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Q2  Integrate the functions x \sin 3x

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Given function is
f(x)=x \sin 3x
We will use integration by parts method
\int x\sin 3x = x.\int \sin 3xdx - \int(\frac{d(x)}{dx}.\int sin 3x dx)dx\\ \\ \int x\sin 3x = x.(\frac{-\cos 3x}{3})- \int (1.(\frac{-\cos 3x}{3}))dx\\ \\ \int x\sin 3x= -\frac{x\cos 3x}{3}+\frac{\sin 3x}{9} + C

Therefore, the answer is -\frac{x\cos 3x}{3}+\frac{\sin 3x}{9} + C
 

Posted by

Gautam harsolia

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