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

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 \( \frac{\pi}{2} \).

Answer 2

Step 1: Analyze the given integral.
We are given the integral: \[ \int_{-1}^{1} \frac{\cos \alpha x}{1 + 3^x} \, dx = \frac{2}{\pi} \] and we need to find the value of \( \alpha \).

Step 2: Symmetry consideration.
Notice that the limits of integration are symmetric, i.e., from \(-1\) to \(1\). A useful strategy for integrals of this form is to examine the symmetry of the integrand.
Let’s consider the transformation \( x \to -x \) in the integral. This gives: \[ I = \int_{-1}^{1} \frac{\cos \alpha x}{1 + 3^x} \, dx \] By the change of variable \( x \to -x \), we get: \[ I = \int_{-1}^{1} \frac{\cos \alpha (-x)}{1 + 3^{-x}} \, (-dx) \] Since \( \cos(-\alpha x) = \cos(\alpha x) \), the integral becomes: \[ I = \int_{-1}^{1} \frac{\cos \alpha x}{1 + 3^{-x}} \, dx \] Now, we add the two expressions for \( I \): \[ 2I = \int_{-1}^{1} \left( \frac{\cos \alpha x}{1 + 3^x} + \frac{\cos \alpha x}{1 + 3^{-x}} \right) dx \] Factor out \( \cos \alpha x \): \[ 2I = \int_{-1}^{1} \cos \alpha x \left( \frac{1}{1 + 3^x} + \frac{1}{1 + 3^{-x}} \right) dx \] Simplify the terms inside the parentheses: \[ \frac{1}{1 + 3^x} + \frac{1}{1 + 3^{-x}} = \frac{(1 + 3^{-x}) + (1 + 3^x)}{(1 + 3^x)(1 + 3^{-x})} \] This simplifies further to: \[ \frac{2 + 3^x + 3^{-x}}{(1 + 3^x)(1 + 3^{-x})} \] Thus, the integral becomes: \[ 2I = \int_{-1}^{1} \cos \alpha x \cdot \frac{2 + 3^x + 3^{-x}}{(1 + 3^x)(1 + 3^{-x})} \, dx \] Now, recall the structure of the original equation, \( \int_{-1}^{1} \frac{\cos \alpha x}{1 + 3^x} \, dx = \frac{2}{\pi} \). By comparing both sides, we can find that \( \alpha = \frac{\pi}{2} \).

Step 3: Conclusion.
The value of \( \alpha \) is \( \boxed{\frac{\pi}{2}} \).