`int(sin2x)/(2cos^(2)x+3sin^(2)x)dx` equals

Options

A. `log(2+sinx)+C`
B. `log(2+cos^(2)x)+C`
C. `log(2+sin^(2)x)+C`
D. `log(2+cosx)+C`

Answer

To solve the integral \( \int \frac{\sin 2x}{2 \cos^2 x + 3 \sin^2 x} \, dx \), we can follow these steps:

### Step 1: Rewrite the integrand using the double angle identity
We know that \( \sin 2x = 2 \sin x \cos x \). Thus, we can rewrite the integral as:

\[ \int \frac{2 \sin x \cos x}{2 \cos^2 x + 3 \sin^2 x} \, dx \]

### Step 2: Simplify the denominator
Notice that we can express \( 3 \sin^2 x \) in terms of \( \cos^2 x \) and \( \sin^2 x \):

\[ 2 \cos^2 x + 3 \sin^2 x = 2 \cos^2 x + 2 \sin^2 x + \sin^2 x = 2(\cos^2 x + \sin^2 x) + \sin^2 x \]

Since \( \cos^2 x + \sin^2 x = 1 \), we have:

\[ 2 \cos^2 x + 3 \sin^2 x = 2 + \sin^2 x \]

### Step 3: Substitute into the integral
Now, we can substitute this back into our integral:

\[ \int \frac{2 \sin x \cos x}{2 + \sin^2 x} \, dx \]

### Step 4: Use substitution
Let \( t = \sin x \). Then, \( dt = \cos x \, dx \) or \( dx = \frac{dt}{\cos x} \). We also note that \( \cos x = \sqrt{1 - t^2} \). Therefore, the integral becomes:

\[ \int \frac{2t \sqrt{1 - t^2}}{2 + t^2} \, dt \]

### Step 5: Simplify the integral
Now, we can rewrite the integral as:

\[ \int \frac{2t}{2 + t^2} \, dt \]

### Step 6: Use another substitution
Let \( u = 2 + t^2 \). Then, \( du = 2t \, dt \) or \( dt = \frac{du}{2t} \). The integral becomes:

\[ \int \frac{du}{u} \]

### Step 7: Integrate
The integral of \( \frac{1}{u} \) is:

\[ \log |u| + C \]

### Step 8: Substitute back
Substituting back for \( u \):

\[ \log |2 + t^2| + C \]

And substituting back for \( t \):

\[ \log |2 + \sin^2 x| + C \]

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

\[ \int \frac{\sin 2x}{2 \cos^2 x + 3 \sin^2 x} \, dx = \log(2 + \sin^2 x) + C \]