Problem 61 - Thomas Calculus

Question

Right, or wrong? Say which for each formula and give a brief reason for each answer. $$ \begin{array}{l}{\text { a. } \int x \sin x d x=\frac{x^{2}}{2} \sin x+C} \\\ {\text { b. } \int x \sin x d x=-x \cos x+C} \\ {\text { c. } \int x \sin x d x=-x \cos x+\sin x+C}\end{array} $$

Step-by-Step Solution

Answer

a. Wrong, it's not integration by parts. b. Wrong, missing \(\sin x\). c. Right, matches integration result.

Step 1: Use Integration by Parts Formula for Part a

To solve \( \int x \sin x \, dx \) using integration by parts, first recall the formula: \( \int u \, dv = uv - \int v \, du \). Let's choose \( u = x \) and \( dv = \sin x \, dx \). Then, derive: \( du = dx \) and \( v = -\cos x \). Using these, the integration becomes:\[ \int x \sin x \, dx = -x \cos x - \int (-\cos x) \, dx = -x \cos x + \int \cos x \, dx = -x \cos x + \sin x + C \]The given formula for part a, \( \frac{x^{2}}{2} \sin x+C \), does not match the calculated one, thus it is incorrect.

Step 2: Evaluate Provided Solution for Part b

Now consider the formula \( \int x \sin x \, dx = -x \cos x + C \). Comparing this with the correct integration by parts result \( -x \cos x + \sin x + C \), it is evident that the term \( \sin x \) is missing in part b's solution. Hence, this formula is incorrect.

Step 3: Evaluate Provided Solution for Part c

Finally, examine the formula \( \int x \sin x \, dx = -x \cos x + \sin x + C \). This matches exactly with the correct integration by parts result. Therefore, this part is correct.

Key Concepts

Definite Integrals

Definite integrals are essential in calculus as they allow us to calculate the exact accumulation of quantities over a specific interval. A definite integral is represented by the symbol \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) denote the limits of integration.
These limits indicate the interval on the x-axis over which we're summing the area under the curve of the function \( f(x) \). Instead of producing another function, like indefinite integrals, definite integrals yield a numeric value.

• The value represents the net area between the function and the x-axis from \( x = a \) to \( x = b \).
• If the area is above the x-axis, the contribution is positive, and if below, it is negative, reflecting the area 'signed' concept.

Understanding definite integrals is pivotal for solving many problems related to areas, volumes, and even differential equations. These integrals form a cornerstone for the Fundamental Theorem of Calculus, linking the concept of antiderivatives and their practical applications.

Indefinite Integrals

Indefinite integrals are used to find antiderivatives, and they are represented without limits of integration as \( \int f(x) \, dx \). They produce a family of functions rather than a numeric value. When integrating terms like \( x \sin x \), we often use techniques like the integration by parts method.
This method is useful when direct integration seems complex or impossible. According to the formula \( \int u \, dv = uv - \int v \, du \), we break down a product into simpler integration tasks.

• Choosing parts wisely, like \( u = x \), gives simpler derivatives, and \( dv = \sin x \, dx \), provides solvable integrals as \( v = -\cos x \).
• Once you determine \( du = dx \) and \( v = -\cos x \), you solve. A term like \( -x \cos x + \sin x + C \) emerges, illustrating the completed indefinite integral form.

The constant \( C \) represents the set of all possible antiderivatives, since an antiderivative is determined up to an additive constant. Mastering indefinite integrals is crucial for tackling a wide range of calculus challenges, including solving differential equations and exploring functions analytically.

Trigonometric Integration

Trigonometric integration involves integrating functions involving trigonometric terms like \( \sin x \) or \( \cos x \). It's a key tool when dealing with oscillatory phenomena in physics and engineering, usually necessitating unique techniques or identities to simplify and solve.
Using trigonometric identities can help simplify integrals. However, sometimes integration by parts, as showcased in the exercise, is the path forward.
For example, integrating \( x \sin x \) goes beyond basic integration rules, and instead uses parts:

• Identify \( u \) and \( dv \). With \( u = x \) and \( dv = \sin x \, dx \), you simplify to find \( du \) and \( v \), where \( v = -\cos x \).
• Apply the formula: \( \int x \sin x \, dx = -x \cos x + \sin x + C \). This demonstrates the interplay between algebraic and trigonometric elements.

Knowing when to use trigonometric identities or the integration by parts method forms the core of mastering trigonometric integrals. This knowledge allows for efficiently solving a variety of calculus problems involving trigonometric functions.