Question

If \( \int x^3 \sin(3x) dx = \frac{1}{27} [f(x)\cos(3x) + g(x)\sin(3x)] + c \) then f(1)+g(1)=

• A. 14
• B. 6
• C. 4
• D. 12

Hint:

Tabular integration (the DI method) is extremely efficient for integrals of the form \(\int P(x)f(x)dx\), where P(x) is a polynomial and f(x) is a function that can be repeatedly integrated (like sin, cos, or e^x). Differentiate the polynomial column down to zero, integrate the other column, and sum the diagonal products with alternating signs.

Answer 1

The Correct Option is C

We use integration by parts, specifically the tabular method for integrals of the form \( \int x^n \sin(ax) dx \).
Let u = \(x^3\) and dv = \(\sin(3x)dx\). We create a table of derivatives of u and integrals of dv.
\begin{tabular}{|c|c|} \hline Derivatives of u & Integrals of dv
\hline \(x^3\) & \(\sin(3x)\)
\hline \(3x^2\) & \(-\frac{1}{3}\cos(3x)\)
\hline \(6x\) & \(-\frac{1}{9}\sin(3x)\)
\hline \(6\) & \(\frac{1}{27}\cos(3x)\)
\hline \(0\) & \(\frac{1}{81}\sin(3x)\)
\hline \end{tabular}
The integral is the sum of the products of the diagonal terms, with alternating signs (+, -, +, -).
\(\int x^3 \sin(3x) dx = (x^3)(-\frac{1}{3}\cos(3x)) - (3x^2)(-\frac{1}{9}\sin(3x)) + (6x)(\frac{1}{27}\cos(3x)) - (6)(\frac{1}{81}\sin(3x)) + c \).
Group the terms with \(\cos(3x)\) and \(\sin(3x)\).
\( = \left(-\frac{x^3}{3} + \frac{6x}{27}\right)\cos(3x) + \left(\frac{3x^2}{9} - \frac{6}{81}\right)\sin(3x) + c \).
\( = \left(-\frac{x^3}{3} + \frac{2x}{9}\right)\cos(3x) + \left(\frac{x^2}{3} - \frac{2}{27}\right)\sin(3x) + c \).
The given form is \( \frac{1}{27} [f(x)\cos(3x) + g(x)\sin(3x)] + c \). Let's factor out \(1/27\).
\( = \frac{1}{27} \left( 27(-\frac{x^3}{3} + \frac{2x}{9}) \cos(3x) + 27(\frac{x^2}{3} - \frac{2}{27}) \sin(3x) \right) + c \).
\( = \frac{1}{27} \left( (-9x^3 + 6x) \cos(3x) + (9x^2 - 2) \sin(3x) \right) + c \).
By comparing with the given form, we have:
\( f(x) = -9x^3 + 6x \) and \( g(x) = 9x^2 - 2 \).
We need to find f(1) + g(1).
\( f(1) = -9(1)^3 + 6(1) = -9+6 = -3 \).
\( g(1) = 9(1)^2 - 2 = 9-2 = 7 \).
\( f(1) + g(1) = -3 + 7 = 4 \).