Answer
See below
Work Step by Step
Given $f_1(x)=\sin \pi x\\
f_2(x)=\sin 2\pi x\\
f_3(x)=\sin 3\pi x$
Obtain:
$\\
=\int^1_{-1}f_m(x)f_n(x)dx\\
=\int^1_{-1} \sin (m\pi x)\sin (n\pi x)dx\\
=\int^1_{-1}\frac{\cos (m\pi x-n\pi x)-\cos (m\pi x+n\pi x)}{2}\\
=\frac{1}{2}[\frac{\sin((m-n)\pi x}{(m-n)\pi}+\frac{\sin ((m+n)\pi x}{(m+n)\pi}]|^1_{-1}\\
=\frac{1}{2}[(\frac{\sin(m-n)\pi }{(m-n)\pi}+\frac{\sin(m+n) \pi }{(m+n)\pi})+(\frac{\sin ((m-n)\pi (-1))}{(m-n)\pi}+\frac{\sin ((m+n)\pi (-1))}{(m+n)\pi}]\\
=0, \forall m,n \in \{1,2,3\}$
Hence, $f_1,f_2$ and $f_3$ are orthogonal.
Obtain:
$||f|_n|\\
=\int^1_{-1}[f_n(x)]^2dx\\
=\int^1_{-1}\sin^2(n\pi x)dx\\
=\int^1_{-1}\frac{1-\cos(2n\pi x)}{2}dx\\
=\frac{1}{2}x-\frac{1}{2}.\frac{\sin (2n\pi )}{2n \pi}|^1_{-1}\\
=(\frac{1}{2}+\frac{1}{2}.\frac{\sin (2n\pi )}{2n \pi}-(\frac{1}{2}(-1)-\frac{1}{2}.\frac{\sin (2n\pi (-1))}{2n \pi})\\
=\frac{1}{2}+\frac{1}{2}\\
=1$
Hence, $f_1, f_2$ and $f_3$ are unit vectors.
Consequently, $f_1,f_2$ and $f_3$ are orthonormal functions.
