`int(sinx)/(3+4cos^(2)x)dx `

Explanation

To solve the integral \( \int \frac{\sin x}{3 + 4 \cos^2 x} \, dx \), we will use the substitution method. Here’s a step-by-step solution:

### Step 1: Set up the integral
Let
\[ I = \int \frac{\sin x}{3 + 4 \cos^2 x} \, dx \]

### Step 2: Use substitution
We will use the substitution \( \cos x = t \). Then, the derivative of \( \cos x \) gives us:
\[ -\sin x \, dx = dt \quad \Rightarrow \quad \sin x \, dx = -dt \]

### Step 3: Substitute in the integral
Substituting \( \cos x = t \) and \( \sin x \, dx = -dt \) into the integral, we have:
\[ I = \int \frac{-dt}{3 + 4t^2} \]
This can be rewritten as:
\[ I = -\int \frac{dt}{3 + 4t^2} \]

### Step 4: Factor out constants
We can factor out the constant from the denominator:
\[ I = -\int \frac{dt}{4\left(\frac{3}{4} + t^2\right)} = -\frac{1}{4} \int \frac{dt}{\frac{3}{4} + t^2} \]

### Step 5: Rewrite the integral in standard form
Recognizing that \( \frac{3}{4} = \left(\frac{\sqrt{3}}{2}\right)^2 \), we rewrite the integral:
\[ I = -\frac{1}{4} \int \frac{dt}{\left(\frac{\sqrt{3}}{2}\right)^2 + t^2} \]

### Step 6: Use the standard integral formula
The integral \( \int \frac{dt}{a^2 + t^2} = \frac{1}{a} \tan^{-1} \left( \frac{t}{a} \right) + C \) applies here, where \( a = \frac{\sqrt{3}}{2} \):
\[ I = -\frac{1}{4} \cdot \frac{1}{\frac{\sqrt{3}}{2}} \tan^{-1} \left( \frac{t}{\frac{\sqrt{3}}{2}} \right) + C \]
This simplifies to:
\[ I = -\frac{1}{2\sqrt{3}} \tan^{-1} \left( \frac{2t}{\sqrt{3}} \right) + C \]

### Step 7: Substitute back for \( t \)
Recall that \( t = \cos x \):
\[ I = -\frac{1}{2\sqrt{3}} \tan^{-1} \left( \frac{2\cos x}{\sqrt{3}} \right) + C \]

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