Problem 54 - Calculus

Question

Use the Substitution Rule for Definite Integrals to evaluate each definite integral. $$ \int_{-\pi / 2}^{\pi / 2}(\cos 3 x+\sin 5 x) d x $$

Step-by-Step Solution

Answer

The value of the definite integral is 0.

Step 1: Identify Substitutions

We need to evaluate the integral \( \int_{-\pi / 2}^{\pi / 2}(\cos 3x + \sin 5x) \, dx \). Notice that the terms can be integrated separately with substitution. Let's start by setting \( u = 3x \) for the term \( \cos 3x \) and \( v = 5x \) for \( \sin 5x \).

Step 2: Compute Differential Elements

Find the differentials: If \( u = 3x \), then \( du = 3 \, dx \) or \( dx = \frac{1}{3} \, du \). If \( v = 5x \), then \( dv = 5 \, dx \) or \( dx = \frac{1}{5} \, dv \).

Step 3: Adjust Limits of Integration

Substitute the limits for \( u \): when \( x = -\pi/2 \), \( u = 3(-\pi/2) = -3\pi/2 \); when \( x = \pi/2 \), \( u = 3(\pi/2) = 3\pi/2 \). Similarly, for \( v \): when \( x = -\pi/2 \), \( v = 5(-\pi/2) = -5\pi/2 \); when \( x = \pi/2 \), \( v = 5(\pi/2) = 5\pi/2 \).

Step 4: Integrate Each Function Separately

Compute \( \int \cos u \, du \). Since \( \int \cos u \, du = \sin u \), the integral with new limits becomes \[ \int_{-3\pi/2}^{3\pi/2} \cos u \, \frac{1}{3} \, du = \frac{1}{3}(\sin u) \bigg|_{-3\pi/2}^{3\pi/2}. \] For the sine term, \( \int \sin v \, dv = -\cos v \), thus the integral is \[ \int_{-5\pi/2}^{5\pi/2} \sin v \, \frac{1}{5} \, dv = -\frac{1}{5}(\cos v) \bigg|_{-5\pi/2}^{5\pi/2}. \]

Step 5: Evaluate the Definite Integrals

For the first integral: \[ \frac{1}{3}(\sin u) \bigg|_{-3\pi/2}^{3\pi/2} = \frac{1}{3}(\sin(3\pi/2) - \sin(-3\pi/2)) = \frac{1}{3}(-1 - (-1)) = 0. \] For the second integral: \[ -\frac{1}{5}(\cos v) \bigg|_{-5\pi/2}^{5\pi/2} = -\frac{1}{5}(\cos(5\pi/2) - \cos(-5\pi/2)) = -\frac{1}{5}(0 - 0) = 0. \]

Step 6: Combine Resulting Integrals

Add the two results: \( 0 + 0 = 0 \). Thus the value of the definite integral is 0.

Key Concepts

Definite Integrals

Definite integrals are a core concept of calculus, representing the net area under a curve from one specific point to another point on the x-axis. This is symbolized as \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the lower and upper limits of integration respectively.
When solving problems involving definite integrals, it is crucial to evaluate the function at both these limits and consider the difference. This process allows us to compute the exact net "accumulated" change over that interval.
In practice, the calculation often entails finding the antiderivative of the function, substituting the bounds into this antiderivative, and subtracting the results. For example, in the integration of trigonometric functions like \( \int_{-\pi / 2}^{\pi / 2} (\cos 3x + \sin 5x) \, dx \), evaluating each integral individually and then summing them helps us find the precise value of the area or the net change we seek.

Trigonometric Substitution

Trigonometric substitution is a specialized technique in integration useful for integrating expressions involving square roots, quadratic polynomials, or trigonometric forms. It leverages trigonometric identities to transform integrands into forms that are easier to integrate.
In cases like \( \cos 3x \) and \( \sin 5x \), substitution allows us to change the variable in a way that simplifies integration, making use of identities such as \( \sin^2 x + \cos^2 x = 1 \). By making suitable substitutions such as \( u = 3x \) and \( v = 5x \), we change the original complex integral into simpler terms, which we can integrate independently.
This conversion might also adjust the limits of integration, converting complex boundaries into more manageable ones, allowing us to compute the definite integrals straightforwardly once substitutions are performed.

Integration Techniques

Integration techniques are various strategies used to solve integrals more efficiently or to handle complex functions.

• **Substitution Rule**: This is a technique where a substitution of variables is used to simplify the integral. It involves identifying a section of the integral that can be replaced with a single variable, thereby transforming it into a basic integral form.
• **Trigonometric Identities**: Using identities allows breaking down trigonometric functions into simpler integrable parts.

In the given exercise, the substitution rule is applied twice, transforming the original function into two separate, simpler integrals. Such approaches, especially with trigonometric integrations, make dealing with products or sums of terms easier.
The substitution rule is particularly effective in definite integrals, as it not only simplifies the main integrand but also aids in recalibrating the limits, transforming them appropriately according to the new variables. This leads to an easier and much cleaner resolution of complex integral problems, ultimately reducing the problem to fundamental integrals we can evaluate directly.