Question

In Exercises \(25-28\) , evaluate the integral analytically. Support your answer using NINT. $$\int_{0}^{\pi / 2} x^{2} \sin 2 x d x$$

Short answer

The value of the definite integral between 0 and \(\pi/2\) is \(\frac{\pi}{4}\).

Step-by-step solution

- Identify the functions to apply the integration by parts formula

In the integral \(\int_{0}^{\pi / 2} x^{2} \sin 2x dx\), the two functions are \(x^2\) and \(\sin 2x\). We will use the method of integration by parts, represented by the formula \(\int udv = uv - \int vdu \). Here, let \(u=x^2\) and \(dv= \sin 2x dx\). We need to calculate the differential \(du\) of \(u\) and an antiderivative \(v\) of \(dv\). Differentiating \(u\) with respect to \(x\) gives \(du=2x dx\). The integral of \(\sin 2x dx\) with respect to \(x\) gives \(v=-\frac{1}{2}\cos 2x\).

- Apply the integration by parts formula

Now we substitute \(u\), \(v\), and \(du\) into the formula \(\int udv = uv - \int vdu\). This gives us: \(-x^{2}\frac{1}{2}\cos 2x - \int -\frac{1}{2}\cos 2x*2x dx\). Simplifying it further gives: \(-\frac{1}{2}x^{2}\cos 2x + \int x\cos 2x dx\).

- Apply integration by parts formula again

Now, look at the new integral term \(\int x\cos 2x dx\). Here, let \(u=x\) and \(dv=\cos 2x dx\). Then \(du= dx\) and \(v=\frac{1}{2}\sin 2x\). Substituting into the integration by parts formula gives \(\int x\cos 2x dx = x*\frac{1}{2}\sin 2x - \int \frac{1}{2}\sin 2x dx\), which simplifies to \(\frac{1}{2}x\sin 2x - \frac{1}{4}\cos 2x\).

- Substitute step 3 into the integral

Substitute \(\int x\cos 2x dx\) from step 3 into the integral from step 2 to get: \(-\frac{1}{2}x^{2}\cos 2x + \left(\frac{1}{2}x\sin 2x - \frac{1}{4}\cos 2x\right)\). This simplifies to: \(-\frac{1}{2}x^{2}\cos 2x + \frac{1}{2}x\sin 2x - \frac{1}{4}\cos 2x\).

- Evaluate the integral from 0 to \(\pi/2\)

Now, evaluate the expression at \(x=\pi/2\) and at \(x=0\) and subtract: \(\left[-\frac{1}{2}\left(\frac{\pi}{2}\right)^{2}\cos 2*\frac{\pi}{2} + \frac{1}{2}*\frac{\pi}{2}\sin 2*\frac{\pi}{2} - \frac{1}{4}\cos 2*\frac{\pi}{2}\right] - \left[-\frac{1}{2}(0)^{2}\cos (0) + \frac{1}{2}*0\sin (0) - \frac{1}{4}\cos (0)\right]\). Calculating each term separately and subtracting gives the answer.

Key concepts

Indefinite Integral

The concept of an indefinite integral refers to the process of finding the original function, often called the antiderivative, from its derivative. In calculus, integration is the inverse operation of differentiation. When we calculate the indefinite integral of a function, we are essentially adding up an infinite number of infinitesimally small pieces to rebuild the original function. The symbol for integration is \( \int \), and unlike definite integrals, indefinite integrals do not have limits and include a constant of integration \(C\). This constant accounts for any constant value that was lost in the differentiation process.
For instance, to find the indefinite integral of \(x^2\), we look for a function whose derivative gives us \(x^2\). In this case, the indefinite integral is \(\frac{1}{3}x^3 + C\), since the derivative of \(\frac{1}{3}x^3\) with respect to \(x\) is \(x^2\). The process involves reversing the rules of derivatives to discover the function behind the derivative behavior.

Trigonometric Integration

Trigonometric integration involves finding the integral of trigonometric functions. This is a common scenario when dealing with periodic functions in calculus. Integrating these functions often requires the use of specific identities and integration techniques, such as integration by parts, substitution, or trigonometric identities.
For example, the integration of \(\sin 2x\) can be approached by using the indefinite integral concept, and it results in \( -\frac{1}{2}\cos 2x + C\). The integral of \(\cos 2x\), similarly, yields \( \frac{1}{2}\sin 2x + C\). These kinds of integrals frequently surface in physics and engineering, particularly in the study of waveforms and harmonic motion.

Integration by Parts

When handling products of algebraic and trigonometric expressions, as in the original exercise, integration by parts is a valuable strategy. This method is based on the formula \( \int u dv = uv - \int v du \), where \(u\) and \(dv\) are parts of the original integrand designated such that \(v\) and \(du\) are more manageable to work with.

Integral Evaluation

The evaluation of an integral is critical to finding the exact area or accumulative quantity represented by a function. For a definite integral, the calculation involves finding the difference between the values of the antiderivative at the upper and lower limits of integration. This process is sometimes referred to as the 'evaluation step' of integration.
In our original exercise, after applying the integration by parts method twice, one would employ the Fundamental Theorem of Calculus to evaluate the integral between the given limits of \(0\) and \(\pi/2\). The result of this step is computed by substituting these limit values into the integrated function and subtracting the lower limit value from the upper limit value to obtain a numerical answer. Integral evaluation is often used to solve practical problems in various fields like physics, economics, and engineering, where it helps in determining quantities like displacement, total cost, or energy.

Calculus Techniques

Calculus techniques encompass a variety of methods used for differentiation and integration, each tailored to tackle specific types of problems. These techniques include the method of substitution, by parts, partial fractions, and the use of trigonometric identities, to name a few.
The method of substitution is useful when dealing with composite functions, while partial fractions can simplify the integration of rational functions. Trigonometric identities help simplify integrands before integration. In the case of our exercise, the integration by parts technique is optimally used to solve an integral that is the product of a polynomial and a trigonometric function.

Choosing the Right Technique

It can be challenging to select the proper calculus technique for a given problem. Practice and familiarity with different types of functions and their intricacies are essential. Sometimes, more than one method may apply to an integral, and finding the most efficient one comes down to experience and understanding the nuances of each technique.