`int(sin x)/(3+4 cos^(2)x)dx =` …………………

Answer

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

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \frac{\sin x}{3 + 4 \cos^2 x} \, dx \]
We can rewrite the denominator:
\[ 3 + 4 \cos^2 x = \sqrt{3}^2 + (2 \cos x)^2 \]

### Step 2: Substitution
Let \( t = 2 \cos x \). Then, we differentiate \( t \) with respect to \( x \):
\[ dt = -2 \sin x \, dx \quad \Rightarrow \quad \sin x \, dx = -\frac{1}{2} dt \]

### Step 3: Substitute into the Integral
Now we substitute \( \sin x \, dx \) and \( t \) into the integral:
\[ I = \int \frac{-\frac{1}{2} dt}{3 + t^2} \]
This simplifies to:
\[ I = -\frac{1}{2} \int \frac{dt}{3 + t^2} \]

### Step 4: Use the Integral Formula
We know the formula for the integral:
\[ \int \frac{1}{a^2 + x^2} \, dx = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + C \]
In our case, \( a^2 = 3 \) (so \( a = \sqrt{3} \)), and \( x = t \):
\[ I = -\frac{1}{2} \cdot \frac{1}{\sqrt{3}} \tan^{-1} \left( \frac{t}{\sqrt{3}} \right) + C \]

### Step 5: Substitute Back for \( t \)
Now we substitute back \( t = 2 \cos x \):
\[ I = -\frac{1}{2\sqrt{3}} \tan^{-1} \left( \frac{2 \cos x}{\sqrt{3}} \right) + C \]

### Final Answer
Thus, the final result of the integral is:
\[ \int \frac{\sin x}{3 + 4 \cos^2 x} \, dx = -\frac{1}{2\sqrt{3}} \tan^{-1} \left( \frac{2 \cos x}{\sqrt{3}} \right) + C \]