Problem 56 - Calculus

Question

Use the Substitution Rule for Definite Integrals to evaluate each definite integral. $$ \int_{-\pi / 2}^{\pi / 2} \cos \theta \cos (\pi \sin \theta) d \theta $$

Step-by-Step Solution

Answer

The value of the integral is 0.

Step 1: Recognize the Need for Substitution

Notice that the expression involves trig functions where direct integration seems complex. We will use a substitution to simplify it. Identify that substitution helps when one function is the derivative of another.

Step 2: Choose a Substitution

Let \( u = \sin \theta \). Then, the derivative \( du = \cos \theta \, d\theta \). This simplifies the integration problem significantly.

Step 3: Change the Limits of Integration

When \( \theta = -\pi / 2 \), \( u = \sin(-\pi/2) = -1 \). When \( \theta = \pi / 2 \), \( u = \sin(\pi/2) = 1 \). Adjust the limits of the integral to these new \( u \) values.

Step 4: Substitute into the Integral

The integral transforms into \( \int_{-1}^{1} \cos(\pi u) \, du \) using the substitution \( u = \sin \theta \) and the new limits.

Step 5: Evaluate the New Integral

Integrate \( \cos(\pi u) \) with respect to \( u \). The antiderivative is \( \frac{1}{\pi} \sin(\pi u) \).

Step 6: Calculate the Definite Integral

Evaluate \( \frac{1}{\pi} \sin(\pi u) \) from \(-1\) to \(1\). At \( u=1 \), it equals \( \frac{1}{\pi} \sin(\pi \times 1) = 0 \), and at \( u=-1 \), it equals \( \frac{1}{\pi} \sin(\pi \times (-1)) = 0 \). Subtract these results.

Step 7: Write the Solution

Since both evaluations resulted in 0, the value of the definite integral is 0.

Key Concepts

Definite Integrals

A definite integral represents the signed area under a curve within a specific interval on the x-axis. In simpler terms, think of definite integrals as the sum of infinite tiny parts that make up the area under a curve. The notation for a definite integral is \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the lower and upper limits of integration, respectively.

Unlike indefinite integrals, which represent a family of functions and include a constant of integration, definite integrals result in a specific number. They are essential in calculating areas, solving rate problems, and are used in various fields like physics and engineering.

• The finite limits \( a \) and \( b \) specify the region over the x-axis we are interested in.
• It gives a single value after evaluation, often related to area or other physical measures.
• The process includes finding the antiderivative or integral of the function and applying the limits to compute the result.

Understanding and applying the substitution rule can simplify the process of solving definite integrals, especially when the integral involves complex functions like trigonometric ones.

Trigonometric Functions

Trigonometric functions such as sine and cosine are fundamental to mathematics, especially in calculus. These functions describe relationships within right-angled triangles and periodic phenomena such as waves.

In calculus, trigonometric functions are often part of integrals, and knowing their properties is crucial. For instance, you will frequently encounter forms like \( \sin(\theta) \) and \( \cos(\theta) \) in integration problems.

• Trigonometric functions have specific periodic properties that allow for simplification of integrals.
• Functions like \( \sin(\theta) \) and \( \cos(\theta) \) are derivatives of each other, facilitating substitution during integration.
• Understanding their symmetry and periodicity helps in setting up proper limits for definite integrals.

These functions are intertwined with many real-world applications, making their mastery essential for tackling complex integral problems.

Integration Techniques

Integration involves finding the integral or antiderivative of a function. It is one of the foundational concepts in calculus. Knowing various techniques can simplify the integration process, especially when dealing with complex functions.

One such technique is the substitution rule, which is particularly useful when dealing with products of functions or compositions that are hard to integrate directly. This technique involves:

• Identifying a part of the integrand that can be transformed or simplified.
• Selecting a substitution that can simplify the integral, often letting \( u = g(x) \) where \( du = g'(x)dx \).
• Changing the limits of integration based on the substitution if it's a definite integral.

For example, in a problem involving \( \cos(\theta) \cos(\pi \sin \theta) \), substitution simplifies the integral by replacing complex trigonometric forms with simpler ones.

The substitution approach reduces the original function's complexity, making it easier to find its antiderivative. This technique is invaluable in both academic exercises and real-world applications.