माना I = $\int$x sin 3x dx x को पहला फलन तथा sin 3x को दूसरा फलन लेकर खण्डशः समाकलन करने पर, I = x $∫$sin 3x dx $-\int\left[\frac{d}{d x}(x) \int \sin 3 x d x\right] d x $ $=\frac{-x \cos 3 x}{3}+\int \frac{\cos 3 x}{3} d x$ ($\because$ $\int$sin ax dx = $\frac{-cos a x}{a}$) $\Rightarrow I=\frac{-x \cos 3 x}{3}$ $+\frac{1}{9} \sin 3 x+C$
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