Problem 46 - Thomas Calculus in SI Units

Question

Find the most general antiderivative or indefinite integral. You may need to try a solution and then adjust your guess. Check your answers by differentiation. $$\int(2 \cos 2 x-3 \sin 3 x) d x$$

Step-by-Step Solution

Answer

\(\sin(2x) + \cos(3x) + C\) is the most general antiderivative.

Step 1: Identify the Functions to Integrate

Look at the integrand: \(2 \cos 2x - 3 \sin 3x\). We recognize two trigonometric functions: \(2 \cos 2x\) and \(-3 \sin 3x\). These can be integrated separately.

Step 2: Integrate the First Term

The first term, \(2 \cos 2x\), can be integrated using the formula \(\int \cos(ax) \, dx = \frac{1}{a} \sin(ax)+C\). So, \(\int 2 \cos 2x \, dx = 2 \times \frac{1}{2} \sin(2x) = \sin(2x)\).

Step 3: Integrate the Second Term

The second term, \(-3 \sin 3x\), uses the formula \(\int \sin(ax) \, dx = -\frac{1}{a} \cos(ax)+C\). So, \(\int -3 \sin 3x \, dx = -3 \times -\frac{1}{3} \cos(3x) = \cos(3x)\).

Step 4: Combine the Results

Combine the integrated results of the two terms: \(\int (2 \cos 2x - 3 \sin 3x) \, dx = \sin(2x) + \cos(3x) + C\), where \(C\) is the constant of integration.

Step 5: Verify by Differentiating

Differentiate \(\sin(2x) + \cos(3x) + C\): \(\frac{d}{dx} [\sin(2x)] = 2\cos(2x)\) and \(\frac{d}{dx} [\cos(3x)] = -3\sin(3x)\). This yields \(2\cos(2x) - 3\sin(3x)\), confirming our antiderivative is correct.

Key Concepts

Antiderivative

The antiderivative, also known as an indefinite integral, is a fundamental concept in calculus. It refers to the process of finding a function whose derivative is the given function. In simpler terms, it reverses the action of differentiation. For instance, if differentiating a function gives you a certain result, the antiderivative is the function you differentiated to get that result.

In the example of \[\int (2\cos 2x - 3\sin 3x) \, dx\]we are looking for the function whose derivative is \(2\cos 2x - 3\sin 3x\). The solution involves recognizing the trigonometric nature of the function and then applying integration rules."
For each part of the expression, integration is performed separately:

• For \(2\cos 2x\), use the integration rule \(\int \cos(ax) \, dx = \frac{1}{a} \sin(ax) + C\).
• For \(-3\sin 3x\), use the integration rule \(\int \sin(ax) \, dx = -\frac{1}{a} \cos(ax) + C\).

Combining these results yields the antiderivative \(\sin(2x) + \cos(3x) + C\), where \(C\) is the constant of integration. This constant represents all possible vertical shifts of the antiderivative function.

Trigonometric Integration

Trigonometric integration involves finding the integral of functions containing trigonometric expressions, such as sine, cosine, and tangent functions. These types of problems often require specific integration techniques due to the periodic and oscillating nature of trigonometric functions.

In our exercise, \[\int (2\cos 2x - 3\sin 3x) \, dx\]we deal with both sine and cosine functions. The process includes:

• Identifying the trigonometric identities or transformations that can simplify the integration.
• Using standard integration formulas, such as \(\int \cos(ax) \, dx = \frac{1}{a} \sin(ax) + C\) and \(\int \sin(ax) \, dx = -\frac{1}{a} \cos(ax) + C\).

These formulas are directly applied here since no additional transformation is required. Recognizing these formulas is crucial for quickly finding and verifying the results of trigonometric integrals. Ultimately, mastering these patterns simplifies complex integration problems involving trigonometric functions.

Differentiation Verification

Once an antiderivative is found, differentiation verification is used to confirm its correctness. This concept ensures that by differentiating the antiderivative, we retrieve the original function or expression we started with.

For the calculated antiderivative \[\sin(2x) + \cos(3x) + C\]differentiating each component helps to check our work:

• Differentiating \(\sin(2x)\) yields \(2\cos(2x)\), based on the chain rule and the derivative of sine being cosine.
• Differentiating \(\cos(3x)\) gives \(-3\sin(3x)\), as the derivative of cosine is negative sine, scaled by the inner function's derivative.

Combining these results, we arrive back at the original expression \(2\cos(2x) - 3\sin(3x)\). This confirms that the integration was performed correctly, as differentiation returns us to the integrand, verifying the solution. Differentiation verification is an essential step in ensuring the accuracy of an antiderivative.