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

Answer: (B)

$\frac{\pi}{2}$

Answer 2

To solve the integral, we note that it is an even function due to the symmetric limits and the even nature of $\cos(\alpha x)$. Therefore, we can simplify 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 the value of this integral equals $\frac{2}{\pi}$, we proceed by testing values of $\alpha$ to match this result.

Through evaluation, it turns out that setting $\alpha = \frac{\pi}{2}$ satisfies this condition, yielding:

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

Therefore, the value of $\alpha$ is $\frac{\pi}{2}$.