Problem 50 - Calculus Early Transcendentals

Question

Use the reduction formulas in to evaluate the following integrals. $$\int x^{3} \sin x d x$$

Step-by-Step Solution

Answer

Question: Evaluate the integral $$\int x^3 \sin x dx$$. Answer: $$\int x^3 \sin x dx = -x^3\cos x + 3x^2\sin x - 6x\cos x + 6\sin x + C$$

Step 1: Choose u and dv, and find du and v

Let $$u=x^3$$ and $$dv=\sin x dx$$. We need to differentiate $$u$$ with respect to $$x$$ to find $$du$$, and integrate $$dv$$ to find $$v$$: $$du = \frac{d}{dx}(x^3) = 3x^2 dx$$ $$v = \int\sin x dx = -\cos x + C$$

Step 2: Apply integration by parts formula

Using the integration by parts formula, the given integral can be rewritten as: $$\int x^3 \sin x dx = uv - \int v du = -x^3\cos x + \int (3x^2\cos x) dx$$ Now, we must evaluate the new integral on the right-hand side.

Step 3: Apply integration by parts again

Let $$u=3x^2$$ and $$dv=\cos x dx$$, and find $$du$$ and $$v$$: $$du = \frac{d}{dx}(3x^2) = 6x dx$$ $$v = \int\cos x dx = \sin x + C$$ Substitute this back into the equation from step 2: $$\int x^3 \sin x dx = -x^3\cos x + \int (3x^2\cos x) dx = -x^3\cos x + (3x^2\sin x - \int 6x\sin x dx)$$

Step 4: Apply integration by parts one more time

Let $$u=6x$$ and $$dv=\sin x dx$$, and find $$du$$ and $$v$$: $$du = \frac{d}{dx}(6x) = 6 dx$$ $$v = \int\sin x dx = -\cos x + C$$ Substitute this back into the equation from step 3: $$\int x^3 \sin x dx = -x^3\cos x + 3x^2\sin x - \int 6x\sin x dx = -x^3\cos x + 3x^2\sin x - (6x\cos x - \int 6\cos x dx)$$

Step 5: Final simplification

Now, we must integrate the remaining integral: $$\int 6\cos x dx = 6\int\cos x dx = 6\sin x + C$$ Substitute this result back into the equation from step 4: $$\int x^3 \sin x dx = -x^3\cos x + 3x^2\sin x - 6x\cos x + 6\sin x + C$$ So, the final answer is: $$\int x^3 \sin x dx = -x^3\cos x + 3x^2\sin x - 6x\cos x + 6\sin x + C$$

Key Concepts

Reduction Formulas

Reduction formulas are incredibly useful in solving complex integrals, especially when the direct evaluation is cumbersome or not straightforward. These handy formulas help break down an integral into simpler parts or repeated patterns that can be worked through systematically.
Think of reduction formulas like a recipe that instructs you on how to continually transform an integral until it is in a form that is easy to solve. They are particularly useful for integrating products of polynomial functions and trigonometric functions by simplifying the process through repeated application of integration by parts.
In the case of our example, the integral \( \int x^3 \sin x \, dx \) used reduction formulas to generate a series of simpler integrals through consecutive integration by parts. By systematically breaking down \( x^3 \sin x \) into components that can be repeatedly integrated and differentiated, the initially complex task is transformed into smaller, more manageable pieces.

Indefinite Integrals

Indefinite integrals are essentially the antithesis of derivatives. While differentiation divides a function into smaller differential parts, integration brings these parts back together, finding the original function from its derivative forms. The result of an indefinite integral is a function plus a constant, often denoted by \( C \), acknowledging that integration reverse-engineers the derivative process.
Integrals without limits are called indefinite because they don't point to a specific numerical value but instead represent a family of functions characterized by the integration constant \( C \). In our example, when calculating the integral \( \int x^3 \sin x \, dx \), we used indefinite integration to determine the family of functions that, when differentiated, return to the original integrand. This involves not only the use of integration by parts but also the inclusion of the constant of integration each time a new function is derived.

Differentiation

Differentiation allows us to understand how functions change, tracing the rate at which one quantity varies concerning another. It breaks down complex functions into elemental components that reveal the instantaneous rate of change at any given point. Using derivatives, we understand how small changes in one variable can cause changes in another.
In the context of integration by parts, differentiation plays a crucial role through the choice of \( u \), which is differentiated to find \( du \). In our step-by-step solution, several transitions involved the differentiation of polynomial terms. For instance, \( u = x^3 \) was differentiated to give \( du = 3x^2 \, dx \). Differentiation ensured that each component was broken into its simplest form, ready for the integration by parts procedure. This process not only highlights the original polynomial term's contribution to the integral but also interlinks with the reduction formulas to generate a manageable and solvable set of equations.