Solution:

$Here \ {\ f}_n(x) = sin \dfrac{(2n+1)\pi{}x}{2l} \\ where \ n = 0,1,2,3,...$

${\ f}_m(x) = sin \dfrac{(2m+1)\pi{}x}{2l}$

$\therefore \int_0^l\ f_m\left(x\right).{\ f}_n(x)dx = \int_0^lsin\ \frac{(2m+1)\pi{}x}{2l}.\ sin\ \frac{(2n+1)\pi{}x}{2l}dx$

$-\dfrac{\ 1}{2}\int_0^l\left[cos\left(\frac{2m+2n+2}{2l}\right)\pi{}x-cos\left(\frac{2m-2n}{2l}\right)\pi{}x\right]dx$

$-\frac{\ 1}{2}\int_0^l\left[cos\left(\frac{m+n+1}{l}\right)\pi{}x-cos\left(\frac{m-n}{l}\right)\pi{}x\right]dx$

$-\dfrac{\ 1}{2}{\left[\dfrac{sin⁡(\dfrac{m+n+1}{l})\pi{}x}{(\dfrac{m+n+1)\pi{}}{l}}-\dfrac{sin⁡(\dfrac{m-n}{l})\pi{}x}{(\dfrac{m-n)\pi{}}{l}}\right]}_0^l$

$=-\dfrac{\ 1}{2}\left[\dfrac{sin⁡(m+n+1)\pi{}}{\frac{(m+n+1)\pi{}}{l}}-\dfrac{sin⁡(m-n)\pi{}}{\frac{(m-n)\pi{}}{l}}\right]$

$When \ m \not=n,\\ \int_{-\pi{}}^{\pi{}}\ f_m\left(x\right).{\ f}_n(x)dx=0 \ ...Since \ m \ and \ …