`int_(pi/2)^(pi/2)((x^2cosx)/(1+pi^x)+(sin^2x+1)/(e^(sin^(2025)x)+1))dx`

Options

A. `pi/2+(3pi)^2/4-2`
B. `pi/4+(3pi)^2/4+2`
C. `pi^2/4+(3pi)/4-2`
D. `pi^2/4+(3pi)/4+2`

Answer

To solve the integral

\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left( \frac{x^2 \cos x}{1 + \pi^x} + \frac{\sin^2 x + 1}{e^{\sin^{2025} x} + 1} \right) dx, \]

we can use properties of definite integrals to simplify our calculations.

### Step 1: Use the property of definite integrals

We can use the property of definite integrals that states:

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

Here, \( a = -\frac{\pi}{2} \) and \( b = \frac{\pi}{2} \), so \( a + b = 0 \). Thus, we have:

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

### Step 2: Substitute \( f(-x) \)

Now we compute \( f(-x) \):

\[ f(-x) = \frac{(-x)^2 \cos(-x)}{1 + \pi^{-x}} + \frac{\sin^2(-x) + 1}{e^{\sin^{2025}(-x)} + 1}. \]

Since \( \cos(-x) = \cos x \) and \( \sin(-x) = -\sin x \), we get:

\[ f(-x) = \frac{x^2 \cos x}{1 + \frac{1}{\pi^x}} + \frac{\sin^2 x + 1}{e^{-\sin^{2025} x} + 1}. \]

### Step 3: Rewrite \( f(-x) \)

Now we can rewrite \( f(-x) \):

\[ f(-x) = \frac{x^2 \cos x}{1 + \frac{1}{\pi^x}} + \frac{\sin^2 x + 1}{\frac{1}{e^{\sin^{2025} x}} + 1} = \frac{x^2 \cos x}{1 + \frac{1}{\pi^x}} + \frac{(\sin^2 x + 1)e^{\sin^{2025} x}}{1 + e^{\sin^{2025} x}}. \]

### Step 4: Combine \( I \) and \( f(-x) \)

Now, we can add \( I \) and \( I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} f(-x) \, dx \):

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

### Step 5: Simplify the expression

Combining the two expressions, we have:

\[ 2I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left( \frac{x^2 \cos x}{1 + \pi^x} + \frac{x^2 \cos x}{1 + \frac{1}{\pi^x}} + \frac{\sin^2 x + 1}{e^{\sin^{2025} x} + 1} + \frac{(\sin^2 x + 1)e^{\sin^{2025} x}}{1 + e^{\sin^{2025} x}} \right) dx. \]

### Step 6: Evaluate the integral

This simplifies to:

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

### Step 7: Use symmetry

Since \( x^2 \cos x \) is an even function and \( \sin^2 x + 1 \) is also even, we can evaluate:

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

### Step 8: Split the integral

Now we can split the integral:

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

### Step 9: Evaluate each integral

1. The first integral \( \int_{0}^{\frac{\pi}{2}} x^2 \cos x \, dx \) can be solved using integration by parts.
2. The second integral \( \int_{0}^{\frac{\pi}{2}} \sin^2 x \, dx = \frac{\pi}{4} \).
3. The third integral \( \int_{0}^{\frac{\pi}{2}} 1 \, dx = \frac{\pi}{2} \).

### Final Step: Combine results

After evaluating, combine the results to find \( I \).