Question

\(\int_{-\pi/2}^{\pi/2}\frac{\cos x}{1+e^{x}}\,dx=\)

• A. 0
• B. 1
• C. \(-1\)
• D. 2

Hint:

Use \(\int_{-a}^{a}f(x)dx=\int_0^a[f(x)+f(-x)]dx\). Here \(f(x)+f(-x)=\cos x\).

Answer 1

The Correct Option is B

Step 1: Approach:
Use the substitution $x\to -x$ on the same integral, then add the two forms of the integral together.

Step 2: Substitute:
Let $I=\int_{-\pi/2}^{\pi/2}\frac{\cos x}{1+e^x}dx$. Replacing $x$ by $-x$ (the limits swap and swap back), $I=\int_{-\pi/2}^{\pi/2}\frac{\cos x}{1+e^{-x}}dx=\int_{-\pi/2}^{\pi/2}\frac{e^x\cos x}{1+e^{x}}dx$.

Step 3: Add the two forms:
\[ 2I=\int_{-\pi/2}^{\pi/2}\frac{(1+e^x)\cos x}{1+e^x}dx=\int_{-\pi/2}^{\pi/2}\cos x\,dx \]

Step 4: Evaluate:
$\int_{-\pi/2}^{\pi/2}\cos x\,dx=\sin\frac{\pi}{2}-\sin\left(-\frac{\pi}{2}\right)=1+1=2$. So $2I=2$ and $I=1$.

Step 5: Choose:
The value 1 is option 2.

Final Answer:
\[ \boxed{1} \]