# Q. $\int x^{3} \sin 3x dx =$

KCET KCET 2019Integrals

Solution:

## $\int x^{3} sin 3x dx=x^{3}\left[\frac{-cos 3x}{3}\right]-3x^{2}\left[\frac{-sin 3x}{9}\right]+6x\left[\frac{cos 3x}{27}\right]-6\left[\frac{sin 3x}{81}\right]+C$ $=-\frac{x^{3}cos 3x}{3}+\frac{x^{2} sin 3x}{3}+\frac{2x cos 3x}{9}-\frac{2 sin 3x}{27}+C$

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