`int_(-pi+4)^(pi//4) (tan^(2)x)/(1+a^(x))dx` is equal to

Options

A. `(pi+4)/(4)`
B. `(pi-4)/(4)`
C. `(api)/(4)`
D. `(a+pi)/(4)`

Answer

To solve the integral \( I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\tan^2 x}{1 + a^x} \, dx \), we can use the property of definite integrals that states:

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

### Step 1: Set up the integral
Let \( I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\tan^2 x}{1 + a^x} \, dx \).

### Step 2: Apply the property of definite integrals
Using the property mentioned, we can rewrite the integral:
\[ I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\tan^2(-x)}{1 + a^{-x}} \, dx \]

### Step 3: Simplify \( \tan^2(-x) \)
Since \( \tan(-x) = -\tan(x) \), we have:
\[ \tan^2(-x) = \tan^2(x) \]
Thus, we can rewrite the integral as:
\[ I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\tan^2 x}{1 + \frac{1}{a^x}} \, dx \]

### Step 4: Simplify the denominator
The term \( 1 + \frac{1}{a^x} \) can be rewritten as:
\[ 1 + \frac{1}{a^x} = \frac{a^x + 1}{a^x} \]
So, the integral becomes:
\[ I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\tan^2 x \cdot a^x}{a^x + 1} \, dx \]

### Step 5: Combine the two expressions for \( I \)
Now we have two expressions for \( I \):
1. \( I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\tan^2 x}{1 + a^x} \, dx \)
2. \( I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\tan^2 x \cdot a^x}{a^x + 1} \, dx \)

Adding these two equations:
\[ 2I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \left( \frac{\tan^2 x}{1 + a^x} + \frac{\tan^2 x \cdot a^x}{a^x + 1} \right) dx \]

### Step 6: Factor out \( \tan^2 x \)
Factoring out \( \tan^2 x \) gives:
\[ 2I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \tan^2 x \left( \frac{1 + a^x}{1 + a^x} \right) dx = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \tan^2 x \, dx \]

### Step 7: Evaluate the integral
The integral \( \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \tan^2 x \, dx \) can be evaluated using the identity:
\[ \tan^2 x = \sec^2 x - 1 \]
Thus:
\[ \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \tan^2 x \, dx = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} (\sec^2 x - 1) \, dx \]
Calculating this gives:
\[ = \left[ \tan x \right]_{-\frac{\pi}{4}}^{\frac{\pi}{4}} - \left[ x \right]_{-\frac{\pi}{4}}^{\frac{\pi}{4}} = \left( 1 - (-1) \right) - \left( \frac{\pi}{4} - (-\frac{\pi}{4}) \right) = 2 - \frac{\pi}{2} \]

### Step 8: Solve for \( I \)
Thus, we have:
\[ 2I = 2 - \frac{\pi}{2} \implies I = 1 - \frac{\pi}{4} \]

### Final Result
The value of the integral is:
\[ I = 1 - \frac{\pi}{4} \]