Problem 67 - Thomas Calculus

Question

Evaluate the integrals in Exercises \(47-70\) $$ \int_{0}^{\ln \sqrt{3}} \frac{e^{x} d x}{1+e^{2 x}} $$

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{\pi}{12} \).

Step 1: Substitution

To simplify the integral, use the substitution method. Let \( u = e^x \). Then the derivative \( du = e^x \, dx \) or \( dx = \frac{du}{u} \). Substituting these into the integral, we have:\[\int \frac{e^x}{1 + e^{2x}} \, dx = \int \frac{u}{1 + u^2} \cdot \frac{du}{u} = \int \frac{du}{1 + u^2}\]

Step 2: Adjust Limits of Integration

Adjust the limits of integration according to the substitution. When \( x = 0 \), \( u = e^0 = 1 \). When \( x = \ln \sqrt{3} \), \( u = e^{\ln \sqrt{3}} = \sqrt{3} \). Therefore, the new limits of integration are from \( 1 \) to \( \sqrt{3} \).

Step 3: Integrate Using Arctangent Function

Recognize that the integral \( \int \frac{du}{1 + u^2} \) is the standard form whose antiderivative is \( \arctan(u) \). Thus, we have:\[\int_1^{\sqrt{3}} \frac{du}{1 + u^2} = \arctan(u) \Big|_1^{\sqrt{3}}\]

Step 4: Evaluate the Definite Integral

Substitute the limits into the antiderivative:\[\arctan(\sqrt{3}) - \arctan(1)\]Knowing the common values of the arctangent function, we find that:- \( \arctan(\sqrt{3}) = \frac{\pi}{3} \)- \( \arctan(1) = \frac{\pi}{4} \)Therefore, the result is:\[\frac{\pi}{3} - \frac{\pi}{4}\]

Step 5: Simplify the Result

Find a common denominator to simplify the subtraction:\[\frac{\pi}{3} = \frac{4\pi}{12}\quad \text{and}\quad \frac{\pi}{4} = \frac{3\pi}{12}\]Thus,\[\frac{\pi}{3} - \frac{\pi}{4} = \frac{4\pi}{12} - \frac{3\pi}{12} = \frac{1\pi}{12} = \frac{\pi}{12}\]

Key Concepts

Substitution Method

The substitution method is a powerful tool in calculus for evaluating definite integrals, especially when dealing with complex expressions. The basic idea is to transform the integral into a simpler form that is easier to evaluate. In this particular exercise, we start by selecting a substitution that simplifies the expression inside the integral.

Here, we let \( u = e^x \). This choice is strategic because it directly simplifies the expression \( 1 + e^{2x} \), which appears in the denominator. After substitution, the differential \( dx \) is replaced with \( \frac{du}{u} \). This transforms the original integral into \[ \int \frac{u}{1 + u^2} \cdot \frac{du}{u} = \int \frac{du}{1 + u^2} \]. This simplification allows us to focus on solving a more manageable integral.

By using substitution, the problem reduces to evaluating a standard integral form, paving the way for much simpler integration techniques.

Arctangent Function

The next step involves recognizing the integral as a standard form associated with the arctangent function. The integral \( \int \frac{du}{1 + u^2} \) directly corresponds to the derivative of the arctangent function.

The antiderivative of this expression is \( \arctan(u) \), which is a classic result in calculus. This association is crucial because knowing the antiderivative allows us to proceed directly to evaluating the definite integral.

• The arctangent function, \( \arctan(u) \), is quite common in problems involving trigonometric substitutions or expressions.
• Recognizing standard integral forms helps speed up the integration process, saving time and reducing complexity.

Using this knowledge, we proceed to apply the integration limits to the antiderivative.

Integration Limits Adjustment

Adjusting the integration limits is a necessary step when using the substitution method. After substituting \( u = e^x \), the limits of integration change according to this substitution.

Initially, our limits in terms of \( x \) were from 0 to \( \ln \sqrt{3} \). With substitution, we calculate the new limits as follows:

• When \( x = 0 \), substituting gives \( u = e^0 = 1 \).
• When \( x = \ln \sqrt{3} \), substituting gives \( u = e^{\ln \sqrt{3}} = \sqrt{3} \).

This conversion is critical because it ensures the transformed integral accurately reflects the original integral's bounds.

Thus, the new integral limits are from 1 to \( \sqrt{3} \), which we use in the subsequent evaluation of the antiderivative.

Simplification of Results

The final step involves simplifying the result of the definite integral. After calculating the antiderivative at the new limits of integration, we have \( \arctan(\sqrt{3}) - \arctan(1) \). Knowing specific values of the arctangent function helps in this simplification.

• \( \arctan(\sqrt{3}) \) is known to be \( \frac{\pi}{3} \), since \( \tan(\frac{\pi}{3}) = \sqrt{3} \).
• \( \arctan(1) \) equals \( \frac{\pi}{4} \), as \( \tan(\frac{\pi}{4}) = 1 \).

After computing these values, the expression becomes \( \frac{\pi}{3} - \frac{\pi}{4} \). To simplify, we find a common denominator, resulting in:

• \( \frac{\pi}{3} = \frac{4\pi}{12} \)
• \( \frac{\pi}{4} = \frac{3\pi}{12} \)

The subtraction gives \( \frac{4\pi}{12} - \frac{3\pi}{12} = \frac{\pi}{12} \). This simplification is crucial as it provides the final answer in the most reduced form possible.