Key Concepts and Formulas

• Partial Fraction Decomposition: Breaking down a rational function into simpler fractions.
• Integration Formulas:
  • $\int \frac{1}{x} dx = \ln|x| + C$
  • $\int \frac{f'(x)}{f(x)} dx = \ln|f(x)| + C$
  • $\int \frac{1}{x^2 + a^2} dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C$
• Trigonometric Identities:
  • $\tan x = \frac{\sin x}{\cos x}$
  • $\sec x = \frac{1}{\cos x}$

Step-by-Step Solution

Step 1: Rewrite the integral in terms of $\tan x$ and $\sec^2 x$.

We are given the integral $I = \int \frac{\sin x}{\sin^3 x + \cos^3 x} dx$ Divide both the numerator and denominator by $\cos^3 x$: $I = \int \frac{\frac{\sin x}{\cos^3 x}}{\frac{\sin^3 x}{\cos^3 x} + \frac{\cos^3 x}{\cos^3 x}} dx = \int \frac{\tan x \sec^2 x}{\tan^3 x + 1} dx$ This step is done to simplify the integral and make it easier to substitute.

Step 2: Substitute $t = \tan x$.

Let $t = \tan x$. Then, $dt = \sec^2 x \, dx$. Substituting this into the integral, we get: $I = \int \frac{t}{t^3 + 1} dt$ This simplifies the integral into a rational function of $t$.

Step 3: Factor the denominator and perform partial fraction decomposition.

We can factor the denominator as $t^3 + 1 = (t+1)(t^2 - t + 1)$. Now, we decompose the integrand into partial fractions: $\frac{t}{t^3 + 1} = \frac{t}{(t+1)(t^2 - t + 1)} = \frac{A}{t+1} + \frac{Bt + C}{t^2 - t + 1}$ Multiplying both sides by $(t+1)(t^2 - t + 1)$, we get: $t = A(t^2 - t + 1) + (Bt + C)(t+1) = A(t^2 - t + 1) + B(t^2 + t) + C(t+1)$ $t = (A+B)t^2 + (-A+B+C)t + (A+C)$ Comparing coefficients, we have the following system of equations: \begin{align*} A+B &= 0 \ -A+B+C &= 1 \ A+C &= 0 \end{align*} From the first equation, $B = -A$. From the third equation, $C = -A$. Substituting into the second equation, we get: $-A - A - A = 1 \Rightarrow -3A = 1 \Rightarrow A = -\frac{1}{3}$ Thus, $B = \frac{1}{3}$ and $C = \frac{1}{3}$. So, the integral becomes: $I = \int \left( \frac{-1/3}{t+1} + \frac{(1/3)t + (1/3)}{t^2 - t + 1} \right) dt = -\frac{1}{3} \int \frac{1}{t+1} dt + \frac{1}{3} \int \frac{t+1}{t^2 - t + 1} dt$

Step 4: Rewrite the second integral.

We want to rewrite the numerator of the second integral to be in the form of the derivative of the denominator plus a constant. The derivative of $t^2 - t + 1$ is $2t - 1$. Thus, we want to rewrite $t+1$ as $\frac{1}{2}(2t - 1) + D$. Then $t + 1 = t - \frac{1}{2} + D$, so $D = \frac{3}{2}$. Then $\frac{t+1}{t^2 - t + 1} = \frac{\frac{1}{2}(2t-1) + \frac{3}{2}}{t^2 - t + 1} = \frac{1}{2} \frac{2t-1}{t^2 - t + 1} + \frac{3}{2} \frac{1}{t^2 - t + 1}$ Thus $I = -\frac{1}{3} \int \frac{1}{t+1} dt + \frac{1}{3} \int \left(\frac{1}{2} \frac{2t-1}{t^2 - t + 1} + \frac{3}{2} \frac{1}{t^2 - t + 1}\right) dt$ $I = -\frac{1}{3} \int \frac{1}{t+1} dt + \frac{1}{6} \int \frac{2t-1}{t^2 - t + 1} dt + \frac{1}{2} \int \frac{1}{t^2 - t + 1} dt$

Step 5: Complete the square in the last integral and evaluate the integrals.

We have $t^2 - t + 1 = (t - \frac{1}{2})^2 + \frac{3}{4} = (t - \frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2$. Thus $\int \frac{1}{t^2 - t + 1} dt = \int \frac{1}{(t - \frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2} dt = \frac{1}{\frac{\sqrt{3}}{2}} \tan^{-1}\left(\frac{t - \frac{1}{2}}{\frac{\sqrt{3}}{2}}\right) + C = \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2t - 1}{\sqrt{3}}\right) + C$ So $I = -\frac{1}{3} \ln|t+1| + \frac{1}{6} \ln|t^2 - t + 1| + \frac{1}{2} \cdot \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2t-1}{\sqrt{3}}\right) + C$ $I = -\frac{1}{3} \ln|t+1| + \frac{1}{6} \ln|t^2 - t + 1| + \frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{2t-1}{\sqrt{3}}\right) + C$

Step 6: Substitute back $t = \tan x$.

Substituting $t = \tan x$, we get: $I = -\frac{1}{3} \ln|\tan x + 1| + \frac{1}{6} \ln|\tan^2 x - \tan x + 1| + \frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{2\tan x - 1}{\sqrt{3}}\right) + C$

Step 7: Identify $\alpha, \beta, \gamma$.

Comparing this with the given form $\alpha \ln|1 + \tan x| + \beta \ln|1 - \tan x + \tan^2 x| + \gamma \tan^{-1}\left(\frac{2\tan x - 1}{\sqrt{3}}\right) + C$ we have $\alpha = -\frac{1}{3}$, $\beta = \frac{1}{6}$, and $\gamma = \frac{1}{\sqrt{3}}$.

Step 8: Calculate $18(\alpha + \beta + \gamma^2)$.

We need to find the value of $18(\alpha + \beta + \gamma^2)$. $18\left(-\frac{1}{3} + \frac{1}{6} + \left(\frac{1}{\sqrt{3}}\right)^2\right) = 18\left(-\frac{1}{3} + \frac{1}{6} + \frac{1}{3}\right) = 18\left(\frac{1}{6}\right) = 3$

Common Mistakes & Tips

• Sign Errors: Be careful with the signs during partial fraction decomposition.
• Completing the Square: Make sure to complete the square correctly to get the correct form for the arctangent integral.
• Checking the Answer: After finding the values of $\alpha, \beta, \gamma$, substitute them back into the original expression to ensure the answer matches.

Summary

We started by converting the integral into a form involving $\tan x$. Then, we used partial fraction decomposition to break down the integrand into simpler terms. We completed the square to evaluate the integral involving the quadratic term. Finally, we identified the values of $\alpha, \beta,$ and $\gamma$ and calculated the expression $18(\alpha + \beta + \gamma^2)$, which resulted in 3.

Final Answer

The final answer is \boxed{3}.