Problem 39 - Calculus: Graphical, Numerical, Algebraic

Question

Multiple Choice \(\int x \sin (5 x) d x=\) (A) \(-x \cos (5 x)+\sin (5 x)+C\) (B) \(-\frac{x}{5} \cos (5 x)+\frac{1}{25} \sin (5 x)+C\) (C) \(-\frac{x}{5} \cos (5 x)+\frac{1}{5} \sin (5 x)+C\) (D) \(\frac{x}{5} \cos (5 x)+\frac{1}{25} \sin (5 x)+C\) (E) \(5 x \cos (5 x)-\sin (5 x)+C\)

Step-by-Step Solution

Answer

Therefore, the integral \(\int x \sin (5 x) d x = -\frac{1}{5}x \cos(5x) + \frac{1}{25}\sin(5x) + C \), which corresponds to option (B).

Step 1: Identify 'u' and 'dv'

The first step in using the integration by parts formula is to identify 'u' and 'dv'. Here, let's take \(u = x\) and \(dv = \sin(5x) dx\) . This choice is made as the derivative of 'x' is simpler, i.e., '1' and easy to integrate \(\sin(5x)\).

Step 2: Compute 'du' and 'v'

Having identified 'u' and 'dv', we now calculate 'du' and 'v'. For \(u = x\), \(du = dx\), and \(v = \int dv = -\frac{1}{5} \cos(5x)\).

Step 3: Apply Integration by Parts

Now we apply the integration by parts formula: \(\int u dv = uv - \int v du\). This results in \(\int x \sin(5x) dx = x (-\frac{1}{5}\cos(5x)) - \int (-\frac{1}{5}\cos(5x)) dx\).

Step 4: Simplify and Calculate the Remaining Integral

Simplifying gives \(-\frac{1}{5}x \cos(5x) + \frac{1}{5}\int \cos(5x) dx\). The integral of \(\cos(5x)\) can be calculated directly which results in \(\frac{1}{5} \sin(5x)\).

Step 5: Combine Outputs

So, our final answer is \(-\frac{1}{5}x \cos(5x) + \frac{1}{25}\sin(5x) + C\) (Remember to add '+ C' to signify the constant of integration.)

Key Concepts

Definite Integrals

Definite integrals represent the accumulation of quantities, and they can be thought of as the total area under a curve between two bounds. Unlike indefinite integrals, they result in a numerical value instead of a function with a constant \(+C\).

When evaluating a definite integral, one applies the fundamental theorem of calculus, which involves first finding an antiderivative of the function, and then calculating the difference between the antiderivative evaluated at the upper limit and the lower limit.

• This process provides a concrete number, giving it real-world application such as calculating distance, area, and other extensive quantities.
• Definite integrals are integral to solving problems involving rates of change and accumulations.

Although the original exercise involves an indefinite integral, understanding definite integrals helps us appreciate the complete toolkit of integration techniques.

Antiderivatives

Antiderivatives, or indefinite integrals, are the reverse of derivatives. They find a function whose derivative is the original function provided. The process of finding an antiderivative is often called 'integration'.

The integration by parts technique is essential in practical calculus. It is a specific method for integrating the product of two functions. For example, in this exercise, we solve for the function by choosing what to differentiate \( (u) \) and what to integrate \( (dv) \).

• Choosing wisely can significantly simplify the process.
• For instance, choosing \( u = x \) makes the differentiation simple, resulting in \( du = dx \).

Ultimately, finding the correct antiderivative involves combining basic integral rules, formulas, and strategic choices to simplify and solve the original problem.

Trigonometric Functions

Trigonometric functions like \( \sin(x) \) and \( \cos(x) \) are often involved in calculus due to their wave-like properties and periodic nature.

In this integration exercise, \( \sin(5x) \) is part of the function being integrated. Understanding these functions' integrals, particularly how they convert between sine and cosine, is crucial.

• For instance, the integral of \( \sin(ax) \) transforms into a negative cosine function divided by the coefficient of \( x \), here \(-\frac{1}{a} \cos(ax)\).
• This property is not just a mathematical shortcut but rather a representation of the trigonometric function's inherent properties.

Trigonometric identities and integrals are pivotal to solving complex calculus problems, enabling the transformation of functions to forms that are simpler to integrate.

Calculus Problem Solving

Problem solving in calculus often involves breaking a problem down into smaller, manageable steps. This exercise demonstrates this through the integration by parts technique.

The step-by-step solution provided effectively illustrates a common calculus strategy:

• First, pinpoint what to select for \( u \) and \( dv \). Here, \( u = x \) and \( dv = \sin(5x) dx \).
• Compute \( du \) and \( v \), which simplifies future steps.
• Use the integration by parts formula, \( \int u dv = uv - \int v du \), to transform the problem.
• Simplify, and solve the resulting integral to get the final solution.

Through this systematic approach, complex calculus problems are demystified, reinforcing mastery through practice and logical reasoning. Understanding the underlying concepts of each step makes solving similar problems in the future much more intuitive.