Need help identifying what property of these integrals make them 0 The integral in question is this one:
$$
A_n=\int_0^1{\sin(n\pi x)\big[\sin(6 \pi x)-\sin(\pi x)\big]\,\mathrm dx}
$$
I had a feeling that this integral would not have non-zero answers for all $n$, so I checked and it turns out that it only has non-zero answers for $n=1,6$. My main question is really just how do I explain why this is the case? I can see clearly when I plot the function that the area underneath the curve sums to $0$ for n $\neq$ 1,6, however explaining that "by looking at the plots..." is most likely not a sufficient answer for this professor. Taking the n=1 case, the integral is: 
$$
A_1=\int_0^1{\sin(\pi x)\big[\sin(6 \pi x)-\sin(\pi x)\big]\,\mathrm dx}
$$
Breaking this up into the difference of two integrals, the integral over $\sin(\pi x)\sin(6 \pi x)$ evaluates to $0$ for a similar reason. I cannot, however, figure out what it is that I am trying to say. I feel like there is some underlying trigonometric property or concept that I am totally spacing on. It seems to have to do with the product of two sine functions of different periods/frequencies that makes the integrals of said functions evaluate to $0$. The kind of answer that this professor has accepted in the past has been one like "clearly $\int_{-\infty}^\infty x\exp(-x^2)\,\mathrm dx$  evaluates to $0$ because the function $x\exp(-x^2)$ is odd." So if I could just explain why the $n=1,6$ cases yield non-zero values for the $A_n$ integral I would be golden! Any help that could be offered would be greatly appreciated! My apologies if this is a silly question.
 A: Hint:
Expand the integrand by distributivity and use for  the result  the linearisation formula
$$\sin a\sin b=\frac12\bigl((\cos(a-b)-\cos(a+b)\bigr),$$
in order to compute the integral.
A: Let us define 
$$I(n)=\int_0^1{\sin(n\pi x)\big[\sin(6 \pi x)-\sin(\pi x)\big]\,\mathrm dx} \tag{1},$$
where $n$ is a real parameter (though in fact, we are interested by integer values of $n$).
A) Theoretical approach : Let us look for values of $u$ giving this integral a $0$ value.
Using the relationship given by Bernard, we can transform $I(n)$ into :
$$I(n):=\frac12\int_0^1 (\cos((n-6)\pi x)-\cos((n+6)\pi x)-\cos((n-1)\pi x)+\cos((n+1)\pi x)dx $$
$$I(n)=\frac12 \left(\frac{\sin((n-6)\pi)}{(n-6)\pi}-\frac{\sin((n+6)\pi)}{(n+6)\pi}-\frac{\sin((n-1)\pi)}{(n-1)\pi}+\frac{\sin((n+1)\pi)}{(n+1)\pi}\right) $$

$$I(n)=\frac12 \left(\text{sinc}(n-6)-\text{sinc}(n+6)-\text{sinc}(n-1)+\text{sinc}(n+1)\right)\tag{2}$$

where $\text{sinc}$ denotes the "cardinal sine" defined in this way :
$$\text{sinc}(x)=\dfrac{\sin \pi x}{\pi x} \  \text{if} \ x \neq 0 \ \ \text{and sinc}(0)=1$$
(there exists a different definition without $\pi$ factors ; see this).
The graphical representation of function sinc is that of a "fading sine" crossing the $x$-axis at integer values of $x$ but for $x=0$. Therefore, it is not surprising to find that : 

$$\text{If} \  n \in \mathbb{N}, I(n)=0, \ \text{but for} \  n=1 \ \text{and} \ n=6$$

(for which they take the resp. values $-\tfrac12$ and $+\tfrac12$.)
B) Experimental approach. Let us take here an unusual way, by considering $n$ as a continuous parameter.
The graphical representation of function of function $I$ (taken under the form (2)) is as follows :

In the interval ($[0,6]$) under consideration, we find six values of $n$ for which $I(n)=0$ ; moreover, the fact that $I(1)=-\tfrac12$ and $I(6)=\tfrac12$ is given a palpable appearance.
There is a special values $n=2.4494897...$ to which it is difficult to attribute a role in the framework of Hilbert space...
Let us end by a "big picture" of the graphical representation of function $I$, which, as a function of $n$ is odd.

Final remark : of course, all our study has had a pedagogical aim. It must be said much more directly : the 2 dimensional space generated by $x \rightarrow \sin \pi x$ and $x \rightarrow \sin 6 \pi x$ is orthogonal to all the other basis functions  $x \rightarrow \sin n \pi x$ ($n \neq 1$ and $n \neq 6$).
A: There is a more fundamental reason. Consider the Sturm-Liouville differential equation $y^{\prime\prime}+\lambda y=0$ with boundary conditions $y(0)=y(1)=0$. With $3$ cases:
Case $1$: $\lambda<0$ then $y=a\cosh\sqrt{-\lambda}\,x+b\sinh\sqrt{-\lambda}\,x$. Then $y(0)=a=0$ and $y(1)=b\sinh\sqrt{-\lambda}=0$ so we don't get a nonzero solution.
Case $2$: $\lambda=0$ then $y=a+bx$, so $y(0)=a=0$ and $y(1)=b=0$ so again no nonzero solution.
Case$3$: $\lambda>0$ then $y=a\cos\sqrt{\lambda}\,x+b\sin\sqrt{\lambda}\,x$, so $y(0)=a=0$ but now $y(1)=b\sin\sqrt{\lambda}=0$ and $\sin n\pi=0$ so we have solutions $y_n=\sin n\pi x$ to the differential equation $y^{\prime\prime}+n^2\pi^2y=0$. Let's evaluate one integral:
$$\begin{align}\int_0^1y_m(x)y_n^{\prime\prime}(x)dx&=\left.y_m(x)y_n^{\prime}(x)\right|_0^1-\int_0^1y_m^{\prime}(x)y_n^{\prime}(x)dx\\
&=-\left.y_m^{\prime}(x)y_n(x)\right|_0^1+\int_0^1y_m^{\prime\prime}(x)y_n(x)dx=\int_0^1y_m^{\prime\prime}(x)y_n(x)dx\end{align}$$
The integrated term vanishing due to the boundary conditions on $y_n(x)$ and $y_m(x)$. Then we can evaluate
$$\begin{align}0&=\int_0^1\left\{y_m(x)\left(y_n^{\prime\prime}(x)+n^2\pi^2y_n(x)\right)-y_n(x)\left(y_m^{\prime\prime}(x)+n^2\pi^2y_m(x)\right)\right\}\\
&=(n^2-m^2)\pi^2\int_0^1y_m(x)y_n(x)dx\end{align}$$
So either $n=m$ or
$$\int_0^1y_m(x)y_n(x)dx=\int_0^1\sin m\pi x\,\sin n\pi x\,dx=0$$
So the orhtogonality follows as a fundamental property of the eigenfunctions of a Sturm-Liouville type differential equation. BTW, when $n=m$ I like to say that the average value of $\sin^2n\pi x$ is $1/2$ and the length of the interval is $1$ so
$$\int_0^1\sin^2n\pi x\,dx=\left(\frac12\right)(1)=\frac12$$
A: The functions $f_k(x) = \sin k\pi x$ and $f_l(x)=\sin l \pi x$ are orthogonal with respect to the inner product $\int_0^1 f(x)g(x) dx.$ 
$$\int_0^1 \sin k\pi x \sin l \pi x = \left\{ \begin{aligned} &0, &k\ne l \\ &\frac{1}{2}, &k=l  \end{aligned} \right. $$
with $\{k,l\}\in \mathbb{Z}.$
