To solve the integral \( \int \frac{\sin x}{\sin^3 x + \cos^3 x} \, dx \), we will follow a series of steps to simplify the expression and find the coefficients \( \alpha \), \( \beta \), and \( \gamma \).
### Step 1: Simplify the Integral
We start with the integral:
\[
\int \frac{\sin x}{\sin^3 x + \cos^3 x} \, dx
\]
We can factor the denominator using the identity for the sum of cubes:
\[
\sin^3 x + \cos^3 x = (\sin x + \cos x)(\sin^2 x - \sin x \cos x + \cos^2 x)
\]
Since \( \sin^2 x + \cos^2 x = 1 \), we have:
\[
\sin^3 x + \cos^3 x = (\sin x + \cos x)(1 - \sin x \cos x)
\]
### Step 2: Rewrite the Integral
Now we can rewrite the integral:
\[
\int \frac{\sin x}{(\sin x + \cos x)(1 - \sin x \cos x)} \, dx
\]
Next, we divide both the numerator and denominator by \( \cos^3 x \):
\[
= \int \frac{\tan x}{\tan^3 x + 1} \sec^2 x \, dx
\]
Let \( t = \tan x \), then \( dt = \sec^2 x \, dx \):
\[
= \int \frac{t}{t^3 + 1} \, dt
\]
### Step 3: Partial Fraction Decomposition
We can decompose \( \frac{t}{t^3 + 1} \):
\[
t^3 + 1 = (t + 1)(t^2 - t + 1)
\]
Thus,
\[
\frac{t}{t^3 + 1} = \frac{A}{t + 1} + \frac{Bt + C}{t^2 - t + 1}
\]
Multiplying through by \( t^3 + 1 \) and equating coefficients will allow us to find \( A \), \( B \), and \( C \).
### Step 4: Solve for Coefficients
From the equation:
\[
t = A(t^2 - t + 1) + (Bt + C)(t + 1)
\]
We can expand and collect like terms to find the coefficients \( A \), \( B \), and \( C \).
1. Coefficient of \( t^2 \): \( A + B = 0 \)
2. Coefficient of \( t \): \( -A + B + C = 1 \)
3. Constant term: \( A + C = 0 \)
Solving these equations gives us:
- From \( A + B = 0 \), we have \( B = -A \).
- Substituting \( B \) into the second equation gives us \( -A - A + C = 1 \) or \( C = 2A + 1 \).
- Substituting \( C \) into the third equation gives \( A + (2A + 1) = 0 \) leading to \( 3A + 1 = 0 \) or \( A = -\frac{1}{3} \).
Thus:
- \( A = -\frac{1}{3} \)
- \( B = \frac{1}{3} \)
- \( C = \frac{1}{3} \)
### Step 5: Integrate Each Term
Now we can integrate:
\[
\int \left( -\frac{1}{3} \frac{1}{t + 1} + \frac{1/3 t + 1/3}{t^2 - t + 1} \right) dt
\]
This results in:
\[
-\frac{1}{3} \ln |t + 1| + \frac{1}{3} \ln |t^2 - t + 1| + \frac{1}{3} \tan^{-1}(2t - 1)
\]
### Step 6: Substitute Back
Substituting back \( t = \tan x \):
\[
-\frac{1}{3} \ln |1 + \tan x| + \frac{1}{3} \ln |1 - \tan x + \tan^2 x| + \frac{1}{3} \tan^{-1}\left(\frac{2\tan x - 1}{\sqrt{3}}\right) + C
\]
### Step 7: Identify Coefficients
From the integral, we identify:
- \( \alpha = -\frac{1}{3} \)
- \( \beta = \frac{1}{3} \)
- \( \gamma = \frac{1}{\sqrt{3}} \)
### Step 8: Calculate \( 18(\alpha + \beta + \gamma^2) \)
Now we compute:
\[
\alpha + \beta + \gamma^2 = -\frac{1}{3} + \frac{1}{3} + \left(\frac{1}{\sqrt{3}}\right)^2 = 0 + \frac{1}{3} = \frac{1}{3}
\]
Thus,
\[
18(\alpha + \beta + \gamma^2) = 18 \cdot \frac{1}{3} = 6
\]
### Final Answer
The final answer is:
\[
\boxed{6}
\]
