Problem 6 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \text { use integration by parts to evaluate each integral. } $$ $$ \int x \sin 2 x d x $$

Step-by-Step Solution

Answer

The integral is \(-\frac{x}{2} \cos 2x + \frac{1}{4} \sin 2x + C \).

Step 1: Identify Parts for Integration by Parts

In the integral \( \int x \sin 2x \, dx \), we need to choose \( u \) and \( dv \). We let \( u = x \) (which makes \( du = dx \)) and \( dv = \sin 2x \, dx \) (so \( v \) will be its integral).

Step 2: Differentiate and Integrate Parts

Differentiate \( u = x \) to get \( du = dx \). Now, integrate \( dv = \sin 2x \, dx \) to get \( v \). The integral of \( \sin 2x \) is \(-\frac{1}{2}\cos 2x \). Thus, \( v = -\frac{1}{2}\cos 2x \).

Step 3: Apply Integration by Parts Formula

Use the integration by parts formula: \( \int u \, dv = uv - \int v \, du \). Substitute \( u, v, du, \) and \( dv \) into the formula. We get: \(\int x \sin 2x \, dx = x \left(-\frac{1}{2}\cos 2x\right) - \int \left(-\frac{1}{2}\cos 2x\right) \, dx \).

Step 4: Simplify and Integrate Remaining Integral

Simplify the expression: \(-\frac{x}{2} \cos 2x + \frac{1}{2} \int \cos 2x \, dx \). Now, integrate \( \cos 2x \, dx \), which is \( \frac{1}{2} \sin 2x \).

Step 5: Substitute Back and Simplify

Substitute back the integral of \( \cos 2x \): \(-\frac{x}{2} \cos 2x + \frac{1}{2} \left( \frac{1}{2} \sin 2x \right) \).Simplify the expression to get the final answer:\(-\frac{x}{2} \cos 2x + \frac{1}{4} \sin 2x + C \), where \( C \) is the constant of integration.

Key Concepts

Integration Techniques

Integration techniques are essential tools in calculus that help solve various integral problems. One such technique is integration by parts, useful when the integration of a product of functions is involved. The integration by parts formula is given by:

• \( \int u \, dv = uv - \int v \, du \)

This formula is derived from the product rule for differentiation and is particularly effective when dealing with integrals of algebraic and trigonometric products. To apply this technique, we choose parts of the integrand to differentiate (\( u \)) and integrate (\( dv \)).
The success of this method largely depends on wisely choosing \( u \) and \( dv \). A common strategy is to let \( u \) be a function that becomes simpler when differentiated, while \( dv \) should be something that can be easily integrated. This helps simplify the problem into a form that can be more straightforward to evaluate.

Definite and Indefinite Integrals

Integrals are fundamental in calculus and come in two main types: definite and indefinite. An indefinite integral, represented as \( \int f(x) \, dx \), provides a family of functions, or antiderivatives, for a given function. The result includes a constant \( C \), since differentiating a constant results in zero, leaving the original function unchanged.

• Indefinite integrals find the general form of an antiderivative.
• The indefinite integral of a function \( f(x) \) is the reverse process of differentiation.

Definite integrals, on the other hand, calculate the net area under a curve within a specific interval \([a, b]\). It is represented by \( \int_{a}^{b} f(x) \, dx \) and does not include a constant \( C \), as it produces a numerical value. The definite integral is key in applications such as finding areas, volumes, and other quantities related to accumulation.

Trigonometric Integrals

Trigonometric integrals involve integrating functions containing trigonometric expressions like sine, cosine, tangent, etc. These integrals might require specific strategies due to the periodic and oscillating nature of trigonometric functions. Basic integral formulas for trigonometric functions can be memorized for simpler calculations:

• \( \int \sin x \, dx = -\cos x + C \)
• \( \int \cos x \, dx = \sin x + C \)

For more complex integrands, integration techniques like substitution, or in our case, integration by parts, may be applied. In the given exercise, the integral of function \( \sin 2x \) was needed, solvable by recognizing that \( \int \sin ax \, dx = -\frac{1}{a} \cos ax + C \).
This understanding enables simplification, reducing the integral to a manageable form, that allows one to proceed with further calculations.