Question

Find the integral by using the simplest method. Not all problems require integration by parts. $$ \int v \sin v d v $$

Short answer

The integral is \( -v \cos v + \sin v + C \).

Step-by-step solution

Identify the Integration Method

The integral \( \int v \sin v \ dv \) does not fit the standard forms or basic integration rules. It requires the use of integration techniques. The product of a polynomial \( v \) and a trigonometric function \( \sin v \) suggests using integration by parts, a method suitable for integrals of products of functions.

Recall the Integration by Parts Formula

The formula for integration by parts is given by \( \int u \, dv = uv - \int v \, du \). We need to choose \( u \) and \( dv \) such that the resulting integral \( \int v \, du \) is simpler to compute. Typically, \( u \) is selected to simplify upon differentiation, and \( dv \) should be easily integrable.

Select \( u \) and \( dv \)

Set \( u = v \), which means \( du = dv \). For \( dv \), choose \( dv = \sin v \ dv \), which gives \( v = -\cos v \) after integration. So we have \( u = v \), \( dv = \sin v \ dv \), \( du = dv \), and \( v = -\cos v \).

Apply Integration by Parts Formula

Substitute the chosen functions into the integration by parts formula: \( \int v \sin v \ dv = uv - \int v \, du \). This becomes \( v(-\cos v) - \int (-\cos v) \, dv \), which simplifies to \( -v \cos v + \int \cos v \, dv \).

Simplify and Integrate

The integral \( \int \cos v \, dv \) is straightforward and equals \( \sin v \). Substitute this back into the equation: \( -v \cos v + \sin v + C \), where \( C \) is the constant of integration.

Conclude the Solution

The solution to the integral is \( -v \cos v + \sin v + C \), where \( C \) is an arbitrary constant representing the family of antiderivatives for the indefinite integral.

Key concepts

Integration Techniques

When tackling integrals, selecting the right method is crucial for simplifying the problem-solving process. There are a variety of integration techniques, but the key is knowing which one to apply based on the function you are dealing with.

For products of functions, especially involving polynomials and trigonometric terms like in the integral \( \int v \sin v \, dv \), one should consider using **Integration by Parts**. This method is particularly effective when a direct integration is not feasible due to the complexity of the function product.

The integration by parts formula, \( \int u \, dv = uv - \int v \, du \), is derived from the product rule of differentiation. It allows us to transform the original integral into a simpler form by wisely choosing parts of the integrand to differentiate or integrate. Choosing \( u \) and \( dv \) effectively hinges on which part will simplify upon differentiation (to form \( du \)) and which part is straightforward to integrate (to form \( v \)).

Practice recognising where integration by parts will be useful, and choose your functions \( u \) and \( dv \) to make the remaining calculations as simple as possible.

Indefinite Integrals

Indefinite integrals represent a family of functions whose derivative is the integrand. They are essential in calculus for solving problems involving area under a curve and accumulation functions. An indefinite integral is expressed as \( \int f(x) \, dx \), and its solution is a function with an arbitrary constant \( C \), reflecting the many possible antiderivatives of \( f(x) \).

The constant \( C \) comes into play because integration is the inverse operation of differentiation, which makes it possible to recover any of the infinite functions that could have led to the same derivative.

In our example, after applying integration by parts to solve \( \int v \sin v \, dv \), we integrate to get \( -v \cos v + \sin v \). Adding \( C \) gives us the complete family of solutions: \[ -v \cos v + \sin v + C \].

Whenever you solve for an indefinite integral, remember to include \( C \) as it signifies the family of all possible solutions, acknowledging shifts vertically in the graph of the antiderivative.

Trigonometric Integration Methods

Trigonometric integration involves integrating functions that contain trigonometric expressions such as \( \sin x \), \( \cos x \), or other related functions. Mastering these techniques requires familiarity with both basic trigonometric identities and calculus rules.

In the problem approach, involving the integral \( \int \sin v \, dv \) is straightforward since the antiderivative of \( \sin v \) is \( -\cos v \). However, when these terms are present in a more complex product, as in \( \int v \sin v \, dv \), a method like integration by parts is needed.

Remember the standard trigonometric identities, such as \( \sin^2 x + \cos^2 x = 1 \), or angle sum and difference formulas, as they can sometimes simplify the calculations or integrands.

It's important to practise solving integrals combining polynomial and trigonometric terms since they frequently appear in practical scenarios like physics and engineering, and other applied mathematics contexts.