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\( \int x^{3} \sin 3 x d x= \)
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Verified Answer
The correct answer is:
\( -\frac{x^{3} \cos 3 x}{3}+\frac{x^{2} \sin 3 x}{3}+\frac{2 x \cos 3 x}{9}-\frac{2 \sin 3 x}{27}+C \)
(D)
\[
\begin{array}{l}
\int x^{3} \sin 3 x d x=x^{3}\left[\frac{-\cos 3 x}{3}\right]+3 x^{2}\left[\frac{-\sin 3 x}{9}\right]+6 x\left[\frac{\cos 3 x}{27}\right]-6\left[\frac{\sin 3 x}{81}\right]+C \\
=-\frac{x^{3} \cos 3 x}{3}+\frac{x^{2} \sin 3 x}{3}+\frac{2 x \cos 3 x}{9}-\frac{2 \sin 3 x}{27}+C
\end{array}
\]
\[
\begin{array}{l}
\int x^{3} \sin 3 x d x=x^{3}\left[\frac{-\cos 3 x}{3}\right]+3 x^{2}\left[\frac{-\sin 3 x}{9}\right]+6 x\left[\frac{\cos 3 x}{27}\right]-6\left[\frac{\sin 3 x}{81}\right]+C \\
=-\frac{x^{3} \cos 3 x}{3}+\frac{x^{2} \sin 3 x}{3}+\frac{2 x \cos 3 x}{9}-\frac{2 \sin 3 x}{27}+C
\end{array}
\]
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