`int_0^pi (x dx)/(a^2 cos^2 x) , (a gt 1)`

Answer

To solve the integral \( I = \int_0^\pi \frac{x \, dx}{a^2 \cos^2 x} \) where \( a > 1 \), we will use the property of definite integrals that states:

\[ \int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx \]

### Step 1: Define the Integral
Let:
\[ I = \int_0^\pi \frac{x}{a^2 \cos^2 x} \, dx \]

### Step 2: Apply the Property of Definite Integrals
Using the property mentioned above, we can rewrite the integral:
\[ I = \int_0^\pi \frac{\pi - x}{a^2 \cos^2(\pi - x)} \, dx \]
Since \( \cos(\pi - x) = -\cos x \), we have:
\[ \cos^2(\pi - x) = \cos^2 x \]
Thus, we can rewrite the integral as:
\[ I = \int_0^\pi \frac{\pi - x}{a^2 \cos^2 x} \, dx \]

### Step 3: Combine the Two Expressions for \( I \)
Now, we have two expressions for \( I \):
1. \( I = \int_0^\pi \frac{x}{a^2 \cos^2 x} \, dx \)
2. \( I = \int_0^\pi \frac{\pi - x}{a^2 \cos^2 x} \, dx \)

Adding these two equations:
\[ 2I = \int_0^\pi \left( \frac{x + (\pi - x)}{a^2 \cos^2 x} \right) \, dx = \int_0^\pi \frac{\pi}{a^2 \cos^2 x} \, dx \]
This simplifies to:
\[ 2I = \frac{\pi}{a^2} \int_0^\pi \sec^2 x \, dx \]

### Step 4: Evaluate the Integral of \( \sec^2 x \)
The integral of \( \sec^2 x \) is:
\[ \int \sec^2 x \, dx = \tan x \]
Evaluating from \( 0 \) to \( \pi \):
\[ \int_0^\pi \sec^2 x \, dx = \tan(\pi) - \tan(0) = 0 - 0 = 0 \]

### Step 5: Substitute Back to Find \( I \)
Substituting back:
\[ 2I = \frac{\pi}{a^2} \cdot 0 = 0 \]
Thus:
\[ I = 0 \]

### Final Answer
\[ \int_0^\pi \frac{x \, dx}{a^2 \cos^2 x} = 0 \]

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