integral into a simpler form. This substitution effectively changes the variable from \( t \) to \( u \), altering the integration limits accordingly. Thus, managing definite integrals often involves simplifying the functions using substitutions, making computations more straightforward.

Trigonometric Functions

Trigonometric functions like sine, cosine, and tangent are essential in many areas of mathematics and physics. In this exercise, the tangent function \( \tan t \) played a crucial role because its derivative, \( \sec^2 t \), appeared in the integrand. Substitution using trigonometric identities can simplify the integration process significantly.
Understanding the derivatives and relationships between these trigonometric functions is key. For instance, knowing that \( \frac{d}{dt} (\tan t) = \sec^2 t \) helps identify a direct correspondence in the integration problem, allowing us to make the substitution \( u = \tan t \), simplifying the integral to a more manageable form. When evaluating definite integrals with trigonometric functions, converting them to exponential terms can sometimes be helpful, albeit not necessary in this exercise.

Exponential Functions

Exponential functions, characterized by expressions like \( a^x \), where \( a \) is a constant, have unique properties: their rate of change is proportional to their current value. In our integral problem, the expression \( \left( \frac{1}{3} \right)^{\tan t} \) is one such exponential function impacted by substitution.
To integrate this, we use a basic rule for integrals of exponential functions: the integral of \( a^u \) is \( \frac{a^u}{\ln a} \). Applying this principle to our substituted integral \( \int_{0}^{1} \left( \frac{1}{3} \right)^u \ du \), we derive the solution by evaluating this antiderivative from the specific bounds. The step by step transformation and simplification illustrate the application of exponential rule in integration, showing how even complex-looking expressions can be resolved through known integral rules.