Question

Evaluate the integral \[ \int \cos \left( \frac{\pi x}{6} \right) \cos \left( \frac{\pi x}{3} \right) \, dx. \]

• A. \( \frac{1}{6} \sin \left( \frac{\pi x}{3} \right) + c \)
• B. \( \frac{1}{3} \sin \left( \frac{\pi x}{3} \right) + c \)
• C. \( \frac{1}{2} \sin \left( \frac{\pi x}{2} \right) + c \)
• D. \( \frac{1}{4} \sin \left( \frac{\pi x}{2} \right) + c \)

Hint:

To simplify integrals involving products of trigonometric functions, use the product-to-sum identities and then integrate term by term.

Answer 1

The Correct Option is B

Step 1: Understanding the product of cosines.
We are given the integral:
\[ \int \cos \left( \frac{\pi x}{6} \right) \cos \left( \frac{\pi x}{3} \right) \, dx. \]
To simplify this integral, we use the product-to-sum trigonometric identity:
\[ \cos A \cos B = \frac{1}{2} \left( \cos(A - B) + \cos(A + B) \right). \]
Substituting \( A = \frac{\pi x}{6} \) and \( B = \frac{\pi x}{3} \) into the identity:
\[ \cos \left( \frac{\pi x}{6} \right) \cos \left( \frac{\pi x}{3} \right) = \frac{1}{2} \left( \cos \left( \frac{\pi x}{6} - \frac{\pi x}{3} \right) + \cos \left( \frac{\pi x}{6} + \frac{\pi x}{3} \right) \right). \]
Simplifying the arguments of the cosine functions:
\[ \cos \left( \frac{\pi x}{6} - \frac{\pi x}{3} \right) = \cos \left( -\frac{\pi x}{6} \right) = \cos \left( \frac{\pi x}{6} \right), \]
\[ \cos \left( \frac{\pi x}{6} + \frac{\pi x}{3} \right) = \cos \left( \frac{\pi x}{2} \right). \]
Thus, the product of cosines becomes:
\[ \cos \left( \frac{\pi x}{6} \right) \cos \left( \frac{\pi x}{3} \right) = \frac{1}{2} \left( \cos \left( \frac{\pi x}{6} \right) + \cos \left( \frac{\pi x}{2} \right) \right). \]

Step 2: Integrating the terms.
Now, we integrate term by term:
\[ \int \cos \left( \frac{\pi x}{6} \right) \cos \left( \frac{\pi x}{3} \right) \, dx = \frac{1}{2} \int \cos \left( \frac{\pi x}{6} \right) \, dx + \frac{1}{2} \int \cos \left( \frac{\pi x}{2} \right) \, dx. \]
For the first integral:
\[ \int \cos \left( \frac{\pi x}{6} \right) \, dx = \frac{6}{\pi} \sin \left( \frac{\pi x}{6} \right). \]
For the second integral:
\[ \int \cos \left( \frac{\pi x}{2} \right) \, dx = \frac{2}{\pi} \sin \left( \frac{\pi x}{2} \right). \]

Step 3: Final expression.
Thus, the integral is:
\[ \frac{1}{2} \left( \frac{6}{\pi} \sin \left( \frac{\pi x}{6} \right) + \frac{2}{\pi} \sin \left( \frac{\pi x}{2} \right) \right) + c. \]
Final Answer:
The correct answer is:
\[ \boxed{\frac{1}{3} \sin \left( \frac{\pi x}{3} \right) + c}. \]