Integrals — Class 12 Maths Solution

Let $I = \int {x\sin 3xdx}$
$= x\left( { - \cfrac{{\cos 3x}}{3}} \right) - \int {\left[ {\cfrac{d}{{dx}}\left( x \right)\left( {\cfrac{{ - \cos 3x}}{3}} \right)} \right]} dx$

$= - \cfrac{{x\cos 3x}}{3} + \cfrac{1}{3}\int {\cos 3xdx} = - \cfrac{{x\cos 3x}}{3} + \cfrac{1}{3} \cdot \cfrac{{\sin 3x}}{3} + C$

$= - \cfrac{1}{3}x\cos 3x + \cfrac{1}{9}\sin 3x + C$