Evaluate the integral.
-$$\int x ^ { 3 } \cos 3 x d x$$
A) $\frac { 1 } { 3 } x ^ { 3 } \cos 3 x + \frac { 1 } { 3 } x ^ { 2 } \sin 3 x - \frac { 2 } { 9 } x \cos 3 x - \frac { 2 } { 27 } \sin 3 x + C$
B) $\frac { 1 } { 3 } x ^ { 3 } \sin 3 x - \frac { 1 } { 3 } x ^ { 2 } \cos 3 x + \frac { 2 } { 9 } x \sin 3 x + \frac { 2 } { 27 } \cos 3 x + C$
C) $\frac { 1 } { 3 } x ^ { 3 } \sin 3 x + 1 x ^ { 2 } \cos 3 x - 2 x \sin 3 x - 2 \cos 3 x + C$
D) $\frac { 1 } { 3 } x ^ { 3 } \sin 3 x + \frac { 1 } { 3 } x ^ { 2 } \cos 3 x - \frac { 2 } { 9 } x \sin 3 x - \frac { 2 } { 27 } \cos 3 x + C$

Question

Quizplus · Accepted Answer

Integrate by parts: $u = x^3$, $du = 3x^2 dx$, $dv = \cos(3x) dx$, $v = \frac{1}{3}\sin(3x)$. Then we have 
$$\int x^3 \cos(3x) dx = \frac{1}{3}x^3\sin(3x) - \frac{1}{3}\int 3x^2 \sin(3x) dx.$$
Integrating by parts again with $u = x^2$, $du = 2x dx$, $dv = \sin(3x) dx$, $v = -\frac{1}{3}\cos(3x)$, we have 
\begin{align*}
\int x^3 \cos(3x) dx &= \frac{1}{3}x^3\sin(3x) + \frac{1}{9}x^2\cos(3x) - \frac{2}{27}\int x \cos(3x) dx\
&= \frac{1}{3}x^3\sin(3x) + \frac{1}{9}x^2\cos(3x) - \frac{2}{27} \left(\frac{1}{3}x\sin(3x) + \frac{1}{9}\cos(3x)\right) + C\
&= \frac{1}{3}x^3\sin(3x) + \frac{1}{9}x^2\cos(3x) - \frac{2}{81}x\sin(3x) - \frac{2}{243}\cos(3x) + C.
\end{align*} Thus our answer is $\boxed{\textbf{(D)}}$.

Evaluate the Integral ∫x3cos⁡3xdx\int x ^ { 3 } \cos 3 x d x∫x3cos3xdx

Evaluate the Integral $\int x ^ { 3 } \cos 3 x d x$