To solve \( \int x^3 \sin(3x) \, dx = \frac{1}{27} [f(x)\cos(3x) + g(x)\sin(3x)] + c \) and find \( f(1) + g(1) \), we can apply the method of integration by parts, which involves using the formula:
\(\int u \, dv = uv - \int v \, du\)
In our case, choose \( u = x^3 \) and \( dv = \sin(3x)\,dx \). This gives us:
Applying integration by parts, we get:
\(\int x^3 \sin(3x) \, dx = -\frac{1}{3} x^3 \cos(3x) + \int \frac{1}{3} \cdot 3x^2 \cos(3x) \, dx\)
This simplifies to:
\(-\frac{1}{3} x^3 \cos(3x) + \int x^2 \cos(3x) \, dx\)
We apply integration by parts again to \(\int x^2 \cos(3x) \, dx\), choosing \( u = x^2 \) and \( dv = \cos(3x) \, dx \).
Therefore:
\(\int x^2 \cos(3x) \, dx = \frac{1}{3} x^2 \sin(3x) - \int \frac{1}{3} \cdot 2x \sin(3x) \, dx\)
Which simplifies to:
\(\frac{1}{3} x^2 \sin(3x) - \int \frac{2}{3} x \sin(3x) \, dx\)
Applying integration by parts one more time to \(\int x \sin(3x) \, dx\), choose \( u = x \) and \( dv = \sin(3x) \, dx \).
Applying the integration by parts formula:
\(\int x \sin(3x) \, dx = -\frac{1}{3} x \cos(3x) + \int \frac{1}{3} \cos(3x) \, dx\)
Simplifying gives:
\(-\frac{1}{3} x \cos(3x) + \frac{1}{9} \sin(3x)\)
Collecting all terms and simplifying, the integral is:
\(f(x) = 9x^2 - 6, \quad g(x) = 6x - x^3\)
Therefore, substituting \( x = 1 \) we have:
\(f(1) = 9(1)^2 - 6 = 3\)
\(g(1) = 6(1) - (1)^3 = 5\)
So, \( f(1) + g(1) = 3 + 5 = 8 \).
Therefore, the correct answer is not as provided in the options. Please check if there is an error in the provided solutions or re-evaluate choices.