Question

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

Short answer

The integral evaluates to 0.

Step-by-step solution

Identify substitution candidates

We have the integral \( \int_{-\pi/2}^{\pi/2}(\cos 3x + \sin 5x) \, dx \). Notice the functions \( \cos 3x \) and \( \sin 5x \) may benefit from substitution due to their periodic arguments. We will tackle each term separately for substitution.

Substitute for \( \cos 3x \) term

Let \( u = 3x \) then \( du = 3\,dx \) or \( dx = \frac{du}{3} \). Changing the limits of integration, when \( x = -\pi/2 \), \( u = -3\pi/2 \), and when \( x = \pi/2 \), \( u = 3\pi/2 \). The integral becomes \( \frac{1}{3} \int_{-3\pi/2}^{3\pi/2} \cos u \, du \).

Evaluate \( \frac{1}{3} \int_{-3\pi/2}^{3\pi/2} \cos u \, du \)

The integral \( \int \cos u \, du = \sin u + C \). Evaluate \( \frac{1}{3} [\sin u]_{-3\pi/2}^{3\pi/2} \). This simplifies to \( \frac{1}{3} [ \sin(3\pi/2) - \sin(-3\pi/2) ] = \frac{1}{3} [ -1 - (-1) ] = 0 \).

Substitute for \( \sin 5x \) term

Let \( v = 5x \) then \( dv = 5\,dx \) or \( dx = \frac{dv}{5} \). Changing the limits of integration, when \( x = -\pi/2 \), \( v = -5\pi/2 \), and when \( x = \pi/2 \), \( v = 5\pi/2 \). The integral becomes \( \frac{1}{5} \int_{-5\pi/2}^{5\pi/2} \sin v \, dv \).

Evaluate \( \frac{1}{5} \int_{-5\pi/2}^{5\pi/2} \sin v \, dv \)

The integral \( \int \sin v \, dv = -\cos v + C \). Evaluate \( \frac{1}{5} [ -\cos v ]_{-5\pi/2}^{5\pi/2} \). This simplifies to \( \frac{1}{5} [ -\cos(5\pi/2) + \cos(-5\pi/2) ] = \frac{1}{5} [ 0 ] = 0 \).

Combine results

Add the results of the two evaluated integrals: \( 0 + 0 = 0 \). Thus, the final value of the original integral \( \int_{-\pi/2}^{\pi/2}(\cos 3x + \sin 5x) \, dx = 0 \).

Key concepts

Definite Integral

A definite integral is a fundamental concept in calculus that represents the net area under a curve between specific limits. In this context, it serves as a method of quantifying the accumulation of quantity over an interval. Definite integrals have limits, usually denoted as the lower and upper bounds of integration. For example, in the exercise, the integral is from \(x = -\pi/2\) to \(x = \pi/2\), which establishes the interval over which the function \(\cos 3x + \sin 5x\) is evaluated.

The result of a definite integral is a number, representing the total area, considering the direction/sign. If the curve lies above the x-axis, the result adds positively, while if below, it contributes negatively. This feature is why when you evaluate the definite integral of functions like \(\cos\) and \(\sin\) over symmetric intervals around zero, they may cancel out, as seen in the self-cancelation in step-by-step solutions.

Trigonometric Substitution

Trigonometric substitution is a technique used to simplify integrals involving trigonometric functions. It's particularly handy when dealing with integrals of periodic functions such as \(\cos\) and \(\sin\). In the problem presented, the trigonometric substitution simplifies the integral evaluation process by allowing us to change the derivative’s complexity.

For instance, with the term \(\cos 3x\), we substituted \(u = 3x\), effectively making the integral easier to manage by reducing the complexity of the periodic function. Similarly, \(\sin 5x\) is addressed by setting \(v = 5x\). These substitutions adjust the function's arguments to more manageable variables, streamlining the integration process.

• This approach often requires changing the limits of integration, as was done in the exercise for both substitutions.
• The substitution method not only simplifies the evaluation process but also highlights the integral's inherent symmetry, especially over symmetric domains.

Change of Variables

The change of variables is another indispensable tool in integration, frequently associated with substitution. It allows us to transform an integral with a difficult expression into one with a simpler variable, hence making the evaluation more straightforward.

In our specific exercise, changing variables involved letting \(u = 3x\) and \(v = 5x\), which simplified the integrals of \(\cos\) and \(\sin\) function respectively. This procedure redefined the function's scope and adjusted limits of integration accordingly:

• For \(u\), when \(-\pi/2\) transforms into \(-3\pi/2\) and \(\pi/2\) into \(3\pi/2\).
• Similarly, \(v\) maps \(-\pi/2\) to \(-5\pi/2\) and \(\pi/2\) to \(5\pi/2\).

By redefining the variable, this shift effectively alters the function scope and can transform more complex integrals into forms where elementary integration rules apply.

Integral Evaluation

Integral evaluation is the process of analytically determining the value of an integral, often utilizing simplifications gleaned from substitutions and symmetry considerations. It involves two key stages: finding the antiderivative (indefinite integration) and applying the bounds for a definite integral.

In this exercise, each term was separately integrated: first for \(\cos 3x\) using substitution to arrive at \[\frac{1}{3} \int_{-3\pi/2}^{3\pi/2} \cos u \, du\]. Once integrated, it simplified to \(\frac{1}{3} [\sin u]_{-3\pi/2}^{3\pi/2}\). This crucial step applies the definite integral evaluation strategy, considering the entire function's symmetry.

When applying the limits, taking \([\sin u]_{-3\pi/2}^{3\pi/2}\), we realized the self-cancelling nature over symmetric limits, where \(\sin\) graphs mirror each other over this interval, thus the result conceptualizes into zero. Similarly, integral evaluation of \(\sin 5x\) simplifies identically, leading to an overall zero when the evaluated integrals are summed, capturing the essence of the definite integral over symmetric intervals yielding unique solutions.