Q106 / 180 · KCET 2019 · Mathematics
\( \int x^{3} \sin 3 x d x= \)
- A\( \frac{x^{3} \cos 3 x}{3}+\frac{x^{2} \sin 3 x}{3}-\frac{2 x \cos 3 x}{9}-\frac{2 \sin 3 x}{27}+C \)
- B\( -\frac{x^{3} \cos 3 x}{3}+\frac{x^{2} \sin 3 x}{3}-\frac{2 x \cos 3 x}{9}-\frac{2 \sin 3 x}{27}+C \)
- C\( -\frac{x^{3} \cos 3 x}{3}-\frac{x^{2} \sin 3 x}{3}+\frac{2 x \cos 3 x}{9}-\frac{2 \sin 3 x}{27}+C \)
- Correct answerD\( -\frac{x^{3} \cos 3 x}{3}+\frac{x^{2} \sin 3 x}{3}+\frac{2 x \cos 3 x}{9}-\frac{2 \sin 3 x}{27}+C \)
Solution
(D)
\[ \begin{array}{l} \int x^{3} \sin 3 x d x=x^{3}\left[\frac{-\cos 3 x}{3}\right]+3 x^{2}\left[\frac{-\sin 3 x}{9}\right]+6 x\left[\frac{\cos 3 x}{27}\right]-6\left[\frac{\sin 3 x}{81}\right]+C \\ =-\frac{x^{3} \cos 3 x}{3}+\frac{x^{2} \sin 3 x}{3}+\frac{2 x \cos 3 x}{9}-\frac{2 \sin 3 x}{27}+C \end{array} \]
\[ \begin{array}{l} \int x^{3} \sin 3 x d x=x^{3}\left[\frac{-\cos 3 x}{3}\right]+3 x^{2}\left[\frac{-\sin 3 x}{9}\right]+6 x\left[\frac{\cos 3 x}{27}\right]-6\left[\frac{\sin 3 x}{81}\right]+C \\ =-\frac{x^{3} \cos 3 x}{3}+\frac{x^{2} \sin 3 x}{3}+\frac{2 x \cos 3 x}{9}-\frac{2 \sin 3 x}{27}+C \end{array} \]
Hence, the correct answer is option: 'D'.