Chapter 5: Problem 29
Find the general antiderivative of the given function. $$ f(x)=2 \sin \left(\frac{\pi}{2} x\right)-3 \cos \left(\frac{\pi}{2} x\right) $$
Short Answer
Expert verified
The antiderivative is \(-\frac{4}{\pi} \cos\left(\frac{\pi}{2} x\right) - \frac{6}{\pi} \sin\left(\frac{\pi}{2} x\right) + C\).
Step by step solution
01
Identify the Components for Integration
The given function is \( f(x) = 2 \sin\left(\frac{\pi}{2} x\right) - 3 \cos\left(\frac{\pi}{2} x\right) \). We need to find the antiderivative of each term separately. For a function \( a \sin(kx) \), the antiderivative is \( -\frac{a}{k} \cos(kx) \). For a function \( b \cos(kx) \), the antiderivative is \( \frac{b}{k} \sin(kx) \). Here, \( a = 2 \), \( b = -3 \), and \( k = \frac{\pi}{2} \).
02
Integrate the Sine Function
The first term is \( 2 \sin\left(\frac{\pi}{2} x\right) \). Its antiderivative is \(-\frac{2}{\frac{\pi}{2}} \cos\left(\frac{\pi}{2} x\right) = -\frac{4}{\pi} \cos\left(\frac{\pi}{2} x\right) \).
03
Integrate the Cosine Function
The second term is \(-3 \cos\left(\frac{\pi}{2} x\right) \). Its antiderivative is \(\frac{-3}{\frac{\pi}{2}} \sin\left(\frac{\pi}{2} x\right) = -\frac{6}{\pi} \sin\left(\frac{\pi}{2} x\right) \).
04
Combine the Antiderivatives
Adding the antiderivatives from Step 2 and Step 3 gives the general antiderivative: \(-\frac{4}{\pi} \cos\left(\frac{\pi}{2} x\right) - \frac{6}{\pi} \sin\left(\frac{\pi}{2} x\right) + C\). Don't forget the constant of integration \( C \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Trigonometric Integration
Trigonometric integration is a fundamental concept in calculus, especially useful when dealing with functions involving trigonometric terms like sine and cosine. It involves finding the antiderivative of functions that include these trigonometric identities. For a function that involves terms like \( a \sin(kx) \) or \( b \cos(kx) \), specific formulas simplify the process. The antiderivative of \( a \sin(kx) \) is \(-\frac{a}{k} \cos(kx) \), making it easier to calculate such integrals. Similarly, the antiderivative of \( b \cos(kx) \) is \(\frac{b}{k} \sin(kx) \). These formulas derive from the basic properties of sine and cosine functions, and their derivatives.
- Understanding these basic antiderivative rules streamlines the integration process, especially when multiple trigonometric terms are involved.
- Applying these rules step-by-step can solve more complex functions efficiently.
Sine and Cosine Functions
Sine and cosine functions are two of the most prominent elements in trigonometry. They represent the x and y coordinates for points on a unit circle, commonly used to describe periodic phenomena. In calculus, they have unique integration and differentiation properties:
- The derivative of \( \sin(x) \) is \( \cos(x) \), and its antiderivative is \(-\cos(x) + C\).
- For \( \cos(x) \), the derivative is \(-\sin(x)\), and its antiderivative is \( \sin(x) + C\).
Calculus Concepts
The overarching field of calculus enables us to deal with change, rates of change, and the accumulation of quantities. It encompasses two primary operations: differentiation and integration. Differentiation focuses on calculating the derivative, which gives the rate at which a quantity changes, while integration, especially antiderivatives, determines the accumulation of quantities over intervals.An antiderivative is a function that 'undoes' differentiation. For instance, if differentiating \( F(x) \) gives you \( f(x) \), then \( F(x) \) is an antiderivative of \( f(x) \). Integrating a function involves finding all possible antiderivatives, typically expressed with a constant of integration \( C \), symbolizing the potential shifts in the function without affecting its derivative.
- Calculus allows us to model and solve real-world problems involving non-linear and dynamic phenomena.
- Understanding the basic properties of functions, including trigonometric ones, expands the range of problems one can solve.