Question

If the value of the integral \[ \int_{-1}^{1} \frac{\cos \alpha x}{1 + 3^x} \, dx = \frac{2}{\pi}, \] then a value of \( \alpha \) is:

• A. \( \frac{\pi}{6} \)
• B. \( \frac{\pi}{2} \)
• C. \( \frac{\pi}{3} \)
• D. \( \frac{\pi}{4} \)

Answer 1

The Correct Option is B

The integral is recognized as an even function due to symmetric limits and the even property of \( \cos(\alpha x) \). This allows simplification by doubling the integral from 0 to 1:

\[ \int_{-1}^1 \frac{\cos \alpha x}{1 + 3^x} \, dx = 2 \int_0^1 \frac{\cos \alpha x}{1 + 3^x} \, dx. \]

Given that this integral's value is \( \frac{2}{\pi} \), we test \( \alpha \) values to achieve this result.

Upon evaluation, \( \alpha = \frac{\pi}{2} \) is found to satisfy this condition, resulting in:

\[ \int_{-1}^1 \frac{\cos \left( \frac{\pi}{2} x \right)}{1 + 3^x} \, dx = \frac{\pi}{2}. \]

Consequently, the value of \( \alpha \) is \( \frac{\pi}{2} \).