The value of `int_(-pi//2)^(pi//2) (dx)/(e^(sin x) +1)` is equal to

Options

A. 0
B. 1
C. `-pi//2`
D. `pi//2`

Answer

To solve the integral

\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{dx}{e^{\sin x} + 1}, \]

we can use a property of definite integrals. This property states that

\[ \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx. \]

In our case, \(a = -\frac{\pi}{2}\) and \(b = \frac{\pi}{2}\), so \(a + b = 0\). Therefore, we can rewrite the integral as:

\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{dx}{e^{\sin(-x)} + 1}. \]

Since \(\sin(-x) = -\sin x\), we have:

\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{dx}{e^{-\sin x} + 1}. \]

Now, we can simplify \(e^{-\sin x}\) using the identity \(e^{-\sin x} = \frac{1}{e^{\sin x}}\):

\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{dx}{\frac{1}{e^{\sin x}} + 1} = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{e^{\sin x}}{1 + e^{\sin x}} \, dx. \]

Now we have two expressions for \(I\):

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

Next, we add these two equations:

\[ 2I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left( \frac{1}{e^{\sin x} + 1} + \frac{e^{\sin x}}{1 + e^{\sin x}} \right) dx. \]

Combining the fractions gives:

\[ 2I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1 + e^{\sin x}}{e^{\sin x} + 1} \, dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 1 \, dx. \]

The integral of 1 over the interval \([- \frac{\pi}{2}, \frac{\pi}{2}]\) is simply the length of the interval:

\[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 1 \, dx = \left[ x \right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = \frac{\pi}{2} - \left(-\frac{\pi}{2}\right) = \pi. \]

Thus, we have:

\[ 2I = \pi \implies I = \frac{\pi}{2}. \]

Therefore, the value of the integral is

\[ \boxed{\frac{\pi}{2}}. \]