Problem 44 - Calculus Single Variable

Question

Simplify the integrand before integrating by parts. $$ \int 4 x \sin (x) \cos (x) d x $$

Step-by-Step Solution

Answer

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

Step 1: Apply Trigonometric Identity

To simplify the integrand, use the trigonometric identity for product-to-sum conversions: \( \sin(x) \cos(x) = \frac{1}{2}\sin(2x) \). Thus, \( 4x \sin(x) \cos(x) \) becomes \( 2x \sin(2x) \).

Step 2: Set Up for Integration by Parts

Integration by parts formula is \( \int u \, dv = uv - \int v \, du \). Choose \( u = 2x \) and \( dv = \sin(2x) \, dx \); then, differentiate and integrate to find \( du \) and \( v \).

Step 3: Differentiate and Integrate

Differentiate \( u = 2x \) to get \( du = 2 \, dx \). Integrate \( dv = \sin(2x) \, dx \) to find \( v = -\frac{1}{2}\cos(2x) \) (note the use of the chain rule for integration).

Step 4: Apply Integration by Parts Formula

Substitute \( u \), \( v \), \( dv \), and \( du \) into the integration by parts formula: \[ \int 2x \sin(2x) \, dx = \left( 2x \cdot -\frac{1}{2}\cos(2x) \right) - \int \left( -\frac{1}{2}\cos(2x) \right) \cdot 2 \, dx. \] Simplify the expression to \[ -x \cos(2x) + \int \cos(2x) \, dx.\]

Step 5: Integrate Remaining Part

Integrate \( \int \cos(2x) \, dx \) to get \( \frac{1}{2}\sin(2x) + C \), where \( C \) is the integration constant. Simplify and combine with earlier result to get: \[ -x \cos(2x) + \frac{1}{2} \sin(2x) + C.\]

Key Concepts

Trigonometric Identities

Trigonometric identities are essential tools in calculus, especially when dealing with integrals involving trigonometric functions. They allow us to simplify expressions, making complex integrals more manageable. In the context of this exercise, we use a specific product-to-sum identity:

• \( \sin(x)\cos(x) = \frac{1}{2}\sin(2x) \)

This identity helps to express the product of sine and cosine as a simpler trigonometric function. Applying this identity, we replace \( 4x \sin(x) \cos(x) \) with \( 2x \sin(2x) \). This transformation is crucial as it reduces the complexity of the integrand, making the integration by parts easier to apply. Understanding and applying trigonometric identities effectively is key to solving a wide range of problems in integral calculus, as they can turn seemingly difficult problems into much simpler ones.

Integral Calculus

Integral calculus focuses on the concept of integration. It involves finding the area under a curve, which is the antithesis of differentiation. For complicated functions, integration by parts is a technique often used to handle products of functions. The integration by parts formula is:\[ \int u \, dv = uv - \int v \, du \]where:

• \( u \) is a chosen function of \( x \)
• \( dv \) is the differential of another function of \( x \)
• \( du \) is the derivative of \( u \)
• \( v \) is the integral of \( dv \)

In this exercise, we have \( u = 2x \) and \( dv = \sin(2x) \ dx \). By differentiating \( u \), we find \( du = 2 \, dx \), and by integrating \( dv \), we find \( v = -\frac{1}{2}\cos(2x) \). Applying these into the integration by parts formula helps us tackle the integral successfully.The technique of integration by parts is especially useful when integrating the product of algebraic and exponential or trigonometric functions, allowing these integrals to be computed in a systematic way.

Product-to-Sum Conversion

The product-to-sum conversion is a powerful trigonometric tool that helps in simplifying expressions involving products of trigonometric functions into easier forms, typically sums. This is particularly helpful in integration, where simplification can drastically ease the process. In the given exercise, the conversion used is:

• \( \sin(x)\cos(x) = \frac{1}{2}\sin(2x) \)

This formula arises from the double-angle identity for sine, which is derived using the angle addition formulas. By converting \( \sin(x) \cos(x) \) into \( \frac{1}{2} \sin(2x) \), the problem simplifies the original integral \( \int 4x \sin(x) \cos(x) \, dx \) into \( \int 2x \sin(2x) \, dx \), a much more approachable form for integration by parts.Understanding how and when to use product-to-sum conversions allows for significant strides in solving complex integrals, making it an invaluable tool in the field of integral calculus.