Problem 13 - Applied Calculus

Question

Find the integrals. $$ \int t \sin t d t $$

Step-by-Step Solution

Answer

The integral of \( \int t \sin t \, dt \) is \(-t \cos t + \sin t + C\).

Step 1: Set up Integration by Parts

For the integral \( \int t \sin t \, dt \), we will use the method of integration by parts. This method is based on the formula: \( \int u \, dv = uv - \int v \, du \). We need to choose \( u \) and \( dv \) from \( t \) and \( \sin t \, dt \).

Step 2: Choose \( u \) and \( dv \)

Let \( u = t \) and \( dv = \sin t \, dt \). This allows us to differentiate and integrate easily. Compute \( du = dt \) and find \( v \) by integrating \( dv \) such that \( v = \int \sin t \, dt = -\cos t \).

Step 3: Apply Integration by Parts Formula

Plug these into the integration by parts formula: \( \int t \sin t \, dt = uv - \int v \, du \). Substitute \( u = t \), \( v = -\cos t \), and \( du = dt \).

Step 4: Simplify the Expression

Substituting the values, we get: \[ \int t \sin t \, dt = t(-\cos t) - \int (-\cos t) \, dt = -t \cos t + \int \cos t \, dt. \]

Step 5: Integrate Remaining Integral

Now, integrate \( \int \cos t \, dt \). We know that \( \int \cos t \, dt = \sin t \).

Step 6: Write Final Expression

Substitute the integral from Step 5 into the expression from Step 4: \[-t \cos t + \sin t + C,\] where \( C \) is the constant of integration.

Key Concepts

Definite Integrals

In calculus, the concept of definite integrals is a powerful tool that allows us to evaluate the net area under a curve between two points. While the exercise at hand is an indefinite integral, it's beneficial to understand how definite integrals extend this idea. Unlike indefinite integrals, which return a family of functions, a definite integral provides a numerical value representing the accumulation of quantities over a specified interval. For example, \( \int_{a}^{b} f(x) \, dx \) calculates the total accumulation or the area under the function \( f(x) \) from \( x = a \) to \( x = b \). Using properties like the Fundamental Theorem of Calculus, definite integrals are particularly useful in practical applications such as computing distances, areas, and more.

• Converts a function's rate of change back to accumulated value
• Requires limits (upper and lower bounds) to find a specific numerical result
• Key concept in determining areas and physical quantities

These characteristics differentiate it from indefinite integrals, which emphasize finding a general antiderivative.

Trigonometric Functions

Trigonometric functions are essential components of calculus and mathematical analysis. They describe relationships in triangles and model periodic phenomena such as sound, light, and motion. In the given exercise, where we need to integrate \( \sin t \), understanding trigonometric functions is crucial. The sine function, \( \sin t \), osculates between \(-1\) and \(1\), making it suitable for modeling cycles.

• Sine and cosine are the foundational trigonometric functions used in integration
• Help express periodic behavior and have derivatives and integrals tied together
• Both have cyclical properties useful in various fields, including physics and engineering

When working with integrals involving trigonometric functions, it is essential to remember their antiderivatives, such as\( \int \sin t \, dt = -\cos t + C \), and corresponding formulas for derivatives. These concepts form various calculus problem-solving strategies.

Calculus Problem-Solving

Calculus problem-solving involves applying mathematical techniques to solve complex expressions, such as those encountered in this exercise about integration by parts. This technique simplifies the integration of products of functions, particularly when standard formulas are insufficient. The integration by parts formula, \( \int u \, dv = uv - \int v \, du \), requires selecting parts of the integrand to differentiate or integrate separately. Knowing when and how to choose \( u \) and\( dv \) is critical for simplifying an integral expression effectively.

• Integration by parts transforms complex integrals into simpler ones
• Choosing the right \( u \) and \( dv \) is key to success
• Step-by-step approach improves precision and understanding

Solving such problems strengthens analytical skills and deepens a student's understanding of calculus concepts. Practice enhances proficiency, allowing more efficient processing of integrals and better tackling of diverse calculus scenarios.