Question

For the following exercises, evaluate the integral using the specified method.\(\int x^{2} \sin (4 x) d x\) using integration by parts

Short answer

\( \int x^2 \sin(4x) \, dx = -\frac{1}{4} x^2 \cos(4x) + \frac{1}{8} x \sin(4x) - \frac{1}{32} \cos(4x) + C \).

Step-by-step solution

Choose 'u' and 'dv'

In integration by parts, we use the formula \( \int u \ dv = uv - \int v \ du \). To apply this method, we need to choose which part of the integrand will be \( u \) and which will be \( dv \). For the integral \( \int x^2 \sin(4x) \, dx \), let \( u = x^2 \) (since its derivative will simplify) and \( dv = \sin(4x) \, dx \).

Differentiate and Integrate

Differentiate \( u = x^2 \) to find \( du \): \( du = 2x \, dx \). Integrate \( dv = \sin(4x) \, dx \) to find \( v \): \( v = -\frac{1}{4} \cos(4x) \) (since the integral of \( \sin(ax) \) is \(-\frac{1}{a}\cos(ax) \)).

Apply Integration by Parts Formula

Substitute \( u \), \( v \), \( du \), and \( dv \) into the integration by parts formula: \[ \int x^2 \sin(4x) \, dx = uv - \int v \, du \] \[ = \left( x^2 \right) \left( -\frac{1}{4} \cos(4x) \right) - \int \left( -\frac{1}{4} \cos(4x) \right) (2x) \, dx \] \[ = -\frac{1}{4} x^2 \cos(4x) + \frac{1}{2} \int x \cos(4x) \, dx \]. This leaves us with a new integral to solve: \( \int x \cos(4x) \, dx \).

Integration by Parts Again

For the integral \( \int x \cos(4x) \, dx \), apply integration by parts again. Choose \( u = x \) and \( dv = \cos(4x) \, dx \). Differentiate to find \( du: du = dx \) and integrate to find \( v: v = \frac{1}{4} \sin(4x) \).

Substitute into Integration by Parts Formula

Substitute the new \( u \), \( v \), \( du \), and \( dv \) into the integration by parts formula: \[ \int x \cos(4x) \, dx = uv - \int v \, du \] \[ = \left(x \right) \left( \frac{1}{4} \sin(4x) \right) - \int \left( \frac{1}{4} \sin(4x) \right) \, dx \] \[ = \frac{1}{4} x \sin(4x) - \frac{1}{16} \cos(4x) + C \].

Combine All Parts

Insert the result from Step 5 back into the expression we found in Step 3 to get the final answer: \[ \int x^2 \sin(4x) \, dx = -\frac{1}{4} x^2 \cos(4x) + \frac{1}{2} \left( \frac{1}{4} x \sin(4x) - \frac{1}{16} \cos(4x) \right) + C \] \[ = -\frac{1}{4} x^2 \cos(4x) + \frac{1}{8} x \sin(4x) - \frac{1}{32} \cos(4x) + C \].

Key concepts

Definite Integrals

Definite integrals calculate the accumulated value of a function over a specific interval. This contrasts with indefinite integrals, which capture a family of functions without defined limits. When evaluating definite integrals, you create a clear numerical picture of the area under a curve between two points, denoted by limits of integration.
An integral like \( \int_{a}^{b} f(x) \, dx \) focuses on finding the value between \( x = a \) and \( x = b \). When you apply methods such as integration by parts to a definite integral, you follow the same steps as with indefinite integrals, but you will ultimately plug the limits into your evaluated antiderivative.

• Best suited for real-world problems involving net area or accumulation.
• Useful in physics and engineering for work, energy, and probability calculations.
• Requires attention to boundaries: don't forget to substitute back and compute.

Remember to maintain consistency with the sign of the area, particularly when the function crosses the x-axis within the defined bounds.

Trigonometric Integrals

Trigonometric integrals involve functions that include sine, cosine, tangent, and other trigonometric functions. These can often involve cyclic or periodic patterns. In calculus, you often encounter combinations of functions like \( \int \sin(ax) \, dx \) or \( \int \cos(ax) \, dx \). For integrals involving trigonometric functions, various techniques such as substitution, trigonometric identities, or integration by parts might be applied.
When you see an integral such as \( \int x^2 \sin(4x) \, dx \), involving a polynomial and a trigonometric function, integration by parts is particularly handy. This method allows you to effectively handle the algebraic-trigonometric mix by breaking it into products and re-evaluating the integral iteratively.

• Understand trigonometric identities: changing functions can simplify integrals.
• Periodicity may help recognize patterns in repetitive sections.
• Integrals usually require more than one technique: be patient and methodical.

Approach each trigonometric integral with a solid understanding of both the algebra and geometry of trigonometric function behavior.

Calculus Methods

Calculus methods, like integration by parts or substitution, form the core toolkit for managing integrals. These methods can transform complex integrals into more manageable equations. Integration by parts, particularly involves recognizing parts of an integral that can be differentiated and integrated easily.
The integration by parts technique uses the formula \( \int u \, dv = uv - \int v \, du \). Choosing \( u \) and \( dv \) can be straightforward: usually, \( u \) is chosen so that its derivative \( du \) is simpler, and \( dv \) is chosen to be integrable easily into \( v \). Reapplying these steps to nested problems is often required, as seen in the example of \( \int x^2 \sin(4x) \, dx \), where the initial result led naturally into another application of the same method.

• Choose your functions \( u \) and \( dv \) wisely: look for simplicity.
• When one technique is not enough, combine methods iteratively.
• Don't hesitate to exploit symmetry or known antiderivatives if it makes life easier.

These powerful techniques are indispensable for exploring deeper mathematical concepts and solving real-world problems effectively.