Problem 48 - Thomas Calculus in SI Units

Question

Evaluate the integrals in Exercises \(29-50.\) $$\int_{0}^{\sqrt{\ln \pi}} 2 x e^{x^{2}} \cos \left(e^{x^{2}}\right) d x$$

Step-by-Step Solution

Answer

The integral evaluates to \(-\sin(1)\).

Step 1: Identify the Integration Technique

The integral \( \int_{0}^{\sqrt{\ln \pi}} 2 x e^{x^{2}} \cos \left(e^{x^{2}}\right) d x \) contains the function \( 2x \cdot e^{x^2} \) with \( \cos(e^{x^2}) \). The presence of \( e^{x^{2}} \) and its derivative \( 2x \) suggests using a substitution method.

Step 2: Choose a Suitable Substitution

Choose the substitution \( u = e^{x^2} \). Then, differentiate to find \( du \). Thus, \( du = 2x e^{x^2} dx \). This matches the structure of the integral and simplifies the expression.

Step 3: Change the Limits of Integration

Convert the limits of integration from \( x \) to \( u \). When \( x = 0, u = e^{0^2} = 1 \). When \( x = \sqrt{\ln \pi}, u = e^{(\sqrt{\ln \pi})^2} = \pi \). Thus, the new limits are from \( u = 1 \) to \( u = \pi \).

Step 4: Compute the Integral

Replace the expression inside the integral with respect to \( u \): \( \int_{0}^{\sqrt{\ln \pi}} 2 x e^{x^{2}} \cos \left(e^{x^{2}}\right) d x = \int_{1}^{\pi} \cos(u) \, du \). The integral of \( \cos(u) \) is \( \sin(u) \). Thus, evaluate \( \sin(u) \) from \( u = 1 \) to \( u = \pi \).

Step 5: Evaluate the Resulting Expression

Compute \( \left[ \sin(\pi) - \sin(1) \right] \). Since \( \sin(\pi) = 0 \) and \( \sin(1) \) will remain as is, the result is \( 0 - \sin(1) = -\sin(1) \).

Key Concepts

Substitution Method

The substitution method is a powerful technique in calculus, especially useful when dealing with complex integrals that are difficult to solve directly. It involves substituting a part of the integral with a new variable, simplifying the scenario. In our example, we have the integral \( \int_{0}^{\sqrt{\ln \pi}} 2x e^{x^2} \cos(e^{x^2}) \, dx \). Observing the expression, we notice \( e^{x^2} \) and its derivative \( 2x \) hint that substitution is apt here.

Here's how it works:

• **Choose a substitution:** We chose \( u = e^{x^2} \).
• **Differentiate to find \( du \):** This gives \( du = 2x e^{x^2} \, dx \).
• **Transform the integral:** Replace all expressions in terms of \( u \) and \( du \).

This replacement turns our initial complicated-looking integral into a simpler form that's easier to integrate, such as \( \int \cos(u) \, du \). After transforming back with original limits, you can solve the integral more straightforwardly.
This method not only simplifies the integration process but also solidifies a connection between calculus and algebraic manipulations.

Definite Integrals

Definite integrals play a critical role in calculus, allowing us to calculate the accumulated quantity represented by the area under the curve of a function. They are integral in scenarios such as determining displacement from velocity graphs or evaluating charge from a current over time.

In our problem, we evaluated the definite integral of a function from \( x = 0 \) to \( x = \sqrt{\ln \pi} \). However, through substitution, the limits were adjusted to \( u = 1 \) to \( u = \pi \).

• **Limits of integration:** Changing the variable means updating the integration limits to match the new variable \( u \).
• **Simplified evaluation:** With simpler integrands post-substitution, calculating definite integrals usually becomes more manageable.

For this case, this method transformed our problem into finding \( \int_{1}^{\pi} \cos(u) \, du \), which is much simpler to evaluate. Ultimately, definite integrals provide a clear answer after evaluating the upper and lower bounds, giving a precise measure of accumulated change.

Trigonometric Functions

Trigonometric functions such as sine and cosine frequently appear in calculus, reflecting their prevalence in various mathematical phenomena related to periodic behavior.
In this exercise, the integral involved a trigonometric function: \( \cos(e^{x^2}) \). After substituting \( u = e^{x^2} \), it became a simpler form: \( \cos(u) \).

Why is \( \cos(u) \) straightforward to integrate?

• **Basic trigonometric identities:** Recognizing that the integral of \( \cos(u) \) is \( \sin(u) \) based on fundamental identities simplifies the problem.
• **Evaluation of trigonometric limits:** Calculating trigonometric function values like \( \sin(\pi) \) and \( \sin(1) \) becomes crucial for final result evaluation.

This simplification provides clear pathways to solving integrals involving trigonometric functions, relying on their periodic nature and well-known integral results. This particular exercise concluded by computing \( [\sin(\pi) - \sin(1)] = -\sin(1) \), underscoring the importance of these functions in integration.