Problem 16 - Calculus: One and Several Variables

Question

Calculate. (If you run out of ideas, use the examples as models.) $$\int_{0}^{\pi / 2} \cos 2 x \sin 3 x d x$$.

Step-by-Step Solution

Answer

The short answer to the question is: The definite integral of the function \(f(x) = \cos (2x) \sin (3x)\) within the interval \([0, \frac{\pi}{2}]\) is calculated as \(\int_{0}^{\pi / 2} \cos (2x) \sin (3x) dx = \frac{7}{10}\).

Step 1: Use the product-to-sum trigonometric identity

Since we have a product of sine and cosine functions, we can use the product-to-sum formula to simplify the expression. Using the product-to-sum identity: \(\sin A \cos B = \frac{1}{2}[\sin(A-B) + \sin(A+B)]\) Replacing A with 3x, and B with 2x, we get: \(\sin(3x) \cos(2x) = \frac{1}{2}[\sin(3x - 2x) + \sin(3x + 2x)]\) Simplifying the expression: \(\sin(3x) \cos(2x) = \frac{1}{2}[\sin(x) + \sin(5x)]\) The integral becomes: \(\int_{0}^{\pi / 2} \frac{1}{2}[\sin(x) + \sin(5x)] dx\)

Step 2: Separate the integral and integrate

Next, we separate the integral into two parts and perform the integration. \(\int_{0}^{\pi / 2} \frac{1}{2}[\sin(x) + \sin(5x)] dx = \frac{1}{2} \int_{0}^{\pi / 2} \sin(x) dx + \frac{1}{2} \int_{0}^{\pi / 2} \sin(5x) dx\) Now let's integrate each term: \(\int \sin(x) dx = -\cos(x)\) and \(\int \sin(5x) dx = -\frac{1}{5}\cos(5x)\) So, the expression becomes: \(\frac{1}{2}[-\cos(x)]\Big|_{0}^{\pi / 2} + \frac{1}{2}[-\frac{1}{5}\cos(5x)]\Big|_{0}^{\pi / 2}\)

Step 3: Evaluate the definite integral

Finally, let's evaluate the integral by substituting the limits of integration into the expression: \(-\frac{1}{2}[\cos(\frac{\pi}{2}) - \cos(0)] - \frac{1}{10}[\cos(5 \cdot \frac{\pi}{2}) - \cos(0)]\) Plug in the values of the trigonometric functions: \(-\frac{1}{2}[0 - 1] - \frac{1}{10}[-1 - 1]\) Simplify the expression: \(\frac{1}{2} + \frac{2}{10} = \frac{1}{2} + \frac{1}{5}\) Add the fractions and get the final answer: \(=\frac{5 + 2}{10} = \frac{7}{10}\) Therefore, \(\int_{0}^{\pi / 2} \cos (2x) \sin (3x) dx = \frac{7}{10}\).

Key Concepts

Product-to-Sum Identity

Understanding and applying the **Product-to-Sum Identity** can significantly simplify complex trigonometric expressions, especially when dealing with integrals. When confronted with a product of sine and cosine functions, this identity transforms it into a sum of trigonometric functions, which are often easier to integrate. The identity is given by:
\[\sin A \cos B = \frac{1}{2}[\sin(A-B) + \sin(A+B)]\]This formula is invaluable as it helps convert the product of a sine and cosine into a sum, unlocking easier pathways for integration.
Applying this to our original problem:
If \(A = 3x\) and \(B = 2x\), we apply the identity:
\[\sin(3x) \cos(2x) = \frac{1}{2}[\sin(3x - 2x) + \sin(3x + 2x)] = \frac{1}{2}[\sin(x) + \sin(5x)]\]By using this identity, the integral simplifies greatly, paving the way for smoother integration steps.

Trigonometric Integration

**Trigonometric Integration** often requires transforming complex trigonometric expressions into forms that are more manageable to integrate. Leveraging identities like the product-to-sum can break down challenging integrals into simpler components.
In our given example:
The transformed integral is:\[\int_{0}^{\pi / 2} \frac{1}{2}[\sin(x) + \sin(5x)] dx\]This breaks down into two separate integrals, making them independent and straightforward.
- The integral of \( \sin(x) \) is: \[ \int \sin(x) \, dx = -\cos(x) + C \]- For \( \sin(5x) \): \[ \int \sin(5x) \, dx = -\frac{1}{5}\cos(5x) + C \]By integrating each part individually, we simplify the process, resulting in manageable calculations that lead us to the solution. It highlights the power of trigonometric identities and knowledge in making integration processes more efficient.

Evaluating Definite Integrals

The process of **Evaluating Definite Integrals** involves calculating the value of an integral within specified limits, revealing the net area under the curve described by the function. After integrating, we substitute the boundaries into the antiderivative.
In our exercise:
The antiderivatives from the integration step were:\[\frac{1}{2}[-\cos(x)]\Big|_{0}^{\pi / 2} + \frac{1}{2}[-\frac{1}{5}\cos(5x)]\Big|_{0}^{\pi / 2}\]To evaluate:

• For \(-\cos(x)\) from \(0\) to \(\frac{\pi}{2}\): \[ -\cos(\frac{\pi}{2}) + \cos(0) = 0 - 1 = -1 \]
• For \(-\frac{1}{5}\cos(5x)\) over the same limits: \[ -\frac{1}{5}[\cos(\frac{5\pi}{2}) - \cos(0)] = -\frac{1}{5}[-1 - 1] = \frac{2}{5} \]

By calculating each part and summing the values, we found:\[\frac{1}{2} + \frac{2}{10} = \frac{1}{2} + \frac{1}{5} = \frac{7}{10}\]This demonstrates the systematic approach needed to solve definite integrals, ensuring a thorough and accurate result.