`int(sec^(2)x)/(sqrt(1-tan^(2)x))dx=?`

Options

A. `sin^(-1)(tanx)+C`
B. `cos^(-1)(sinx)+C`
C. `tan^(-1)(cosx)+C`
D. `tan^(-1)(sinx)+C`

Explanation

To solve the integral \( \int \frac{\sec^2 x}{\sqrt{1 - \tan^2 x}} \, dx \), we can follow these steps:

### Step 1: Rewrite the integral
We start with the integral:
\[ I = \int \frac{\sec^2 x}{\sqrt{1 - \tan^2 x}} \, dx \]

### Step 2: Use substitution
Let \( u = \tan x \). Then, the derivative of \( u \) with respect to \( x \) is:
\[ \frac{du}{dx} = \sec^2 x \quad \Rightarrow \quad du = \sec^2 x \, dx \]
This implies that \( dx = \frac{du}{\sec^2 x} \).

### Step 3: Substitute in the integral
Now, substituting \( u \) into the integral, we get:
\[ I = \int \frac{\sec^2 x}{\sqrt{1 - u^2}} \cdot \frac{du}{\sec^2 x} \]
The \( \sec^2 x \) terms cancel out:
\[ I = \int \frac{1}{\sqrt{1 - u^2}} \, du \]

### Step 4: Recognize the integral
The integral \( \int \frac{1}{\sqrt{1 - u^2}} \, du \) is a standard integral, which evaluates to:
\[ I = \arcsin(u) + C \]

### Step 5: Substitute back for \( u \)
Recalling that \( u = \tan x \), we substitute back:
\[ I = \arcsin(\tan x) + C \]

### Final Answer
Thus, the final result of the integral is:
\[ \int \frac{\sec^2 x}{\sqrt{1 - \tan^2 x}} \, dx = \arcsin(\tan x) + C \]