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Integrate the function
$x \sin 3 x$
$x \sin 3 x$
Solution:
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Verified Answer
$\int x \sin 3 x d x=x\left(-\frac{\cos 3 x}{3}\right)-\int 1 \cdot\left(\frac{-\cos 3 x}{3}\right) d x$
$=-\frac{1}{3} x \cos 3 x+\frac{1}{9} \sin 3 x+\mathrm{C}$
$=-\frac{1}{3} x \cos 3 x+\frac{1}{9} \sin 3 x+\mathrm{C}$
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