Chapter 7 - Techniques of Integration - 7.1 Integration by Parts - 7.1 Exercises - Page 516: 6

$-\frac{1}{\pi}(x-1) cos \pi x + \frac{1}{\pi^2} sin \pi x +C$

Let $u = x-1, dv = sin \pi x dx$ and $du = dx, v = -\frac{1}{\pi} cos \pi x$ then $\int (x-1) sin \pi x dx$ $= -\frac{1}{\pi}(x-1) cos \pi x - \int -\frac{1}{\pi} cos \pi x dx$ $= -\frac{1}{\pi} (x-1) cos \pi x + \frac{1}{\pi} \int cos \pi x dx$ $=-\frac{1}{\pi}(x-1) cos \pi x + \frac{1}{\pi^2} sin \pi x +C$