Problem 17 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-28, perform the indicated integrations. \(\int x \cos ^{2} x \sin x d x\) Hint: Use integration by parts.

Step-by-Step Solution

Answer

Use integration by parts with substitution to solve.

Step 1: Identify the Parts for Integration by Parts

We start by identifying functions for integration by parts, which is based on the formula:\[\int u \, dv = uv - \int v \, du.\]Here, set \(u = x\) and \(dv = \cos^2 x \sin x \, dx\). Differentiating \(u\), we get \(du = dx\).

Step 2: Simplify the Function dv

Before proceeding, we simplify \(dv = \cos^2 x \sin x \, dx\). Using the identity \(\cos^2 x = \frac{1 + \cos 2x}{2}\), we substitute:\[\cos^2 x = \frac{1 + \cos 2x}{2}\] so that \(dv = \frac{1 + \cos 2x}{2} \sin x \, dx\).

Step 3: Perform Integration by Parts

Now we compute \(v\) by integrating our simplified \(dv\). The integral becomes:\[v = \int \frac{1 + \cos 2x}{2} \sin x \, dx.\]Let's integrate the simpler form: - For \(\int \sin x \, dx = -\cos x + C.\) - For \(\cos 2x\), perform further simplification into two parts later.

Step 4: Handle the Component \(\int \cos 2x \sin x \, dx\)

We substitute \(dv = \frac{1}{2}(\sin x + \sin x \cos 2x)\) and focus on \(\int \cos 2x \sin x \, dx\). Use another substitution: let \(t = \cos x\), \(dt = -\sin x \, dx\), to finish the integral.

Step 5: Complete Evaluations and Substitute Back

After performing integration and substitution, revisit the integration by parts formula to substitute back the parts where applicable: \[\int x \cos^2 x \sin x \, dx = xV - \int V \, dx.\]This combines performed evaluations for the complete answer.

Key Concepts

Trigonometric Integrals

Trigonometric integrals involve the integration of functions that include trigonometric functions like sine, cosine, tangent, and their powers. In the problem at hand, we have the integral \( \int x \cos^2 x \sin x \, dx \). This type of integral is approached by leveraging the properties and identities of trigonometric functions.
To simplify such integrals, trigonometric identities are often employed. One common identity used here is \( \cos^2 x = \frac{1 + \cos 2x}{2} \). This transforms the problem into a more manageable form by reducing the complexity of the trigonometric terms involved.
Careful identification and simplification using these identities is an essential step in solving trigonometric integrals, especially when higher powers or combinations of different trigonometric functions are involved.

Calculus Problem Solving

When tackling calculus problems like integration by parts, it's crucial to first identify the right functions to assign within the formula \( \int u \, dv = uv - \int v \, du \). This equation guides the integration by parts process.
In the given problem, the choice is \( u = x \) and \( dv = \cos^2 x \sin x \, dx \). Choosing \( u \) as the polynomial part of the product and \( dv \) as the trigonometric expression helps streamline further steps.
Problem solving in calculus also requires attention to detail. Breaking down the trigonometric expression with a suitable substitution aids in completing the integral transformation. A logical sequence of simplifications and substitutions makes the process systematic. It’s these small, strategic decisions that can make complex integrals become much more manageable.

Integral Calculus Techniques

Integral calculus employs a variety of techniques depending on the type of function involved. For this problem, integration by parts is the primary technique used, which is especially useful when dealing with products of polynomial and trigonometric functions.
Another technique applied here is the substitution method, helpful in transforming trigonometric integrals into integrable forms. This often involves using trigonometric identities or an alternate substitution like \( t = \cos x \) for deeper nesting functions.
Recognizing which calculus techniques to apply in a given problem is crucial as it can simplify the process and minimize errors. Balancing between integration by parts and substitution allows for breaking down the integral into solvable pieces, eventually yielding the correct solution efficiently.