The value of the integral `int_(0)^(pi) (1)/(e^(cosx)+1)dx`, is

Options

A. `pi`
B. 0
C. `2pi`
D. `(pi)/(2)`

Explanation

To solve the integral

\[ I = \int_{0}^{\pi} \frac{1}{e^{\cos x} + 1} \, dx, \]

we can use a property of definite integrals. Specifically, we can use the substitution \( x = \pi - t \). This gives us:

\[ dx = -dt. \]

When \( x = 0 \), \( t = \pi \) and when \( x = \pi \), \( t = 0 \). Thus, we can rewrite the integral as:

\[ I = \int_{\pi}^{0} \frac{1}{e^{\cos(\pi - t)} + 1} (-dt) = \int_{0}^{\pi} \frac{1}{e^{-\cos t} + 1} \, dt. \]

Now, we know that \( \cos(\pi - t) = -\cos t \). Therefore, we can simplify the integral:

\[ I = \int_{0}^{\pi} \frac{1}{e^{-\cos t} + 1} \, dt = \int_{0}^{\pi} \frac{1}{\frac{1}{e^{\cos t}} + 1} \, dt = \int_{0}^{\pi} \frac{e^{\cos t}}{1 + e^{\cos t}} \, dt. \]

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

1. \( I = \int_{0}^{\pi} \frac{1}{e^{\cos x} + 1} \, dx \)
2. \( I = \int_{0}^{\pi} \frac{e^{\cos x}}{1 + e^{\cos x}} \, dx \)

Adding these two equations gives:

\[ 2I = \int_{0}^{\pi} \left( \frac{1}{e^{\cos x} + 1} + \frac{e^{\cos x}}{1 + e^{\cos x}} \right) dx. \]

The expression inside the integral simplifies as follows:

\[ \frac{1}{e^{\cos x} + 1} + \frac{e^{\cos x}}{1 + e^{\cos x}} = \frac{1 + e^{\cos x}}{e^{\cos x} + 1} = 1. \]

Thus, we have:

\[ 2I = \int_{0}^{\pi} 1 \, dx = \pi. \]

Dividing both sides by 2 gives:

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

Therefore, the value of the integral is:

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