`int_(-pi//2)^(pi//2) (pi^(sin x))/(1+ pi^(sin x))dx`=

Options

A. 0
B. `pi//4`
C. `pi//2`
D. `pi`

Answer

To solve the integral

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

we can utilize a property of definite integrals. This property states that:

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

### Step 1: Define the Integral
Let
\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\pi^{\sin x}}{1 + \pi^{\sin x}} \, dx. \]

### Step 2: Apply the Property
We will apply the property mentioned above. Here, \( a = -\frac{\pi}{2} \) and \( b = \frac{\pi}{2} \), so \( a + b = 0 \). Thus, we can replace \( x \) with \( -x \):

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

### Step 3: Simplify the Expression
Using the property of sine, we have \( \sin(-x) = -\sin x \). Therefore, we can rewrite the integral as:

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

### Step 4: Rewrite the Integral
Now, we can simplify \( \frac{\pi^{-\sin x}}{1 + \pi^{-\sin x}} \):

\[ \frac{\pi^{-\sin x}}{1 + \pi^{-\sin x}} = \frac{1}{\pi^{\sin x}} \cdot \frac{1}{1 + \frac{1}{\pi^{\sin x}}} = \frac{1}{\pi^{\sin x} + 1}. \]

Thus, we have:

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

### Step 5: Combine the Two Integrals
Now we have two expressions for \( I \):

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

Adding these two expressions together:

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

### Step 6: Simplify the Sum
Notice that:

\[ \frac{\pi^{\sin x}}{1 + \pi^{\sin x}} + \frac{1}{\pi^{\sin x} + 1} = 1. \]

Thus, we have:

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

### Step 7: Evaluate the Integral
The integral of 1 from \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \) is simply the length of the interval:

\[ 2I = \left[ x \right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = \frac{\pi}{2} - \left(-\frac{\pi}{2}\right) = \pi. \]

### Step 8: Solve for \( I \)
Now, divide both sides by 2:

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

### Final Answer
Thus, the value of the integral is:

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

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