If `I= int (sin 2x)/((3+4cosx)^(3))dx,` then `I` equals

Options

A. `(3cos x+8)/((3+4cosx)^(2))+C`
B. `(3+8cos x)/(16(3+4cosx)^(2))+C`
C. `(3+cos x)/((3+4cosx)^(2))+C`
D. `(3-8cos x)/(16(3+4cosx)^(2))+C`

Explanation

To solve the integral \( I = \int \frac{\sin 2x}{(3 + 4\cos x)^3} \, dx \), we can follow these steps:

### Step 1: Rewrite the integrand
We know that \( \sin 2x = 2 \sin x \cos x \). Thus, we can rewrite the integral as:
\[ I = \int \frac{2 \sin x \cos x}{(3 + 4\cos x)^3} \, dx \]

### Step 2: Substitution
Let \( t = 3 + 4\cos x \). Then, we differentiate \( t \) with respect to \( x \):
\[ dt = -4 \sin x \, dx \quad \Rightarrow \quad \sin x \, dx = -\frac{1}{4} dt \]
Now, we can substitute \( t \) and \( \sin x \, dx \) into the integral:
\[ I = \int \frac{2(-\frac{1}{4} dt) \cos x}{t^3} \]
This simplifies to:
\[ I = -\frac{1}{2} \int \frac{\cos x}{t^3} \, dt \]

### Step 3: Express \( \cos x \) in terms of \( t \)
From our substitution \( t = 3 + 4\cos x \), we can express \( \cos x \) as:
\[ \cos x = \frac{t - 3}{4} \]
Substituting this back into the integral gives:
\[ I = -\frac{1}{2} \int \frac{\frac{t - 3}{4}}{t^3} \, dt \]
This simplifies to:
\[ I = -\frac{1}{8} \int \frac{t - 3}{t^3} \, dt = -\frac{1}{8} \int \left( \frac{1}{t^2} - \frac{3}{t^3} \right) \, dt \]

### Step 4: Integrate
Now we can integrate term by term:
\[ I = -\frac{1}{8} \left( -\frac{1}{t} + \frac{3}{2t^2} \right) + C \]
This simplifies to:
\[ I = \frac{1}{8t} - \frac{3}{16t^2} + C \]

### Step 5: Substitute back for \( t \)
Recall that \( t = 3 + 4\cos x \). Substituting back gives:
\[ I = \frac{1}{8(3 + 4\cos x)} - \frac{3}{16(3 + 4\cos x)^2} + C \]

### Final Answer
Thus, the final answer for the integral is:
\[ I = \frac{1}{8(3 + 4\cos x)} - \frac{3}{16(3 + 4\cos x)^2} + C \]