$Let\ fn(x)\ =\sin{\left(2n+1\right)x\ \ \ \ \ \ \ \ \ \ \ n=0,1,2,3…}\ \\[2ex] \displaystyle \ \ \ \ \ \ \ fm(x)=\sin{\left(2m+1\right)x\ \ \ \ \ \ \ \ \ \ \ m=0,1,2,3…} \\[2ex] \displaystyle Case\ i:\ m\not=n \\[2ex] \displaystyle \int_a^b\ fm(x)\bullet{}\ fn(x)dx=\int_0^{\frac{\pi{}}{2}}\sin{\left(2n+1\right)x\bullet{}\sin{\left(2m+1\right)x\ \times{}\frac{2}{2}}}\ dx \\[2ex] $

$\displaystyle =\frac{1}{2}\int_0^{\frac{\pi{}}{2}}\left[\left(\cos{\left(2n+1-2m-1\right)}\right)x-cos\left(2n+1+2m+1\right)x\right]dx \\[2ex] \displaystyle =\frac{1}{2}\int_0^{\frac{\pi{}}{2}}\left[\cos{\left(n-m\right)\bullet{}2x-\cos{\left(n+m+1\right)\bullet{}2x}}\right]dx \\[2ex] \displaystyle =\frac{1}{2}{\left[\frac{\sin{\left(n-m\right)\bullet{}2x}}{\left(n-m\right)\bullet{}2}-\frac{\sin{\left(n+m+1\right)2x}}{\left(n+m+1\right)2}\right]}_0^{\frac{\pi{}}{2}} \\[2ex] \displaystyle =\frac{1}{2}\left\{\left[0-0\right]-\left[0-0\right]\right\} \\[4ex] \displaystyle =0 \\[2ex] $

$ \displaystyle Case\ ii:\ m=n \\[2ex] \displaystyle \int_a^b{\left[fn(x)\right]}^2dx=\int_0^l{sin}^2{\left(2n+1\right)xdx}\ \\[2ex] \displaystyle =\int_0^{\frac{\pi{}}{2}}\frac{1-\cos{2\left(2n+1\right)x}}{2}dx \\[2ex] \displaystyle =\frac{1}{2}{\left[x-\frac{\sin{2(2n+1)}}{2(2n+1)}\right]}_0^{\frac{\pi{}}{2}} \\[2ex] \displaystyle =\frac{1}{2}\left\{\left(\frac{\pi{}}{2}+0\right)-(0-0)\right\} \\[2ex] \displaystyle =\frac{\pi{}}{4}\ \ \ \\[2ex] $