`int(dx)/(1+2sinx+cosx)`

Options

A. `log(2sinx+cosx)=`
B. `log(2+2tan""(x)/(2))+c`
C. `(1)/(2)log(1-2tan""(x)/(2))+c`
D. `(1)/(2)log(1+2tan""(x)/(2))+c`

Answer

To solve the integral \( I = \int \frac{dx}{1 + 2 \sin x + \cos x} \), we can follow these steps:

### Step 1: Rewrite the Denominator
We start with the expression in the denominator:
\[ 1 + 2 \sin x + \cos x \]
Using the identities for sine and cosine, we can express this in terms of \( \tan \frac{x}{2} \):
\[ \sin x = \frac{2 \tan \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} \quad \text{and} \quad \cos x = \frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} \]
Substituting these into the denominator gives:
\[ 1 + 2 \left( \frac{2 \tan \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} \right) + \left( \frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} \right) \]

### Step 2: Simplify the Denominator
Combining the terms in the denominator:
\[ = \frac{(1 + \tan^2 \frac{x}{2}) + 4 \tan \frac{x}{2} + (1 - \tan^2 \frac{x}{2})}{1 + \tan^2 \frac{x}{2}} \]
This simplifies to:
\[ = \frac{2 + 4 \tan \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} \]

### Step 3: Rewrite the Integral
Now our integral becomes:
\[ I = \int \frac{1 + \tan^2 \frac{x}{2}}{2 + 4 \tan \frac{x}{2}} \, dx \]

### Step 4: Use Substitution
Let \( t = \tan \frac{x}{2} \). Then, we know:
\[ dx = \frac{2}{1 + t^2} dt \]
Substituting this into the integral gives:
\[ I = \int \frac{1 + t^2}{2 + 4t} \cdot \frac{2}{1 + t^2} dt = 2 \int \frac{1}{2 + 4t} dt \]

### Step 5: Integrate
Now we can integrate:
\[ I = 2 \cdot \frac{1}{4} \ln |2 + 4t| + C = \frac{1}{2} \ln |2 + 4t| + C \]

### Step 6: Back Substitute
Substituting back \( t = \tan \frac{x}{2} \):
\[ I = \frac{1}{2} \ln |2 + 4 \tan \frac{x}{2}| + C \]

### Step 7: Final Result
We can express this as:
\[ I = \frac{1}{2} \ln |1 + 2 \tan \frac{x}{2}| + C \]