Question

The value of \(\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\cos x}{1+e^x}\ dx\) is

• A. 2
• B. 0
• C. 1
• D. -2

Answer 1

The Correct Option is C

Let:

$$ I = \int_{-\pi/2}^{\pi/2} \frac{\cos x}{1+e^x} dx. $$

We can use the property $ \int_a^b f(x) dx = \int_a^b f(a+b-x) dx $. 

Applying this property:

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

Since $ \cos(-x) = \cos x $, we have:

$$ I = \int_{-\pi/2}^{\pi/2} \frac{\cos x}{1+e^{-x}} dx. $$

Rewriting $ 1+e^{-x} $ as $ \frac{e^x+1}{e^x} $, the integral becomes:

$$ I = \int_{-\pi/2}^{\pi/2} \frac{\cos x}{\frac{e^x+1}{e^x}} dx = \int_{-\pi/2}^{\pi/2} \frac{e^x \cos x}{1+e^x} dx. $$

Now, add the original expression for $ I $ and this new expression:

$$ 2I = \int_{-\pi/2}^{\pi/2} \frac{\cos x}{1+e^x} dx + \int_{-\pi/2}^{\pi/2} \frac{e^x \cos x}{1+e^x} dx. $$

Combine the fractions under a common denominator:

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

Evaluate $ \int_{-\pi/2}^{\pi/2} \cos x dx $:

$$ \int_{-\pi/2}^{\pi/2} \cos x dx = \sin x \Big|_{-\pi/2}^{\pi/2} = \sin(\pi/2) - \sin(-\pi/2) = 1 - (-1) = 2. $$

Thus:

$$ 2I = 2 \implies I = 1. $$

Final Answer: The final answer is $ \boxed{1} $.

Answer 2

We need to find the value of \(\int_{-\pi/2}^{\pi/2} \frac{\cos x}{1 + e^x} \, dx\).

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

We use the property \(\int_a^b f(x) dx = \int_a^b f(a+b-x) dx\). 

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

Since \(\cos(-x) = \cos x\), we have: \(I = \int_{-\pi/2}^{\pi/2} \frac{\cos x}{1 + e^{-x}} \, dx = \int_{-\pi/2}^{\pi/2} \frac{\cos x}{\frac{e^x+1}{e^x}} \, dx = \int_{-\pi/2}^{\pi/2} \frac{e^x \cos x}{1 + e^x} \, dx\).

Now we have two expressions for I:

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

Adding these two expressions for I, we get:

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

\(2I = \int_{-\pi/2}^{\pi/2} \cos x \, dx\)

\(2I = [\sin x]_{-\pi/2}^{\pi/2} = \sin(\pi/2) - \sin(-\pi/2) = 1 - (-1) = 2\)

\(I = \frac{2}{2} = 1\)

Therefore, the correct option is (C) 1.