Here's the integration by parts formula \int f(x) g'(x) dx = f(x)g(x) - \int f'(x) g(x) dx.
I would suggest not relying on mnemonics such as the "LIATE rule", and instead think about what's going on in the formula. Note that on the left hand side of the above formula, you need to find an antiderivative of f(x)g'(x) while on the right hand side, you need to find an antiderivative of f'(x) g(x).
Let's look at your integral now. It might be best to find the indefinite integral first as it reduces chance of making errors, and then compute the definite integral.
When using integration by parts to evaluate $ \int x\sin(\pi x)\,dx, $
you essentially say
"well, finding an antiderivative of $x\sin(\pi x)$ is hard. But if I differentiate one part and antidifferentiate the other, could I deal with what results (that is, would the integration by parts formula prove useful)?"
Let's see. There are two choices:
\ \ \underbrace{x\vphantom{(}}_f \underbrace{\sin(\pi x)}_{g'} \longrightarrow \underbrace{\vphantom{(}1}_{f'}\cdot\underbrace{{-\cos(\pi x)\over\pi}}_g
and
\ \ \underbrace{ x}_{g'} \underbrace{\sin(\pi x)}_f \longrightarrow \underbrace{{x^2\over2}}_g \cdot\underbrace{\pi \cos(\pi x)}_{f'}
The former option is the best, we can integrate ${-\cos(\pi x)\over\pi} $.
So, using the integration by parts formula with $f(x)=x$ and g'(x)=\sin(\pi x):
\eqalign{ \int \underbrace{x\vphantom{( }}_f \,\underbrace{\sin(\pi x)}_{g'} &= \underbrace{x\vphantom{(1\over2}}_{f} \underbrace{-\cos (\pi x)\over \pi}_{g} -\int \underbrace{1\vphantom{(1\over2}}_{f'} \cdot\underbrace{-\cos (\pi x)\over \pi}_{g}\, dx\cr &={-x\cos(\pi x)\over\pi}+{1\over\pi}\cdot{1\over\pi}\sin(\pi x)+C. } So $ \int_0^1 x\sin(\pi x)\, dx= \Bigl[{-x\cos(\pi x)\over\pi}+{1\over\pi}\cdot{1\over\pi}\sin(\pi x)\Bigr]_{x=0}^{x=1} ={1\over\pi}-{0}={1\over\pi}. $
Note: either I made a mistake somewhere, or for your problem, you'd say "none of the above".