What could be the answer for this $\int \limits_{-1}^{1}\big(x\sin{\pi}x\big)dx$?? $\displaystyle\int \limits_{-1}^{1}\big(x\sin{\pi}x\big)dx $?
a) 2 
b) -2
c)1
d)0
I've reduced it to
$=2\displaystyle\int \limits_{0}^{1}\big(x\sin{\pi}x\big)dx$ as it is an even function.
Do I have to apply LIATE rule from here?
Update 1: After applying integration by part..
$$ = 2\left[ {x\int\limits_0^1 {\sin } \pi xdx - \int\limits_0^1 {\left( {\frac{d}{{dx}}x\int\limits_0^1 {\sin } \pi xdx} \right)} dx} \right]$$
 A: 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".
A: That you have reduced your integral to what you have given is a good job. I'll illustrate a technique, you often will have to use.
Set $I$ equals to the integral at hand. (i.e.)
$$I=\int_0^1{x \sin(\pi x)}$$ 
Now, I leave it to you (You can ask me how to prove this after some try!) to prove, 

$$\int_a^b{f(x)dx}=\int_a^b{f(a+b-x)dx}$$

Now what would this property mean for this integral?
This means, 
$$I=\int_0^1(1-x)\sin[\pi(1-x)]dx$$
Now, simplifying, you'll have that $$I=\int_0^1(1-x)\sin(\pi-\pi x)dx$$
That $\sin(\pi-y)=\sin y$, gives you, $$I=\int_0^1\sin(\pi x) dx-\int_0^1x\sin(\pi x) dx$$ $$2I=\dfrac{2}{\pi}$$ giving you $\boxed{I=\dfrac{1}{\pi}}$. 
So, the answer to your question is $\dfrac{2}{\pi}$. which is your (e) None of these =)
