Question

Evaluate the definite integral: $$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos x}{1 + e^x}\ dx$$

• A. 1
• B. 2
• C. -1
• D. 0

Hint:

Whenever you see an integral with symmetric limits $[-a, a]$ and an exponential roadblock function like $\frac{1}{1 + e^x}$ or $\frac{1}{1 + a^x}$ multiplied by an even function, the exponential part always cancels out upon applying King's property. The integral simply reduces to half the width: $\int_{0}^{a} \text{even}(x)\ dx$.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The problem requires computing the value of a definite integral involving a trigonometric cosine term and an exponential function over the symmetric limits $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

Step 2: Key Formula or Approach:
We apply the definitive integral property universally known as King's Property:
$$\int_{a}^{b} f(x)\ dx = \int_{a}^{b} f(a + b - x)\ dx$$ Here, our limits yield $a + b = -\frac{\pi}{2} + \frac{\pi}{2} = 0$. Therefore, the substitution simplifies to replacing $x$ with $-x$.

Step 3: Detailed Explanation:
Let the initial integral be $I$:
$$I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos x}{1 + e^x}\ dx \quad \text{--- (Equation 1)}$$ Applying the property $x \rightarrow -x$:
$$I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos(-x)}{1 + e^{-x}}\ dx$$ Since $\cos(-x) = \cos x$ and $e^{-x} = \frac{1}{e^x}$, rewrite the integrand:
$$I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos x}{1 + \frac{1}{e^x}}\ dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{e^x \cos x}{1 + e^x}\ dx \quad \text{--- (Equation 2)}$$ Now, add Equation 1 and Equation 2 together:
$$2I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos x}{1 + e^x}\ dx + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{e^x \cos x}{1 + e^x}\ dx$$ Combine the numerators over the common denominator:
$$2I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos x (1 + e^x)}{1 + e^x}\ dx$$ The $(1 + e^x)$ terms cancel out perfectly:
$$2I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos x\ dx$$ Since $\cos x$ is an even function, we can simplify the symmetric limits:
$$2I = 2 \int_{0}^{\frac{\pi}{2}} \cos x\ dx \implies I = \int_{0}^{\frac{\pi}{2}} \cos x\ dx$$ Integrate and apply the bounds:
$$I = \Big[ \sin x \Big]_{0}^{\frac{\pi}{2}} = \sin\left(\frac{\pi}{2}\right) - \sin(0) = 1 - 0 = 1$$

Step 4: Final Answer:
The value of the definite integral is 1, matching option (A).