Prove that $\{\sin x, \sin 2x, ... , \sin nx\}$ is linearly independent.
The short solution that I do not understand is as follow: For p and q are positive integer, we have $$ \int\limits_{0}^{\pi}\sin{px}\sin{qx} dx=\left\{\begin{array}{l}0\qquad \textrm{if } \, p\ne q\\ \dfrac{\pi}{2}\qquad \textrm{if }\, p=q\ne0\\ \end{array} \right. $$ Applying this result to show that if $\sum\limits_{k=1}^{n}\alpha_k\sin kx=0$ then $\alpha_k=0,\, k=1,\dots,n$
The solution is too short for me too understand. I would be grateful if you could explain this problem more in detail for me.
Take an arbitrary linear combination $\sum_{i=1}^n\alpha_{i}\sin{p_ix}=0$.Multiply the sum with $\sin{p_jx}$ and integrate the outcome between $0$ and $\pi$. One gets
$$\sum_{i=1}^n\alpha_i\int_0^{\pi}\sin{p_ix}\sin{p_jx}dx=\alpha_j\frac{\pi}{2}=0$$
$j$ being arbitrary, This means all $\alpha_i=0$ and the family is independant