Question

, use the Substitution Rule for Definite Integrals to evaluate each definite integral. $$ \int_{-\pi / 2}^{\pi / 2} x^{2} \sin ^{2}\left(x^{3}\right) \cos \left(x^{3}\right) d x $$

Short answer

The integral evaluates to \( \frac{2}{9} \sin^3(\pi^3/8) \).

Step-by-step solution

Identify the Substitution

The integral given is \( \int_{-\pi / 2}^{\pi / 2} x^{2} \sin^{2}\left(x^{3}\right) \cos \left(x^{3}\right) dx \). Notice that it involves \( \sin^2(x^3) \) and \( \cos(x^3) \). This suggests using substitution. Let \( u = x^3 \). Then, \( du = 3x^2 \, dx \), or equivalently, \( dx = \frac{du}{3x^2} \).

Change Limits of Integration

With \( u = x^3 \), the limits of integration will change accordingly. When \( x = -\pi/2 \), \( u = (-\pi/2)^3 = -\pi^3/8 \). Similarly, when \( x = \pi/2 \), \( u = (\pi/2)^3 = \pi^3/8 \).

Substitute and Simplify the Integral

Substitute \( x^3 = u \), thus converting \( \sin^2(x^3) \) to \( \sin^2(u) \) and \( \cos(x^3) \) to \( \cos(u) \). Since \( du = 3x^2 \, dx \), solve for \( dx = \frac{du}{3x^2} \). Therefore, the expression \( x^2 \sin^2(u) \cos(u) \, dx \) becomes \( \frac{1}{3} \sin^2(u) \cos(u) \, du \).

Integrate Using Trigonometric Identity

The integral is now \( \frac{1}{3} \int_{-\pi^3/8}^{\pi^3/8} \sin^2(u) \cos(u) \, du \). Using the substitution \( v = \sin(u) \), \( dv = \cos(u) \, du \), the integral becomes \( \frac{1}{3} \int_{\sin(-\pi^3/8)}^{\sin(\pi^3/8)} v^2 \, dv \).

Evaluate the Simplified Integral

Since the function \( v^2 \) is even and its limits of integration are symmetric around zero, the integral from \( -a \) to \( a \) of an even function is twice the integral from \( 0 \) to \( a \). So, \( \int_{-\sin(\pi^3/8)}^{\sin(\pi^3/8)} v^2 \, dv \) simplifies to \( 2 \times \int_{0}^{\sin(\pi^3/8)} v^2 \, dv = 2 \times \left[\frac{v^3}{3}\right]_0^{\sin(\pi^3/8)} \).

Compute Final Answer

Calculate the definite integral \( \frac{1}{3} \times 2 \times \left[ \frac{1}{3} \sin^3(\pi^3/8) \right] \). Evaluating this gives \( \frac{2}{9} \sin^3(\pi^3/8) \) as the area under the curve from \( -\pi/2 \) to \( \pi/2 \).

Key concepts

Definite Integrals

Definite integrals are a foundational concept in calculus, signifying the total area under a curve between two specific limits on the x-axis. They are expressed in the form \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the lower and upper limits of integration, respectively. This operation calculates the accumulation of quantities, such as areas, mass, or volume, depending on the context of the problem.
One of the integral's most notable features is its application of the Fundamental Theorem of Calculus. This theorem bridges the concept of differentiation and integration, allowing us to evaluate the integral by finding an antiderivative of the function, \( F(x) \), and computing \( F(b) - F(a) \).
Definite integrals are extremely useful in various fields of science and engineering for calculating diverse entities like probabilities, work done by a force, and total distance traveled by an object. The process often involves transforming complex expressions into more manageable forms through techniques like substitution or partial fraction decomposition, ensuring the evaluation is both accurate and efficient.

Trigonometric Substitution

Trigonometric substitution is a valuable method in integration, used primarily when encountering integrals involving radicals or trigonometric expressions. This method transforms challenging integrals into simpler forms by substituting a trigonometric function for a variable, aiding in evaluating the integral more effortlessly.
Here's how it works:

• Identify the type of integral you're dealing with. Common scenarios involve expressions like \( \sqrt{a^2 - x^2} \), \( \sqrt{a^2 + x^2} \), or \( \sqrt{x^2 - a^2} \).
• Choose an appropriate trigonometric substitution. For example:
  • For \( \sqrt{a^2 - x^2} \), use \( x = a \sin(\theta) \).
  • For \( \sqrt{a^2 + x^2} \), use \( x = a \tan(\theta) \).
  • For \( \sqrt{x^2 - a^2} \), use \( x = a \sec(\theta) \).
• Replace \( x \) and \( dx \) in the integral with the corresponding expressions involving the trigonometric function.
• Solve the resulting trigonometric integral.
• Convert back to the original variable once the integration is performed.

In the context of our exercise, although the substitution \( u = x^3 \) was not exactly a standard trigonometric substitution, trigonometric identities played a crucial role in simplifying and evaluating the integral by transforming it into a simpler polynomial form. This illustrates the flexibility of substitution methods and their necessity when dealing with intricate integrals.

Integration Techniques

Various integration techniques exist to tackle different types of integral expressions. Some of these are specific to certain kinds of functions, and learning how to apply them effectively is key to mastering calculus.
Some of the most useful techniques include:

• **Substitution**: As demonstrated in our exercise, this is particularly helpful when the integral involves a composite function. By letting \( u = g(x) \), one can transform the integral of \( f(g(x))g'(x) \) into the much simpler \( \int f(u) \, du \).
• **Integration by Parts**: Useful for products of functions, employing the formula \( \int u \, dv = uv - \int v \, du \).
• **Partial Fraction Decomposition**: Ideal for rational functions, breaking a complex fraction into simpler fractions that are easier to integrate.
• **Trigonometric Identities**: Simplifying integrals using identities such as \( \sin^2(x) + \cos^2(x) = 1 \) can substantially reduce the complexity.

In the original exercise, identifying the most suitable technique depended heavily on analyzing the structure of the integral. The use of substitution simplified the integral's initial complexity, facilitating its evaluation. These techniques are fundamental for not only simplifying but also ensuring the precise solution of integral problems.