## Calculus: Early Transcendentals 8th Edition

$$\int (x-1)\sin\pi xdx=\frac{(1-x)\cos\pi x}{\pi}+\frac{\sin(\pi x)}{\pi^2}+C$$
$$A=\int (x-1)\sin\pi xdx$$ We would choose $u=x-1$ and $dv=\sin\pi xdx$ For $u=x-1$, $du=dx$ For $dv=\sin\pi xdx$, we analyze as follows $$\int\sin\pi x dx=\int\frac{1}{\pi}\sin(\pi x)d(\pi x)=-\frac{1}{\pi}\cos(\pi x)+C$$ Therefore, for $dv=\sin\pi xdx$, $v=-\frac{1}{\pi}\cos\pi x$ Apply Integration by Parts to A, we have $$A=uv-\int vdu$$ $$A=-\frac{1}{\pi}(x-1)\cos\pi x-(-\frac{1}{\pi})\int\cos\pi xdx$$ $$A=\frac{-(x-1)\cos\pi x}{\pi}+\frac{1}{\pi}\int\frac{1}{\pi}\cos (\pi x)d(\pi x)$$ $$A=\frac{(1-x)\cos\pi x}{\pi}+\frac{1}{\pi^2}\int\cos(\pi x)d(\pi x)$$ $$A=\frac{(1-x)\cos\pi x}{\pi}+\frac{\sin(\pi x)}{\pi^2}+C$$