Problem 65 - Calculus: Early Transcendental Functions

Question

Find an antiderivative by reversing the chain rule, product rule or quotient rule. $$\int\left(x \sin 2 x+x^{2} \cos 2 x\right) d x$$

Step-by-Step Solution

Answer

\(\frac{1}{2}x^{2}\sin(2x) + C\)

Step 1: Identify the function and its derivative

By looking at the expression, it can be determined that \(x\sin(2x)\) and \(x^2\cos(2x)\) are the product of a function and its derivative. The function in this case is \(f(x) = x^2\) and its derivative is \(f'(x) = 2x\). We can observe that there is a \(2x\) in both \(x\sin(2x)\) and \(x^2\cos(2x)\), and a \(x\) in \(x\sin(2x)\), which are part of the derivative of \(f(x)\).

Step 2: Use the reverse of product rule

The reverse of the product rule states that the integral of a function and its derivative is equal to the function itself. Therefore, the integral of the expression can be denoted as follows: \(\int \left(x \sin 2 x+x^{2} \cos 2 x\right) d x = \frac{1}{2}x^{2}\sin(2x) + C\), where \(C\) is the constant of integration.

Key Concepts

Chain Rule

The chain rule is a critical concept in calculus, particularly when working with composite functions. A composite function occurs when one function is nested within another. For example, if you have a function which is the sine of a doubled variable, like \( \sin(2x) \), you're dealing with a composition of functions: the sine function and the function \( 2x \).

The chain rule states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function. In equation form, if \( f(x) = g(h(x)) \), then \( f'(x) = g'(h(x)) \cdot h'(x) \). In the given exercise, although we are looking for an antiderivative, understanding the chain rule helps you identify parts of the integrand that are the result of applying this rule, which in turn helps reverse the process for integration.

Product Rule

The product rule is essential for finding the derivative of two multiplied functions. Suppose we have two differentiable functions \( u(x) \) and \( v(x) \). Their derivative, when multiplied, isn't just \( u'(x) \cdot v'(x) \), as one might initially guess. Instead, the product rule tells us that \( (u(x)v(x))' = u'(x)v(x) + u(x)v'(x) \).

For our particular exercise, we're dealing with the reverse situation. You're integrating an expression that could have come from applying the product rule. This means that you can sometimes 'split' the integrated function into two parts that resemble \( u(x) \) and \( v(x) \) and figure out the original function. In this case, the expression \( x \sin(2x) + x^2 \cos(2x) \) when integrated, suggests a function whose differentiation via the product rule would yield these terms.

Quotient Rule

The quotient rule is a derivative rule used for a function divided by another function. It's somewhat similar to the product rule but is applied when division is involved. Given two differentiable functions \( u(x) \) and \( v(x) \) where \( v(x) \) is nonzero, the quotient rule is given by \( \left(\frac{u(x)}{v(x)}\right)' = \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2} \).

In the context of our exercise, the quotient rule isn't directly applicable. However, understanding how it differs from the product rule can help students avoid confusion when determining which rule to reverse when finding an antiderivative. It's crucial to recognize the structure of the terms to apply the correct 'reverse rule' for integration.

Integration by Parts

Integration by parts is a technique based on the product rule of differentiation. Essentially, it is the product rule in reverse and is used to integrate products of functions where standard integration methods do not apply. The formula for integration by parts is given by \( \int u dv = uv - \int v du \), where \( u \) and \( dv \) are parts of the original function chosen strategically.

Albeit not directly used in this exercise, integration by parts can be invaluable when dealing with integrals of products where each part of the product is not simply a derivative of another function. Knowledge of this technique also broadens understanding of how product rules operate in integration, allowing for better insight into solving complex integrals.