Problem 64 - Thomas Calculus in SI Units

Question

Evaluate the integrals. $$\int_{1}^{e^{\pi / 4}} \frac{4 d t}{t\left(1+\ln ^{2} t\right)}$$

Step-by-Step Solution

Answer

The value of the integral is \( \pi \).

Step 1: Recognize the Integral Form

The given integral is \( \int_{1}^{e^{\pi / 4}} \frac{4 d t}{t(1+\ln^2 t)} \). This integral suggests a substitution since the presence of \( t \) and \( \ln t \) in the denominator indicates a logarithmic form can be simplified using a substitution.

Step 2: Choose a Substitution

Let \( u = \ln t \). Then \( \frac{d u}{d t} = \frac{1}{t} \), which implies \( d u = \frac{d t}{t} \). The bounds change according to \( u = \ln t \): \( t = 1 \) gives \( u = 0 \) and \( t = e^{\pi/4} \) gives \( u = \frac{\pi}{4} \).

Step 3: Substitute and Simplify the Integral

Replace \( \ln t \) by \( u \) and \( \frac{d t}{t} \) by \( d u \) in the integral. The integral becomes \( 4 \int_{0}^{\pi/4} \frac{d u}{1 + u^2} \).

Step 4: Recognize the Resulting Integral

The integral \( \int \frac{d u}{1 + u^2} \) is known to be \( \tan^{-1} u + C \), where \( C \) is the constant of integration.

Step 5: Evaluate the Integral

Now, evaluate \( 4 \left[ \tan^{-1} u \right]_{0}^{\pi/4} \), which becomes \( 4 \left[ \tan^{-1} \frac{\pi}{4} - \tan^{-1} 0 \right] \). Knowing \( \tan^{-1} 0 = 0 \) and \( \tan^{-1} \frac{\pi}{4} = \frac{\pi}{4} \), calculate the result.

Step 6: Final Calculation

Substitute the values to find \( 4 \left( \frac{\pi}{4} - 0 \right) = \pi \).

Key Concepts

Substitution Method

The substitution method plays a pivotal role in solving integrals that seem complex at first glance. This technique turns analytically difficult integrals into simpler ones by substituting variables. When you look at the given integral \( \int_{1}^{e^{\pi / 4}} \frac{4 d t}{t(1+\ln^2 t)} \), the key is to notice patterns that suggest substitution. Here, both \( t \) and \( \ln t \) are in the denominator.

**Why Use Substitution?**

• To simplify the integration process by introducing a new variable.
• To transform the original variable into a form that resembles standard integral formulas.
• To change the bounds of integration accordingly, making the computation straightforward.

**How Does Substitution Work?**

• Identify a function or expression within the integral that makes the process complex.
• Introduce a new variable, say \( u \), that simplifies this expression.
• Change the differential accordingly, like transforming \( dt \) to \( du \).
• Adjust limits of integration when dealing with definite integrals.

In our exercise, setting \( u = \ln t \) simplifies the integral to a form involving \( du \), easing the process of solving the integral.

Definite Integrals

Definite integrals go beyond just finding anti-derivatives; they give you the area under a curve within specific bounds. In context, you are asked to evaluate the integral from \( t = 1 \) to \( t = e^{\pi/4} \).

**Understanding Definite Integrals**

• The integral is evaluated over an interval \([a, b]\).
• The result is a real number representing the net area under the curve, considering any part below the x-axis as negative.
• Definite integrals require limits of integration, transforming a general solution into a specific value.

**Steps to Solve Definite Integrals**

• Perform the integration process, considering the presence of any transformations like substitution.
• Evaluate the anti-derivative at the upper limit and subtract the value at the lower limit.
• This subtraction yields the net area under the curve over the specified interval.

The substitution changes the limits to \( u = 0 \) and \( u = \frac{\pi}{4} \), making it easier to evaluate the definite integral \( 4 \left[ \tan^{-1} u \right]_{0}^{\pi/4} \), ultimately giving us the result \( \pi \).

Inverse Trigonometric Functions

Inverse trigonometric functions like \( \tan^{-1} u \) arise naturally in the integration of specific types of functions. They help in solving integrals where standard algebraic or exponential antiderivatives are insufficient.

**Importance in Integration**

• These functions provide antiderivatives for certain rational functions.
• Recognizing when an inverse trigonometric form is involved can simplify the integration considerably.
• Such integrals often directly relate back to arc length problems or problems involving angles in mathematics.

In the exercise, the integral simplifies to \( \tan^{-1} u \), leveraging the known derivative \( \frac{d}{dx}(\tan^{-1}(x)) = \frac{1}{{1+x^2}} \). Knowing this derivative helps us recognize the integral \( \int \frac{du}{1 + u^2} \) is equal to \( \tan^{-1} u + C \).

**Using Inverse Functions Effectively**

• Understand the derivatives of inverse trigonometric functions to recognize potential substitutions or forms during integration.
• Keep in mind the domain and range restrictions, which are crucial for ensuring meaningful solutions.
• Leverage the inverse functions to revert back transformations made during substitution.

Ultimately, substituting back and understanding these functions allow us to efficiently solve and evaluate the integral.