Problem 63 - Thomas Calculus in SI Units

Question

Evaluate the integrals. $$\int_{0}^{\ln \sqrt{3}} \frac{e^{x} d x}{1+e^{2 x}}$$

Step-by-Step Solution

Answer

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

Step 1: Simplify the integrand

Consider the integral \[ \int_{0}^{\ln \sqrt{3}} \frac{e^{x} dx}{1+e^{2x}} \]To simplify, notice that \( e^{2x} = (e^x)^2 \), so rewrite as\[ \frac{e^x}{1+(e^x)^2}\] This setup is reminiscent of the derivative of \( \tan^{-1}(x) \), which is \( \frac{1}{1+x^2} \).

Step 2: Use substitution

Perform a substitution to simplify the integral. Let \( u = e^x \), so \( du = e^x \, dx \). This means \[ dx = \frac{du}{u} \]Change the limits of integration: when \( x = 0 \), \( u = e^0 = 1 \); when \( x = \ln \sqrt{3} \), \( u = \sqrt{3} \).Thus, the integral becomes\[ \int_{1}^{\sqrt{3}} \frac{1}{1+u^2} du \]

Step 3: Identify the antiderivative

Recognize that the integral \[ \int \frac{1}{1+u^2} du \]is a standard form, which equals \( \tan^{-1}(u) + C \) where \( C \) is the integration constant.

Step 4: Evaluate the definite integral

Evaluate the definite integral using the antiderivative found in Step 3:\[ \left[ \tan^{-1}(u) \right]_{1}^{\sqrt{3}} = \tan^{-1}(\sqrt{3}) - \tan^{-1}(1)\]Recall the standard angle values: \( \tan^{-1}(\sqrt{3}) = \frac{\pi}{3} \) and \( \tan^{-1}(1) = \frac{\pi}{4} \).

Step 5: Compute the final result

Subtract the two angles to find the result:\[ \frac{\pi}{3} - \frac{\pi}{4} = \frac{4\pi}{12} - \frac{3\pi}{12} = \frac{\pi}{12} \]Conclude that the value of the integral is \( \frac{\pi}{12} \).

Key Concepts

Definite Integral

The concept of a definite integral is fundamental in calculus, as it allows us to find the exact area under a curve over a specified interval. This is different from an indefinite integral, which represents a family of functions. A definite integral has

• specific upper and lower limits of integration
• a definite numerical value as its result

When we evaluate a definite integral like \[\int_{0}^{\ln \sqrt{3}} \frac{e^{x} dx}{1+e^{2x}}\]the process involves several steps, including finding an antiderivative and using the limits of integration to compute the exact value.

In this example, we applied the substitution method to simplify the process before finding the antiderivative. Once the antiderivative was identified, we evaluated it between the limits determined by substituting the original bounds. The final result was calculated by subtracting the antiderivative's values at these bounds.

Substitution Method

The substitution method, also known as u-substitution, is a common technique used in calculus for simplifying the process of integration. It involves changing the variable of integration to make the integral easier to solve. Here's how it works:

• Select a substitution variable; in our example, we chose \( u = e^x \).
• Express \( dx \) in terms of \( du \), as \( dx = \frac{du}{u} \).
• Transform the integral from the original variable \( x \) to \( u \).
• Don't forget to change the limits of integration based on the substitution.
  
  

After substitution, the integral simplification often results in a standard form that can be easily integrated. In this case, it reduced the integral to the recognizable form \( \int \frac{1}{1+u^2} \, du \), which is the antiderivative of \( \tan^{-1}(u) \). This method relies on identifying expressions within the integrand that match the derivative of a composed function.

Trigonometric Integration

Trigonometric integration involves using trigonometric identities and inverse trigonometric functions to evaluate integrals. Often these techniques are necessary when dealing with rational functions involving quadratic expressions in the denominator or expressions like sine and cosine.

• In this exercise, the integral \( \int \frac{1}{1+u^2} \, du \) is recognized as the derivative of the \( \tan^{-1}(u) \) function.
• This recognition allows us to directly identify the antiderivative, thus simplifying integration.

When performing trigonometric integration, it's helpful to remember standard integrals:

• For \( \int \frac{1}{1+u^2} \, du \), the result is \( \tan^{-1}(u) + C \), where \( C \) is the constant of integration.
• Knowing such forms can significantly speed up the process of integration by substitution or direct pattern recognition.
  
  

In our worked example, applying the trigonometric inverse function made it straightforward to compute the definite integral by using its known values at specific bounds, resulting in \( \frac{\pi}{12} \).