Problem 28 - AP Calculus with 8 Practice Tests 15th

Question

\(\int x \cos x d x=\) (A) \(x \sin x+C\) (B) \(x \sin x+\cos x+C\) (C) \(x \sin x-\cos x+C\) (D) \(\cos x-x \sin x+C\)

Step-by-Step Solution

Answer

The answer is (B) \(x \sin x + \cos x + C\).

Step 1: Identify Integration Technique

The integral involves the product of a polynomial function, \(x\), and a trigonometric function, \(\cos x\). This is a strong indication that integration by parts is a suitable technique to use here.

Step 2: Set Up Integration by Parts

The formula for integration by parts is \(\int u \, dv = uv - \int v \, du\). For this exercise, let us choose \(u = x\) and \(dv = \cos x \, dx\).

Step 3: Differentiate and Integrate

Differentiate \(u = x\) to find \(du = dx\). Integrate \(dv = \cos x \, dx\) to find \(v = \sin x\).

Step 4: Apply the Integration by Parts Formula

Substitute \(u\), \(v\), \(du\), and \(dv\) into the formula: \(\int x \cos x \, dx = x \sin x - \int \sin x \, dx\).

Step 5: Integrate Remaining Integral

The remaining integral is \(\int \sin x \, dx\). The integral of \(\sin x\) is \(-\cos x\). Thus, \(\int \sin x \, dx = -\cos x\).

Step 6: Simplify and Add Constant of Integration

Substitute back to get the final expression: \(x \sin x - \int \sin x \, dx = x \sin x + \cos x + C\).

Step 7: Verify the Solution

The integral \(\int x \cos x \, dx\) results in \(x \sin x + \cos x + C\). This matches option (B) in the multiple-choice answers.

Key Concepts

Trigonometric Integrals

Trigonometric integrals involve functions such as sine, cosine, tangent, and other trigonometric functions. They are a foundational concept in calculus due to the periodic nature and oscillating behavior of these functions. When dealing with integrals that involve trigonometric functions, it is important to recognize patterns and utilize appropriate techniques for simplification.

For instance, in our exercise, we are integrating the product of a linear function and a trigonometric function: - \[\int x \cos x \, dx\] - This requires us to choose the best method to integrate the function efficiently. Trigonometric identities may also facilitate integration if they transform the integrals into more manageable forms.

Calculus Problem Solving

Solving calculus problems often requires strategic thinking and the application of various techniques. The integration by parts technique is a method that comes in handy while dealing with products of functions seen in problems like this one.

Integration by parts is based on the product rule for differentiation, and it can be stated as follows:

• \[ \int u \, dv = uv - \int v \, du \]

By choosing the right parts for \(u\) and \(dv\), such as in our exercise where \(u = x\) and \(dv = \cos x \, dx\), you can transform a complex integral into one that is simpler to solve. This gets us to solve the integral step-by-step efficiently.

Once the formula is applied, the problem morphs into the integration of simpler parts that can be managed with basic integration knowledge, as illustrated in the original problem's step-by-step solution.

Integral Calculus Techniques

Integral calculus consists of numerous techniques that apply to different classes of functions. Techniques like substitution, integration by parts, and partial fraction decomposition allow us to find integrals that would otherwise be difficult to compute.

In the case of this exercise, integration by parts is pivotal:- We initially transform the integral's form by selecting suitable components \(u\) and \(dv\) from \(u = x\) and \(dv = \cos x \, dx\).- Differentiation of \(u\) gives \(du = dx\), and integration of \(dv\) gives \(v = \sin x\).- Applying the parts formula reduces the original complex integral to simpler forms, which we proceed to handle with direct integration.

This shows that choosing and applying the right method based on the functions involved can significantly ease problem solving and integration approaches, providing a more seamless calculus experience.