Evaluate `int(dx)/((sinx+a s e c x)^2)` when `|a|> 1/2.`

Answer

To evaluate the integral

\[ I = \int \frac{dx}{(\sin x + a \sec x)^2} \]

given that \(|a| > \frac{1}{2}\), we can follow these steps:

### Step 1: Rewrite the Integral
We start by rewriting the integral in a more manageable form. We can multiply the numerator and denominator by \(\sec^2 x\):

\[ I = \int \frac{\sec^2 x \, dx}{(\sin x \sec x + a \sec^2 x)^2} \]

### Step 2: Simplify the Expression
Using the identity \(\sec x = \frac{1}{\cos x}\) and \(\sin x \sec x = \tan x\), we can simplify the expression:

\[ I = \int \frac{\sec^2 x \, dx}{(\tan x + a)^2} \]

### Step 3: Substitution
Let us make the substitution \(t = \tan x\). Then, we have:

\[ dt = \sec^2 x \, dx \quad \Rightarrow \quad dx = \frac{dt}{\sec^2 x} \]

Thus, the integral becomes:

\[ I = \int \frac{dt}{(t + a)^2} \]

### Step 4: Evaluate the Integral
The integral \(\int \frac{dt}{(t + a)^2}\) can be evaluated using the formula:

\[ \int \frac{dx}{(x + b)^n} = -\frac{1}{(n-1)(x + b)^{n-1}} + C \quad \text{for } n \neq 1 \]

In our case, \(n = 2\):

\[ I = -\frac{1}{(t + a)} + C = -\frac{1}{(\tan x + a)} + C \]

### Step 5: Final Result
Thus, the final result of the integral is:

\[ I = -\frac{1}{\tan x + a} + C \]