Problem 27 - Calculus for Biology and Medicine

Question

In Problems 1-40, find the general antiderivative of the given function. $$ f(x)=\sin \left(\frac{x}{3}\right)+\cos \left(\frac{x}{3}\right) $$

Step-by-Step Solution

Answer

The antiderivative is \( F(x) = -3\cos\left(\frac{x}{3}\right) + 3\sin\left(\frac{x}{3}\right) + C \).

Step 1: Recall the Basic Integration Formulas

Remember that the antiderivative of \( \sin(x) \) is \( -\cos(x) \) and the antiderivative of \( \cos(x) \) is \( \sin(x) \). These basic integration rules will help us integrate \( \sin\left(\frac{x}{3}\right) \) and \( \cos\left(\frac{x}{3}\right) \).

Step 2: Apply the Chain Rule for Integration

Since our function involves \( \frac{x}{3} \), a substitution can be helpful. Let \( u = \frac{x}{3} \), then \( du = \frac{1}{3}dx \) or \( dx = 3du \). We can rewrite the integral of \( f(x) = \sin(u) + \cos(u) \) with respect to \( x \) as \( \int (\sin(u) + \cos(u)) \cdot 3\, du \).

Step 3: Integrate Each Term

Integrate each term separately: \( \int 3\sin(u)\, du = -3\cos(u) + C_1 \) and \( \int 3\cos(u)\, du = 3\sin(u) + C_2 \). Combine constants \( C_1 \) and \( C_2 \) to a general constant \( C \).

Step 4: Substitute Back in Terms of \( x \)

Substitute \( u = \frac{x}{3} \) back into the antiderivatives: \( -3\cos\left(\frac{x}{3}\right) + 3\sin\left(\frac{x}{3}\right) + C \). Thus, the general antiderivative of \( f(x) \) is \( F(x) = -3\cos\left(\frac{x}{3}\right) + 3\sin\left(\frac{x}{3}\right) + C \).

Key Concepts

Trigonometric Functions

The function given in the exercise contains trigonometric functions, specifically \( \sin \left( \frac{x}{3} \right) \) and \( \cos \left( \frac{x}{3} \right) \). These functions are paramount in calculus, and understanding their behaviors is crucial. Trigonometric functions describe relationships between angles and sides in right-angled triangles, but in calculus, they extend to periodic functions that repeat over specific intervals. For example, the sine and cosine functions repeat every \( 2\pi \). This periodicity is vital when calculating integrals and derivatives, as it influences how these functions behave over various intervals.When dealing with antiderivatives of sine and cosine:

• The antiderivative of \( \sin(x) \) is \( -\cos(x) \)
• The antiderivative of \( \cos(x) \) is \( \sin(x) \)

Therefore, recognizing and manipulating these properties is key to solving integration problems involving trigonometric functions.

Chain Rule

The chain rule is a fundamental principle in calculus used to differentiate composite functions. However, it equally plays a role in integration when inverse operations, like finding antiderivatives, are required.In this problem, the function involves \( \frac{x}{3} \). This implies that it's a composite function because \( x \) is part of another function \( \sin \) or \( \cos \). To integrate such functions, a useful approach is substitution, which simplifies computations.

• You set \( u = \frac{x}{3} \). This handles the inner function.
• Then, differentiate \( u \) to find \( du = \frac{1}{3}dx \). Solving for \( dx \) gives \( dx = 3du \).

This substitution allows us to express the function in terms of \( u \), effectively simplifying the integration process. By applying substitution properly, integrating sometimes complex-looking functions becomes manageable.

Integration Techniques

Integration can sometimes be straightforward, but often requires specific techniques to handle different types of functions. Here, recognizing how to apply certain techniques makes solving the given problem feasible and comprehensible.In the given problem, we applied substitution, which is especially handy when dealing with composite functions, to simplify the integrands of \( \sin(u) \) and \( \cos(u) \). By multiplying each by 3, due to \( dx = 3du \), we integrated the simplified expressions separately:

• \( \int 3\sin(u)\, du = -3\cos(u) + C_1 \)
• \( \int 3\cos(u)\, du = 3\sin(u) + C_2 \)

These steps highlight two common integration techniques:

• Substitution – making the function easier to integrate by changing variables.
• Basic antiderivatives – using known antiderivative formulas.

Finally, combine and substitute back the original variable to yield the general solution. This problem-solving approach demonstrates how integration techniques are applied in calculus effectively.