`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

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

we can follow these steps:

### Step 1: Rewrite the denominator
We can rewrite the denominator \(2 \cos^2 x + 3 \sin^2 x\) as follows:

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

Using the Pythagorean identity \(\cos^2 x + \sin^2 x = 1\), we have:

\[ 2 \cos^2 x + 3 \sin^2 x = 2(1) + \sin^2 x = 2 + \sin^2 x. \]

### Step 2: Substitute for \(\sin 2x\)
Recall that \(\sin 2x = 2 \sin x \cos x\). Thus, we can rewrite the integral as:

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

### Step 3: Use substitution
Let \(t = \sin x\). Then, \(dt = \cos x \, dx\), or \(dx = \frac{dt}{\cos x}\). Since \(\cos^2 x = 1 - \sin^2 x = 1 - t^2\), we have \(\cos x = \sqrt{1 - t^2}\). Therefore, the integral becomes:

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

### Step 4: Integrate
Now we can integrate:

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

Using the substitution \(u = 2 + t^2\), we have \(du = 2t \, dt\) or \(dt = \frac{du}{2t}\). Thus, the integral becomes:

\[ I = 2 \cdot \frac{1}{2} \int \frac{1}{u} \, du = \int \frac{1}{u} \, du = \ln |u| + C = \ln |2 + t^2| + C. \]

### Step 5: Substitute back
Now substituting back \(t = \sin x\):

\[ I = \ln |2 + \sin^2 x| + C. \]

### Final Answer
Thus, the final answer is:

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