`int(2sinx)/(3+sin2x)\ dx`

Options

A. `(1)/(2) "In"|(2+sinx-cosx)/(2-sinx+cosx)|-(1)/(sqrt(2))tan^(-1)((sinx+cosx)/(sqrt(2)))+c`
B. `(1)/(2) "In"|(2+sinx-cosx)/(2-sinx+cosx)|-(1)/(2sqrt(2))tan^(-1)((sinx+cosx)/(sqrt(2)))+c`
C. `(1)/(4) "In"|(2+sinx-cosx)/(2-sinx+cosx)|-(1)/(sqrt(2))tan^(-1)((sinx+cosx)/(sqrt(2)))+c`
D. none of these

Explanation

To solve the integral \( \int \frac{2 \sin x}{3 + \sin 2x} \, dx \), we will follow a systematic approach.

### Step-by-Step Solution:

1. **Rewrite the Integral:**
Start with the original integral:
\[ I = \int \frac{2 \sin x}{3 + \sin 2x} \, dx \]
We can use the identity \( \sin 2x = 2 \sin x \cos x \) to rewrite \( \sin 2x \):
\[ I = \int \frac{2 \sin x}{3 + 2 \sin x \cos x} \, dx \]

2. **Split the Integral:**
We can express \( 2 \sin x \) as \( \sin x + \cos x + \sin x - \cos x \):
\[ I = \int \left( \frac{\sin x + \cos x}{3 + 2 \sin x \cos x} + \frac{\sin x - \cos x}{3 + 2 \sin x \cos x} \right) \, dx \]

3. **Rewrite the Denominator:**
We can rewrite \( 3 + 2 \sin x \cos x \) in a more convenient form:
\[ 3 + 2 \sin x \cos x = 4 - 1 + 2 \sin x \cos x = 4 - (1 - 2 \sin x \cos x) \]

4. **Substitution:**
Let \( t = \sin x + \cos x \) and \( u = \sin x - \cos x \). Then, we differentiate:
\[ dt = (\cos x - \sin x) \, dx \quad \text{and} \quad du = (\cos x + \sin x) \, dx \]

5. **Change of Variables:**
Substitute \( t \) and \( u \) into the integral:
\[ I = \int \frac{dt}{4 - t^2} - \int \frac{du}{\sqrt{2^2 + u^2}} \]

6. **Integrate Each Part:**
The first integral can be solved using the formula for the integral of \( \frac{1}{a^2 - x^2} \):
\[ \int \frac{dt}{4 - t^2} = \frac{1}{4} \ln \left| \frac{2 + t}{2 - t} \right| + C_1 \]
The second integral can be solved using the formula for the integral of \( \frac{1}{a^2 + x^2} \):
\[ \int \frac{du}{\sqrt{2^2 + u^2}} = \frac{1}{2} \tan^{-1} \left( \frac{u}{2} \right) + C_2 \]

7. **Combine Results:**
Combine the results from both integrals:
\[ I = \frac{1}{4} \ln \left| \frac{2 + (\sin x + \cos x)}{2 - (\sin x + \cos x)} \right| - \frac{1}{2} \tan^{-1} \left( \frac{\sin x - \cos x}{2} \right) + C \]

8. **Final Result:**
Therefore, the final result of the integral is:
\[ I = \frac{1}{4} \ln \left| \frac{2 + \sin x + \cos x}{2 - \sin x - \cos x} \right| - \frac{1}{2} \tan^{-1} \left( \frac{\sin x - \cos x}{2} \right) + C \]