`int_(0)^(pi//2)(e^(sin x))/(e^(sin x)+e^(cos x))dx=`

Options

A. `e^(pi)`
B. 0
C. `(pi)/(2)`
D. `(pi)/(4)`

Answer

To solve the integral

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{e^{\sin x}}{e^{\sin x} + e^{\cos x}} \, dx, \]

we can use a property of definite integrals. Specifically, we can use the substitution \( x = \frac{\pi}{2} - t \), which gives us:

\[ dx = -dt. \]

When \( x = 0 \), \( t = \frac{\pi}{2} \), and when \( x = \frac{\pi}{2} \), \( t = 0 \). Thus, we can rewrite the integral as:

\[ I = \int_{\frac{\pi}{2}}^{0} \frac{e^{\sin\left(\frac{\pi}{2} - t\right)}}{e^{\sin\left(\frac{\pi}{2} - t\right)} + e^{\cos\left(\frac{\pi}{2} - t\right)}} (-dt). \]

This simplifies to:

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{e^{\cos t}}{e^{\cos t} + e^{\sin t}} \, dt. \]

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

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

Adding these two equations gives:

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

The denominators are the same, so we can combine the fractions:

\[ 2I = \int_{0}^{\frac{\pi}{2}} \frac{e^{\sin x} + e^{\cos x}}{e^{\sin x} + e^{\cos x}} \, dx = \int_{0}^{\frac{\pi}{2}} 1 \, dx. \]

Calculating the integral:

\[ \int_{0}^{\frac{\pi}{2}} 1 \, dx = \left[ x \right]_{0}^{\frac{\pi}{2}} = \frac{\pi}{2}. \]

Thus, we have:

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

Dividing both sides by 2 gives:

\[ I = \frac{\pi}{4}. \]

Therefore, the value of the integral is:

\[ \boxed{\frac{\pi}{4}}. \]