Problem 23 - Calculus Single Variable

Question

Integrate by parts successively to evaluate the given indefinite integral. $$ \int x^{2} \cos (x) d x $$

Step-by-Step Solution

Answer

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

Step 1: Identify u and dv

In the integration by parts formula, we choose one part of the integral to be differentiated (u) and the other to be integrated (dv). For \(\int x^2 \cos(x) \, dx\), we choose \(u = x^2\) and \(dv = \cos(x) \, dx\).

Step 2: Differentiate u and Integrate dv

Differentiate \(u\) to find \(du\) and integrate \(dv\) to find \(v\).- \( u = x^2 \Rightarrow du = 2x \, dx\)- \( dv = \cos(x) \, dx \Rightarrow v = \sin(x)\)

Step 3: Apply Integration by Parts Formula

Use the integration by parts formula \( \int u \, dv = uv - \int v \, du\).\[ \int x^2 \cos(x) \, dx = x^2 \sin(x) - \int 2x \sin(x) \, dx \]

Step 4: Integrate the Remaining Integral

Apply integration by parts again to \( \int 2x \sin(x) \, dx \).Choose \( u = 2x \) and \( dv = \sin(x) \, dx \).- \( u = 2x \Rightarrow du = 2 \, dx\)- \( dv = \sin(x) \, dx \Rightarrow v = -\cos(x)\)

Step 5: Apply Integration by Parts Again

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

Step 6: Final Integration

Compute \( \int 2 \cos(x) \, dx \).\[ \int 2 \cos(x) \, dx = 2 \sin(x) \]

Step 7: Combine and Simplify

Combine the results from previous integrations:\[ \int 2x \sin(x) \, dx = -2x \cos(x) + 2 \sin(x) \]Substitute back into the original integration by parts:\[ \int x^2 \cos(x) \, dx = x^2 \sin(x) - (-2x \cos(x) + 2 \sin(x)) + C \]Simplify:\[ \int x^2 \cos(x) \, dx = x^2 \sin(x) + 2x \cos(x) - 2\sin(x) + C \]

Step 8: Conclusion

After performing integration by parts twice, we have evaluated the original integral. The final answer is:\[ x^2 \sin(x) + 2x \cos(x) - 2 \sin(x) + C \]

Key Concepts

Indefinite Integral

An indefinite integral, unlike a definite integral, does not have upper and lower limits. It represents a family of functions, and its antiderivative is expressed with an integration constant, typically denoted as **C**. This constant is essential because when we differentiate the integral to check our answer, the derivative of a constant is zero. Thus, the indefinite integral includes all possible shifts upwards or downwards of the function along the Y-axis. It provides a general solution rather than a specific value.

• The notation for an indefinite integral is usually in the form \( \int f(x) \, dx \), where \( f(x) \) is the function we're integrating.
• When you find the indefinite integral of a function, you are basically reversing the process of differentiation.
• Adding the constant \( C \) accounts for any original constant that might have been present before the function was differentiated.

In this exercise, we are evaluating the indefinite integral \( \int x^2 \cos(x) \, dx \) using a specific technique: integration by parts.

Integral Calculus

Integral calculus is the study of integrals and their properties. It is one half of calculus, the other being differential calculus. Integral calculus helps us understand areas, volumes, displacement, accumulation, and many other concepts. It is primarily concerned with two types of operations: determining the area under curves (definite integral) and finding antiderivatives (indefinite integral).

• Integral: This can be thought of as the opposite of a derivative. If differentiation is about breaking things down into pieces, integration is about adding them up.
• Applications: Integral calculus is used in various fields such as physics, engineering, economics, biology, and many more to solve complex problems.

In the provided exercise, integral calculus is employed to find an antiderivative of a function involving a polynomial and a trigonometric component. Integration by parts is particularly useful here because it simplifies the process of integrating a product of two functions.

Trigonometric Integration

Trigonometric integration involves finding the antiderivatives of functions involving trigonometric functions such as sine, cosine, and tangent. These integrations can sometimes be straightforward, but often require specific techniques like substitution or integration by parts to solve efficiently.

• Basic Integers: The basic indefinites involving trigonometric functions are \( \int \sin(x) \, dx = -\cos(x) + C \) and \( \int \cos(x) \, dx = \sin(x) + C \).
• Complex Functions: When trigonometric functions are part of more complex expressions, integration may require iterative techniques as seen in the original exercise.

The integral \( \int x^2 \cos(x) \, dx \) in the provided solution showcases trigonometric integration. The cosine function is integrated as part of applying the integration by parts method multiple times to simplify the problem.