Problem 43 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 37-48, apply integration by parts twice to evaluate each integral (see Examples 5 and 6). $$ \int x^{2} \cos x d x $$

Step-by-Step Solution

Answer

The integral is \( x^2 \sin x + 2x \cos x - 2\sin x + C \).

Step 1: Identify Parts for Integration by Parts

We need to apply integration by parts, which requires identifying parts for the formula \( \int u \, dv = uv - \int v \, du \). Set \( u = x^2 \) and \( dv = \cos x \, dx \). This choice maximizes the reduction in powers of \( x \).

Step 2: Differentiate \( u \) and Integrate \( dv \)

Differentiate \( u = x^2 \) to find \( du = 2x \, dx \). Integrate \( dv = \cos x \, dx \) to obtain \( v = \sin x \).

Step 3: Apply the Integration by Parts Formula

Apply the formula \( \int u \, dv = uv - \int v \, du \):\[ \int x^2 \cos x \, dx = x^2 \sin x - \int \sin x (2x) \, dx \]

Step 4: Simplify the Remaining Integral

Rewrite the remaining integral:\[ \int 2x \sin x \, dx \]This integral requires a second application of integration by parts.

Step 5: Second Application - Choose New Parts

For the integral \( \int 2x \sin x \, dx \), choose \( u = 2x \) and \( dv = \sin x \, dx \). Differentiate to get \( du = 2 \, dx \) and integrate to get \( v = -\cos x \).

Step 6: Apply the Integration by Parts Formula Again

Apply the integration by parts formula:\[ \int 2x \sin x \, dx = 2x (-\cos x) - \int (-\cos x) (2) \, dx \]Simplifying, we have:\[ = -2x \cos x + 2 \int \cos x \, dx \]

Step 7: Complete the Final Integration

Integrate \( \int \cos x \, dx \) to obtain \( \sin x \). Substituting back:\[ -2x \cos x + 2 \sin x \]

Step 8: Combine Results from Both Parts

Substitute the result from Step 7 back into the expression from Step 3:\[ \int x^2 \cos x \, dx = x^2 \sin x + 2x \cos x - 2\sin x + C \]

Key Concepts

Calculus problem solving

Calculus problem solving often involves breaking down complex problems into more manageable steps. One of the key approaches in calculus is integration by parts, a method useful for solving integrals involving the product of two functions. This technique is based on the formula \( \int u \, dv = uv - \int v \, du \) and helps in finding the antiderivative of products by selecting appropriate functions for \( u \) and \( dv \).
To effectively solve a calculus problem using this technique, it's crucial to pick functions \( u \) and \( dv \) that simplify the integration process. For quadratic terms like \( x^2 \), paired with trigonometric functions like \( \cos x \), choosing \( u = x^2 \) is strategic because differentiating decreases its degree, simplifying subsequent integrals.
Breaking down problems, as seen in our example, helps in systematically addressing and simplifying each part of the calculus challenge. The choice of parts should aim to make the remaining integrals easier to evaluate, sometimes requiring multiple applications of integration by parts.

Definite integrals

Definite integrals represent the signed area under a curve between two points on a graph. Unlike indefinite integrals, which aim to find the general form of the antiderivative, definite integrals evaluate this expression over a specific interval, providing a numerical result.
When applying integration by parts to definite integrals, the boundaries of the integration \( a \) and \( b \) are included in the calculus, modifying the integration by parts formula to: \( \int_{a}^{b} u \, dv = \left. uv \right|_{a}^{b} - \int_{a}^{b} v \, du \).
The process involves calculating the product \( uv \) at both endpoints and subtracting the integral of \( v \, du \) over the interval. This ensures the original integral is fully evaluated within the given range, crucial for problems that apply integration to real-world scenarios like physics and engineering.

Calculus integration techniques

In calculus, mastering various integration techniques is essential for tackling different kinds of functions. Integration by parts is one of these techniques, ideal for integrating products. This technique's central idea is transforming a difficult integral into two simpler parts, which might separately be easier to integrate.
By repeatedly applying integration by parts, particularly for integrals like \( \int x^2 \cos x \, dx \), complex expressions gradually reduce into manageable calculations. It’s vital to recognize when multiple applications of the technique are necessary, as it can simplify expressions that initially appear intractable.

• Identify clear \( u \) and \( dv \) parts where \( u \) typically simplifies upon differentiation, and \( dv \) easily integrates.
• Understand that some integrals may require more than one application to solve completely.
• Practice is key: the more problems you solve using different techniques, the more intuitive choosing parts and simplifying becomes.

The breadth of calculus integration techniques, from substitution to partial fractions, complements an understanding of when and how to apply these tools effectively in various contexts.