Question

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$\int\limits {(x-1)sin3x} \, dx$

Answer 1

 $u=x-1 \\ dv=sin3x \ dx \\ du=dx \\ v= -\frac{cos3x}{3} \\\\ \int\limits { (x-1) sin3x } \, dx = \frac{(1-x)cos3x}{3} - \int\limits { \frac{-cos3x} {3}} \, dx \\\\ \int\limits { -\frac{cos3x}{3}} \, dx \\ 3x=t\\ 3dx=dt\\\\ \int\limits { - \frac{cost}{9}} \, dt = -\frac{sint}{9} = - \frac{ sin3x}{9}\\\\ \\ \frac{(1-x)cos3x}{3} + \frac{sin3x}{9 } = \frac{ (3-3x)cos3x + sin3x}{9}$