Problem 67 - Calculus: Early Transcendental Functions

Question

Find an antiderivative by reversing the chain rule, product rule or quotient rule. $$\int \frac{x \cos x^{2}}{\sqrt{\sin x^{2}}} d x$$

Step-by-Step Solution

Answer

The antiderivative of the given function is \(-2\sqrt{\sin (x^2)} + C\).

Step 1: Identifying Inner Function

Firstly, identify the inner function. Here, it's noted that \(x^2\) is the inner function. It's present both in cosine and sine functions.

Step 2: Calculating the Derivative of Inner Function

The derivative of the inner function \(x^2\) is \(2x\). Notice that this is the factor that multiplies the cosine function in the integral.

Step 3: Setting Up a Substitution

Since \(u = x^2\) is the inner function, make a substitution such that \(u = x^2\). The corresponding derivative is now \(du = 2x dx\), which is part of the integrand.

Step 4: Substituting in the Integral

Replace \(x^2\) by \(u\), \(2x dx\) by \(du\) and divide the integral by 2 since \(du\) was defined above as equal to \(2x dx\) to get: \(\frac{1}{2}\int \frac{\cos(u)}{\sqrt{\sin(u)}} du\)

Step 5: Calculating the Integral

The integral can now be solved using standard methods. The integral of \(\frac{\cos(u)}{\sqrt{\sin(u)}}\) is \(-2\sqrt{\sin(u)} + C\), where \(C\) is the constant of integration.

Step 6: Substituting \(u\) back in terms of \(x\)

Finally, substitute the \(u\) back into the integral. Therefore, the antiderivative of the original function is \(-2\sqrt{\sin (x^2)} + C\).

Key Concepts

Exploring Integration Techniques

Integration is a fundamental concept in calculus, used to find antiderivatives or the area under a curve. When dealing with integrals, we often apply different techniques, including:

• By parts: Useful when integrating the product of two functions.
• Substitution: Primarily when a function's derivative appears within the integrand.
• Partial fractions: Effective for rational functions.

In our exercise, we apply the **substitution method**, one of the core techniques. This involves finding a part of the integrand whose derivative is also present. After substitution, the integral becomes simpler. We'll explore this further in the next section.

Understanding the Chain Rule

The chain rule is crucial when dealing with derivatives of composite functions. If you have a function composed of two or more smaller functions, the chain rule helps differentiate it. For a function in the form \(f(g(x))\), the derivative is \(f'(g(x)) \, g'(x)\).
In reverse, working back to the antiderivative using the chain rule involves finding the original function from its derivative. In our example, \(x^2\) is the inner function embedded in a composite structure with cosine and sine. Recognizing this is vital as it guides us on how to simplify the integration process using substitution.

Decoding Substitution Method

The substitution method is a pivotal integration technique. It involves changing variables in an integral to reach a simpler form. Here's the step-by-step approach:

• Identify the inner function in the integrand, often recognizable via its derivative also appearing in the integral.
• Set the identified inner function as \(u\), and derive \(du\) in relation to \(x\).
• Transform the integrand using these substitutions.
• Remember to revert your substitution once the integral is evaluated.

In our problem, \(u = x^2\). The presence of \(2x\), the derivative of \(x^2\), in the integral, simplifies this approach. The method effectively extracts the complexity, resulting in manageable calculations.

The Bigger Picture of Calculus

Calculus fundamentally deals with change, through differentiation and integration. It provides solutions for various mathematical models in physics, engineering, and economics.

• **Differentiation** helps us understand rates of change and slopes of curves.
• **Integration** aids in finding areas under curves and reconstructing functions from their rates of change.

The tools and methods like the chain rule or the substitution method allow tackling complex integration problems. Mastering these concepts offers profound insights into the workings of dynamic systems and functions. Integrals, as seen in our example, are not just abstract concepts but are deeply interwoven in real-life applications.