Problem 29 - Calculus for Biology and Medicine

Question

Find the general antiderivative of the given function. $$ f(x)=2 \sin \left(\frac{\pi}{2} x\right)-3 \cos \left(\frac{\pi}{2} x\right) $$

Step-by-Step Solution

Answer

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 1: 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 \( k = \frac{\pi}{2} \).

Step 2: Integrate the Sine Function

The first term is \( 2 \sin\left(\frac{\pi}{2} x\right) \). Its antiderivative is \(-\frac{2}{\frac{\pi}{2}} \cos\left(\frac{\pi}{2} x\right) = -\frac{4}{\pi} \cos\left(\frac{\pi}{2} x\right) \).

Step 3: Integrate the Cosine Function

The second term is \(-3 \cos\left(\frac{\pi}{2} x\right) \). Its antiderivative is \(\frac{-3}{\frac{\pi}{2}} \sin\left(\frac{\pi}{2} x\right) = -\frac{6}{\pi} \sin\left(\frac{\pi}{2} x\right) \).

Step 4: 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 \( C \).

Key Concepts

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.

Taking the example given, one can separate the sine and cosine terms to individually integrate each, significantly easing the computation and understanding.

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 \( \sin(x) + C\).

These functions oscillate between -1 and 1, defining smooth, wave-like curves that repeat every \(2\pi\). In the context of trigonometric integration, these properties are incredibly useful for resolving problems efficiently, just as seen in the provided solution. The solution of the exercise also showcases the importance of the constant multiplier \( k \) inside the function, which affects integration by altering the amplitude and phase of the resulting antiderivative.

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.

Through calculus, especially via trigonometric integration, one connects fundamental elements such as sine and cosine with broader mathematical practices, building toward a deeper understanding of both theoretical and practical applications.