Answer
$\frac{1}{\pi}\sin(\pi x-1)-\frac{1}{3\pi}\sin^{3}(\pi x-1)+C$
Work Step by Step
$\int \cos^3(\pi x-1)dx$
$=\int \cos^2(\pi x-1)\cos(\pi x-1)dx$
$=\int (1-\sin^2(\pi x-1))(\cos(\pi x-1))dx$
$=\int (\cos(\pi x-1)dx -\int\sin^2(\pi x-1)\cos(\pi x-1)dx$
Use u-substitution for both parts (they have different u's)
Let $u= \pi x-1$
$\frac{du}{dx}= \pi$
$dx=\frac{1}{\pi}du$
Let $u=\sin(\pi x-1)$
$\frac{du}{dx}=\pi \cos(\pi x -1)$
$dx=\frac{1}{\pi \cos(\pi x -1)}du$
$=\int \frac{1}{\pi}\cos udu-\int \frac{1}{\pi}u^{2}du$
$=\frac{1}{\pi}\sin(\pi x-1)-\frac{1}{3\pi}\sin^{3}(\pi x-1)+C$
