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

Explanation

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

### Step 1: Substitution
Let \( t = \cos x \). Then, the derivative \( dt = -\sin x \, dx \) or \( \sin x \, dx = -dt \).

### Step 2: Rewrite the Integral
Substituting \( t \) into the integral, we have:
\[ \int \frac{\sin x}{3 + 4 \cos^2 x} \, dx = \int \frac{-dt}{3 + 4t^2} \]

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

### Step 4: Recognize the Integral Form
The integral \( \int \frac{dt}{a^2 + t^2} \) has the solution \( \frac{1}{a} \tan^{-1} \left( \frac{t}{a} \right) + C \). Here, \( a^2 = \frac{3}{4} \) implies \( a = \frac{\sqrt{3}}{2} \).

### Step 5: Solve the Integral
Using the formula:
\[ -\frac{1}{4} \int \frac{dt}{\frac{3}{4} + t^2} = -\frac{1}{4} \cdot \frac{1}{\frac{\sqrt{3}}{2}} \tan^{-1} \left( \frac{t}{\frac{\sqrt{3}}{2}} \right) + C \]
This simplifies to:
\[ = -\frac{1}{2\sqrt{3}} \tan^{-1} \left( \frac{2t}{\sqrt{3}} \right) + C \]

### Step 6: Substitute Back
Now, substituting back \( t = \cos x \):
\[ = -\frac{1}{2\sqrt{3}} \tan^{-1} \left( \frac{2\cos x}{\sqrt{3}} \right) + C \]

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

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