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

Answer

To solve the integral \( \int \frac{1}{2 + \sin^2 x} \, dx \), we will follow a systematic approach. Here’s the step-by-step solution:

### Step 1: Rewrite the Integral
We start with the integral:
\[ \int \frac{1}{2 + \sin^2 x} \, dx \]

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

### Step 3: Use the Identity \( \sec^2 x = 1 + \tan^2 x \)
We know that \( \sec^2 x = 1 + \tan^2 x \). Thus, we can rewrite the integral as:
\[ \int \frac{\sec^2 x}{2 \sec^2 x} \, dx = \int \frac{1}{2 + \tan^2 x} \, dx \]

### Step 4: Substitute \( \tan x = t \)
Let \( t = \tan x \). Then, \( dt = \sec^2 x \, dx \) or \( dx = \frac{dt}{\sec^2 x} \). Hence, we can rewrite our integral:
\[ \int \frac{1}{2 + t^2} \cdot \frac{dt}{\sec^2 x} \]
Since \( \sec^2 x = 1 + t^2 \), we can substitute:
\[ dx = \frac{dt}{1 + t^2} \]
Thus, the integral becomes:
\[ \int \frac{1}{2 + t^2} \cdot \frac{dt}{1 + t^2} \]

### Step 5: Simplify the Integral
Now, we can simplify the integral:
\[ \int \frac{dt}{(2 + t^2)(1 + t^2)} \]

### Step 6: Partial Fraction Decomposition
We can use partial fraction decomposition:
\[ \frac{1}{(2 + t^2)(1 + t^2)} = \frac{A}{2 + t^2} + \frac{B}{1 + t^2} \]
Multiplying through by the denominator \( (2 + t^2)(1 + t^2) \) and solving for \( A \) and \( B \) gives us:
\[ 1 = A(1 + t^2) + B(2 + t^2) \]

### Step 7: Solve for Coefficients
Setting up equations for coefficients:
1. For \( t^2 \): \( A + B = 0 \)
2. For constant term: \( 2B = 1 \) → \( B = \frac{1}{2} \)

From \( A + B = 0 \), we have \( A = -\frac{1}{2} \).

### Step 8: Rewrite the Integral
Now we can rewrite the integral:
\[ \int \left( -\frac{1}{2(2 + t^2)} + \frac{1/2}{1 + t^2} \right) dt \]

### Step 9: Integrate Each Term
Integrating each term separately:
\[ -\frac{1}{2} \int \frac{1}{2 + t^2} dt + \frac{1}{2} \int \frac{1}{1 + t^2} dt \]
Using the known integrals:
\[ -\frac{1}{2} \cdot \frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{t}{\sqrt{2}}\right) + \frac{1}{2} \tan^{-1}(t) + C \]

### Step 10: Substitute Back for \( t \)
Substituting back \( t = \tan x \):
\[ -\frac{1}{2\sqrt{2}} \tan^{-1}\left(\frac{\tan x}{\sqrt{2}}\right) + \frac{1}{2} \tan^{-1}(\tan x) + C \]
Since \( \tan^{-1}(\tan x) = x \):
\[ -\frac{1}{2\sqrt{2}} \tan^{-1}\left(\frac{\tan x}{\sqrt{2}}\right) + \frac{x}{2} + C \]

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