`int(1)/(2+sin^(2) x)dx`

Answer

To solve the integral \( \int \frac{1}{2 + \sin^2 x} \, dx \), we will follow these steps:

### Step 1: Rewrite the integral
We start with the integral:
\[ I = \int \frac{1}{2 + \sin^2 x} \, dx \]
To simplify the expression, we can divide the numerator and denominator by \( \cos^2 x \):
\[ I = \int \frac{\frac{1}{\cos^2 x}}{\frac{2}{\cos^2 x} + \frac{\sin^2 x}{\cos^2 x}} \, dx = \int \frac{\sec^2 x}{\frac{2}{\cos^2 x} + \tan^2 x} \, dx \]

### Step 2: Use the substitution \( t = \tan x \)
Let \( t = \tan x \), then \( dx = \frac{1}{\cos^2 x} \, dt = \sec^2 x \, dt \). Thus, we can rewrite the integral as:
\[ I = \int \frac{1}{2 + t^2} \, dt \]

### Step 3: Recognize the standard integral form
The integral \( \int \frac{1}{a^2 + x^2} \, dx \) has a standard result:
\[ \int \frac{1}{a^2 + x^2} \, dx = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + C \]
In our case, \( a^2 = 2 \) (so \( a = \sqrt{2} \)), and \( x = t \):
\[ I = \frac{1}{\sqrt{2}} \tan^{-1} \left( \frac{t}{\sqrt{2}} \right) + C \]

### Step 4: Substitute back for \( t \)
Since \( t = \tan x \), we substitute back:
\[ I = \frac{1}{\sqrt{2}} \tan^{-1} \left( \frac{\tan x}{\sqrt{2}} \right) + C \]

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

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