Problem 43 - Calculus Early Transcendentals: Pearson New International Edition

Question

apply integration by parts twice to evaluate each integral. $$ \int x^{2} \cos x $$

Step-by-Step Solution

Answer

\( \int x^2 \cos x \, dx = x^2 \sin x + 2x \cos x - 2 \sin x + C \).

Step 1: Identify Parts for Integration by Parts

Integration by parts is applied using the formula \( \int u \, dv = uv - \int v \, du \). First, select \( u = x^2 \) and \( dv = \cos x \, dx \) so that \( du = 2x \, dx \) and \( v = \sin x \) (since \( \int \cos x \, dx = \sin x \)).

Step 2: Apply First Integration by Parts

Substitute the chosen \( u \) and \( v \) into the integration by parts formula:\[\int x^2 \cos x \, dx = x^2 \sin x - \int \sin x (2x) \, dx\]This results in: \[x^2 \sin x - 2 \int x \sin x \, dx\]

Step 3: Identify Parts for Second Integration by Parts

Now, apply integration by parts again to the remaining integral \( \int x \sin x \, dx \). Choose \( u = x \) and \( dv = \sin x \, dx \). Then, \( du = dx \) and \( v = -\cos x \) (since \( \int \sin x \, dx = -\cos x \)).

Step 4: Apply Second Integration by Parts

Substitute the new \( u \) and \( v \) into the integration by parts formula:\[\int x \sin x \, dx = -x \cos x + \int \cos x \, dx\]This simplifies to: \[-x \cos x + \sin x\]

Step 5: Combine Results of Both Applications

Substitute the result of the second integration back into the expression from the first part:\[x^2 \sin x - 2(-x \cos x + \sin x)\]Simplify this to obtain the final result:\[x^2 \sin x + 2x \cos x - 2\sin x + C\]

Step 6: Final Answer

The evaluated integral is: \[\int x^2 \cos x \, dx = x^2 \sin x + 2x \cos x - 2 \sin x + C\] where \( C \) is the constant of integration.

Key Concepts

Integral Calculus

Integral calculus is an essential part of mathematics that focuses on finding the total sizes, values, or quantities by summing infinitely small factors. In simpler terms, it's about adding up lots of tiny slices to find a whole. This branch of mathematics deals with two main types of problems:

• Finding the area under a curve, which is known as integration.
• Recovering a function from its rate of change, known as the antiderivative or indefinite integral.

Integration by parts, as seen in this exercise, is a specific method within integral calculus that helps us solve integrals involving products of functions. By choosing specific parts of a function to derive (decompose) and integrate, one can transform the integral into a simpler form, eventually reaching a solution. Once the integral calculation is completed, you usually add an arbitrary constant, denoted as 'C', because of the indefinite nature of integrating a function.

Techniques of Integration

Integration by parts is just one of many techniques used in solving integrals, especially when dealing with products of functions. This technique is a powerful tool in integral calculus and is derived from the product rule of differentiation. Here's the integration by parts formula:\[ \int u \, dv = uv - \int v \, du \]The goal is to choose functions \( u \) and \( dv \) such that the ensuing integral \( \int v \, du \) is easier to solve. In our example:

• Firstly, we select \( u = x^2 \) and \( dv = \cos x \, dx \). This choice simplifies the derivative \( du \) to \( 2x \, dx \) and integrates \( dv \) to obtain \( v = \sin x \).
• This leads to the first application of integration by parts which simplifies the problem, but introduces a new integral.
• We repeat integration by parts for the new integral \( \int x \sin x \, dx \), selecting \( u = x \) and \( dv = \sin x \, dx \).

This strategy helps break down complex problems into manageable steps. Each choice of \( u \) and \( dv \) requires practice and insight, honed through experience with integrals.

Trigonometric Integrals

Trigonometric integrals involve functions that include trigonometric expressions like \( \sin x \) or \( \cos x \). These types of integrals are found frequently when solving problems in physics or engineering. Such integrals can be challenging due to the oscillatory nature of trigonometric functions. Often, integration requires applying identities or specific techniques to reduce the integral to a familiar form. In our specific exercise:

• We dealt with \( \cos x \) as part of the original integral \( \int x^2 \cos x \, dx \).
• The integral of \( \cos x \) is straightforward, \( \sin x \), allowing us to find \( v \) in integration by parts.
• The subsequent integral, \( \int x \sin x \, dx \), further uses trigonometric integration techniques by converting \( \sin x \) into \(-\cos x \).

The combination of direct integration and identity use is key in handling trigonometric integrals effectively. By becoming familiar with these repetitive processes and identities, tackling these integrals becomes manageable.