Answer
See below
Work Step by Step
Given $f_1(x)=1\\
f_2(x)=\sin \pi x\\
f_3(x)=\cos \pi x$
Obtain:
$=\int ^1_{-1}f_1(x)f_2(x)f_3(x)dx\\
=\int^1_{-1}\sin \pi x dx\\
=(-\frac{\cos \pi x}{\pi})|^1_{-1}\\
=-\frac{\cos \pi }{\pi}-(-\frac{\cos (-\pi )}{\pi})\\
=0$
$=\int ^1_{-1}f_1(x)f_3(x)dx\\
=\int^1_{-1}\cos \pi xdx\\
=(-\frac{\sin \pi x}{\pi})|^1_{-1}\\
=\frac{\sin \pi }{\pi}-(\frac{\sin (-\pi )}{\pi})\\
=0$
$=\int ^1_{-1}f_2(x)f_3(x)dx\\
=\int^1_{-1}\sin \pi x\cos \pi x dx\\
=\int^1_{-1}\frac{\sin 2\pi x}{y} dx\\
=(-\frac{\cos 2\pi x}{4\pi})|^1_{-1}\\
=-\frac{\cos 2\pi }{4\pi}+\frac{\cos 2\pi }{4\pi}\\
=0$
We have:
$||f_1||=\sqrt \\
=\sqrt \int^1_{-1}f_1(x)f_1(x)dx\\
=\sqrt \int^1_{-1}1dx\\
=\sqrt x|^1_{-1}\\
=\sqrt 1-(-1)\\
=\sqrt 2$
$||f_2||=\sqrt \\
=\sqrt \int^1_{-1}f_2(x)f_2(x)dx\\
=\sqrt \int^1_{-1}\sin^2\pi xdx\\
=\sqrt \int^1_{-1}\frac{1-\cos 2\pi x}{2}dx\\
=\sqrt \frac{x}{2}-\frac{\sin 2\pi x}{4\pi}|^1_{-1}\\
=\sqrt (\frac{1}{2}-\frac{\sin 2\pi }{4\pi})-(-\frac{1}{2}-\frac{\sin (-2\pi) }{4\pi})\\
=\sqrt \frac{1}{2}-(-\frac{1}{2})\\
=1$
$||f_3||=\sqrt \\
=\sqrt \int^1_{-1}f_3(x)f_3(x)dx\\
=\sqrt \int^1_{-1}\cos^2\pi xdx\\
=\sqrt \int^1_{-1}\frac{1+\cos 2\pi x}{2}dx\\
=\sqrt \frac{x}{2}+\frac{\sin 2\pi x}{4\pi}|^1_{-1}\\
=\sqrt (\frac{1}{2}+\frac{\sin 2\pi }{4\pi})-(-\frac{1}{2}+\frac{\sin (-2\pi) }{4\pi})\\
=\sqrt \frac{1}{2}-(-\frac{1}{2})\\
=1$
