`int(sin^2x*sec^2x+2tanx*sin^(- 1)x*sqrt(1-x^2))/(sqrt(1-x^2)(1+tan^2x))dx`

Options

A. `(sin^(-1)x)(cos^2x)+C`
B. `(sin^(-1)x)(sin^2x)+C`
C. `(cos^(-1)x)(sin^2x)+C`
D. `-sin^(-1)x(sin^2x)+C`

Answer

To solve the integral

\[ \int \frac{\sin^2 x \sec^2 x + 2 \tan x \sin^{-1} x \sqrt{1 - x^2}}{\sqrt{1 - x^2} (1 + \tan^2 x)} \, dx, \]

we can break it down step by step.

### Step 1: Rewrite the integral
First, we can separate the terms in the integral:

\[ \int \left( \frac{\sin^2 x \sec^2 x}{\sqrt{1 - x^2} (1 + \tan^2 x)} + \frac{2 \tan x \sin^{-1} x \sqrt{1 - x^2}}{\sqrt{1 - x^2} (1 + \tan^2 x)} \right) \, dx. \]

### Step 2: Simplify the first term
Recall that \( \sec^2 x = 1 + \tan^2 x \). Therefore, we can rewrite the first term:

\[ \frac{\sin^2 x \sec^2 x}{\sqrt{1 - x^2} (1 + \tan^2 x)} = \frac{\sin^2 x (1 + \tan^2 x)}{\sqrt{1 - x^2} (1 + \tan^2 x)} = \frac{\sin^2 x}{\sqrt{1 - x^2}}. \]

### Step 3: Simplify the second term
Now, simplify the second term:

\[ \frac{2 \tan x \sin^{-1} x \sqrt{1 - x^2}}{\sqrt{1 - x^2} (1 + \tan^2 x)} = \frac{2 \tan x \sin^{-1} x}{1 + \tan^2 x}. \]

### Step 4: Combine the terms
Now, we can combine the two simplified terms:

\[ \int \left( \frac{\sin^2 x}{\sqrt{1 - x^2}} + \frac{2 \tan x \sin^{-1} x}{1 + \tan^2 x} \right) \, dx. \]

### Step 5: Rewrite \(\tan x\) and \(1 + \tan^2 x\)
We know that \( \tan x = \frac{\sin x}{\cos x} \) and \( 1 + \tan^2 x = \frac{1}{\cos^2 x} \). Thus, we can rewrite:

\[ \frac{2 \tan x \sin^{-1} x}{1 + \tan^2 x} = 2 \sin x \sin^{-1} x \cos x. \]

### Step 6: Final integration
Now we have:

\[ \int \left( \frac{\sin^2 x}{\sqrt{1 - x^2}} + 2 \sin x \sin^{-1} x \cos x \right) \, dx. \]

The first term can be recognized as part of the derivative of \( \sin^2 x \sin^{-1} x \).

### Step 7: Differentiate to confirm
To confirm, we differentiate \( \sin^2 x \sin^{-1} x \):

\[ \frac{d}{dx} (\sin^2 x \sin^{-1} x) = \sin^2 x \cdot \frac{1}{\sqrt{1 - x^2}} + 2 \sin x \cos x \sin^{-1} x. \]

This matches our integral, confirming that:

\[ \int \left( \frac{\sin^2 x}{\sqrt{1 - x^2}} + 2 \sin x \sin^{-1} x \cos x \right) \, dx = \sin^2 x \sin^{-1} x + C. \]

### Final Answer
Thus, the final result is:

\[ \sin^2 x \sin^{-1} x + C. \]