Problem 57 - Essential Calculus

Question

In Exercises \(57-62,\) use integration by parts to verify the formula. (For Exercises \(57-60,\) assume that \(n\) is a positive integer. \()\) $$ \int x^{n} \sin x d x=-x^{n} \cos x+n \int x^{n-1} \cos x d x $$

Step-by-Step Solution

Answer

The proof shows that the original formula \(\int x^n sinx dx = -x^n cosx + n \int x^{n-1} cosx dx\) is correct.

Step 1: Define the Function Parts

Integration by parts follows the formula \(\int u dv=uv−\int v du\), where \(u\) is one part of the function and \(dv\) is the other part. In our integration, \(x^n\) is chosen as \(u\), and \(sinx dx\) as \(dv\). So differentiate \(u\) to get \(du = n*x^{n-1} dx\) and integrate \(dv\) to get \(v =-cosx\).

Step 2: Apply Integration by Parts Formula

Now plug \(u\), \(du\), \(v\), and \(dv\) into the integration by parts formula \(\int u dv=uv−\int v du\). This leads to: \(\int x^n sinx dx = -x^n* cosx - \int -cosx * n*x^{n-1} dx\), which simplifies to: \(-x^n cosx + n \int x^{n-1} cosx dx\).

Step 3: Compare Result with Original Integration

Compare the resulting formula with the original formula given in the question: \(-x^n cosx + n \int x^{n-1} cosx dx = -x^n cosx + n \int x^{n-1} cosx dx\). The two formulas match, therefore the verification is done. We just demonstrated that: \(\int x^n sinx dx = -x^n cosx + n \int x^{n-1} cosx dx\).

Key Concepts

Integration Techniques

Integration techniques are tools that help us solve complex integrals that aren't solvable by basic methods alone.One of the versatile techniques is 'Integration by Parts'. It is particularly useful for integrating products of functions, such as polynomials multiplied by trigonometric, exponential, or logarithmic functions.
When integrating by parts, we use the formula: \[\int u \ dv = uv - \int v \ du\]The choice of which part of the integrand becomes \(u\) and which becomes \(dv\) can massively simplify the problem. For example:

• Choose \(u\) such that its derivative, \(du\), simplifies the equation.
• Choose \(dv\) that easily integrates to \(v\).
  

In our problem, with \(x^n \sin x\), we selected \(x^n\) as \(u\). It becomes simpler when derived. We select \(\sin x \, dx\) as \(dv\) because its integration to \(-\cos x\) is straightforward. This choice yields an elegantly simplified solution.

Definite and Indefinite Integrals

Integrals can be classified into two main types: definite and indefinite.An indefinite integral, typically denoted without upper and lower limits, represents a general form of the area under a curve as a function plus a constant \(C\). This constant represents an infinite family of curves, all shifted vertically.
For example: \[\int x^n \ dx = \frac{x^{n+1}}{n+1} + C\]Definite integrals have specified upper and lower limits.They result in a numerical value representing the actual area under a curve between two points.While the exercise uses indefinite integrals, understanding both types deepens comprehension.It's the indefinite form that we apply in the integration by parts process.

Calculus Problem Solving

Solving calculus problems often involves selecting the right method for the given problem. Whenever you face an integral, think of your set of tools like substitution or integration by parts.
Remember these strategies:

• Break down complex problems into parts.
• Try to simplify functions or identify patterns.
  
• Verify solutions by comparing them, as we did in our exercise to check against the formula.

Effective problem-solving in calculus revolves around practice and understanding when and how to apply techniques. In our example, picking pieces of the function for integration by parts required knowledge of how derivatives and integrals work with each piece.
Continued practice will make recognizing these opportunities second nature.