`int(1)/(3+4 sin x)dx`

Answer

To solve the integral \( \int \frac{1}{3 + 4 \sin x} \, dx \), we will follow a systematic approach. Here’s the step-by-step solution:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \frac{1}{3 + 4 \sin x} \, dx \]

### Step 2: Use the Identity for \(\sin x\)
We can use the identity for \(\sin x\):
\[ \sin x = \frac{2 \tan \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} \]
Let \( t = \tan \frac{x}{2} \). Then, we have:
\[ \sin x = \frac{2t}{1 + t^2} \]
Substituting this into the integral gives:
\[ I = \int \frac{1}{3 + 4 \left(\frac{2t}{1 + t^2}\right)} \, dx \]

### Step 3: Simplify the Integral
The integral becomes:
\[ I = \int \frac{1 + t^2}{3(1 + t^2) + 8t} \, dx \]
Now, we need to express \(dx\) in terms of \(dt\). The relationship between \(dx\) and \(dt\) is given by:
\[ dx = \frac{2}{1 + t^2} \, dt \]

### Step 4: Substitute \(dx\) in the Integral
Substituting \(dx\) into the integral:
\[ I = \int \frac{1 + t^2}{3(1 + t^2) + 8t} \cdot \frac{2}{1 + t^2} \, dt \]
This simplifies to:
\[ I = 2 \int \frac{1}{3 + 8t + 3t^2} \, dt \]

### Step 5: Complete the Square
Next, we need to complete the square in the denominator:
\[ 3t^2 + 8t + 3 = 3\left(t^2 + \frac{8}{3}t + 1\right) \]
Completing the square:
\[ t^2 + \frac{8}{3}t + 1 = \left(t + \frac{4}{3}\right)^2 - \frac{16}{9} + 1 = \left(t + \frac{4}{3}\right)^2 - \frac{7}{9} \]
Thus, we rewrite the integral:
\[ I = \frac{2}{3} \int \frac{1}{\left(t + \frac{4}{3}\right)^2 - \left(\frac{\sqrt{7}}{3}\right)^2} \, dt \]

### Step 6: Use the Integral Formula
This integral can be solved using the formula:
\[ \int \frac{1}{x^2 - a^2} \, dx = \frac{1}{2a} \ln \left| \frac{x - a}{x + a} \right| + C \]
Applying this:
\[ I = \frac{2}{3} \cdot \frac{1}{2 \cdot \frac{\sqrt{7}}{3}} \ln \left| \frac{t + \frac{4}{3} - \frac{\sqrt{7}}{3}}{t + \frac{4}{3} + \frac{\sqrt{7}}{3}} \right| + C \]
Simplifying gives:
\[ I = \frac{1}{\sqrt{7}} \ln \left| \frac{t + \frac{4 - \sqrt{7}}{3}}{t + \frac{4 + \sqrt{7}}{3}} \right| + C \]

### Step 7: Substitute Back for \(t\)
Recall that \( t = \tan \frac{x}{2} \):
\[ I = \frac{1}{\sqrt{7}} \ln \left| \frac{\tan \frac{x}{2} + \frac{4 - \sqrt{7}}{3}}{\tan \frac{x}{2} + \frac{4 + \sqrt{7}}{3}} \right| + C \]

### Final Answer
Thus, the final result for the integral is:
\[ \int \frac{1}{3 + 4 \sin x} \, dx = \frac{1}{\sqrt{7}} \ln \left| \frac{\tan \frac{x}{2} + \frac{4 - \sqrt{7}}{3}}{\tan \frac{x}{2} + \frac{4 + \sqrt{7}}{3}} \right| + C \]