$\displaystyle Let\\[2ex] c=0\ and\ c+2l=2\pi{} \\[2ex] \displaystyle \therefore{}0+2l=2\pi{} \\[2ex] \displaystyle \therefore{}l=\pi{} \\[2ex] \displaystyle a_0=\frac{1}{l}\int_c^{c+2l}f\left(x\right)dx \\[2ex] \displaystyle =\frac{1}{\pi{}}\int_0^{2\pi{}}\cos{px\ dx} \\[2ex] \displaystyle =\frac{1}{\pi{}}{\left[\frac{\sin{px}}{p}\right]}_0^{2\pi{}} \\[2ex] \displaystyle =\frac{1}{\pi{}p}\left[\sin{2\pi{}p-\sin0}\right] \\[2ex] \displaystyle =\frac{\sin{2\pi{}p}}{\pi{}p} \\[2ex] $

$\displaystyle a_n=\frac{1}{l}\int_c^{c+2l}f\left(x\right)\cos{\frac{n\pi{}x}{l}}dx \\[2ex] \displaystyle =\frac{1}{\pi{}}\int_0^{2\pi{}}\cos{px\bullet{}}\cos{\frac{n\pi{}x}{l}}dx \\[2ex] \displaystyle =\frac{1}{2\pi{}}\int_0^{2\pi{}}\left[\cos{\left(px+nx\right)+\cos{\left(px-nx\right)\ }}\right]dx \\[2ex] \displaystyle =\frac{1}{2\pi{}}{\left[\frac{\sin{\left(p+n\right)x}}{\left(p+n\right)}+\frac{\sin{\left(p-n\right)x}}{\left(p-n\right)}\right]}_0^{2\pi{}} \\[2ex] \displaystyle =\frac{1}{2\pi{}}\left\{\left.\left[\frac{\sin{\left(p+n\right)2\pi{}}}{\left(p+n\right)}+\frac{\sin{\left(p-n\right)2\pi{}}}{\left(p-n\right)}\right]-\left[0+0\right]\right\}\right. \\[2ex] $

$\displaystyle Consider,\\[2ex] \sin{\left(p\pm{}n\right)2\pi{}=\sin{\left(2\pi{}p\pm{}2\pi{}n\right)}} \\[2ex] \displaystyle =\sin{2\pi{}p\cos{2\pi{}n\pm{}\sin{2\pi{}n\cos{2\pi{}p}}}} \\[2ex] \displaystyle =\sin\ 2\pi{}p\bullet{}1\pm{}0 \\[2ex] \displaystyle =\sin\ 2\pi{}p \\[2ex] $

$\displaystyle therefore \\[2ex] \displaystyle a_n=\left[\frac{\sin{2\pi{}p}}{\left(p+n\right)}+\frac{\sin{2\pi{}p}}{\left(p-n\right)}-0+0\right] \\[2ex] \displaystyle a_n=\frac{2\pi{}p}{2\pi{}}\left[\frac{1}{p+n}+\frac{1}{p-n}\right] \\[2ex] $

$\displaystyle b_n=\frac{1}{l}\int_c^{c+2l}f\left(x\right)\sin{\frac{n\pi{}x}{l}}dx \\[2ex] \displaystyle b_n=\frac{1}{\pi{}}\int_c^{c+2l}\cos{px\bullet{}}\sin{\frac{n\pi{}x}{l}}dx \\[2ex] \displaystyle b_n=\frac{1}{2\pi{}}\int_0^{2\pi{}}\left[\sin{\left(px+nx\right)+\sin{\left(px-nx\right)\ }}\right]dx \\[2ex] \displaystyle b_n=\frac{1}{2\pi{}}{\left[\frac{-\cos{\left(p+n\right)x}}{\left(p+n\right)}-\frac{\cos{\left(p-n\right)x}}{\left(p-n\right)}\right]}_0^{2\pi{}} \\[2ex] \displaystyle b_n=\frac{1}{2\pi{}}\left\{\left.\left[\frac{-\cos{\left(p+n\right)2\pi{}}}{\left(p+n\right)}-\frac{\cos{\left(p-n\right)2\pi{}}}{\left(p-n\right)}\right]-\left[\frac{-cos\ 0}{p+n}-\frac{cos\ 0}{p-n}\right]\right\}\right. \\[2ex] $

$\displaystyle Consider,\\[2ex] \cos{\left(p\pm{}n\right)2\pi{}\\[2ex] =\cos{\left(2\pi{}p\pm{}2\pi{}n\right)}} \\[2ex] \displaystyle =\cos{2\pi{}p\cos{2\pi{}n\pm{}\cos{2\pi{}n\cos{2\pi{}p}}}} \\[2ex] \displaystyle =\cos\ 2\pi{}p\bullet{}1\pm{}0 \\[2ex] \displaystyle =\cos\ 2\pi{}p \\[2ex] $

$\displaystyle bn\ =\frac{1}{2\pi{}}\left[\frac{-\cos{2\pi{}p}}{p+n}-\frac{\cos{2\pi{}p}}{p-n}+\frac{1}{p+n}-\frac{1}{p-n}\right] \\[2ex] \displaystyle b_n=\frac{1}{2\pi{}}\left.\left\{-\cos{2\pi{}p\left[\frac{1}{p+n}+\frac{1}{p-n}\right]}\right.+1\left[\frac{1}{p+n}+\frac{1}{p-n}\right]\right\} \\[2ex] \displaystyle b_n=\frac{1}{2\pi{}}\bullet{}\left(-\cos{2\pi{}p+1}\right)\left[\frac{1}{p+n}+\frac{1}{p-n}\right] \\[2ex] $

$\displaystyle In\ Fourier\ series, \\[2ex] \displaystyle f(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty{}}a_n\cos\frac{n\pi{}x}{l}+\sum_{n=1}^{\infty{}}b_n\sin\frac{n\pi{}x}{l}\ \\[2ex] \displaystyle \therefore{}\cos{px=\frac{\sin{2\pi{}p}}{2\pi{}p}+\sum_{n=1}^{\infty{}}\frac{\sin{2\pi{}p}}{2\pi{}p}}\left[\frac{1}{p+n}+\frac{1}{p-n}\right]\cos\frac{n\pi{}x}{\pi{}}\\[4ex] \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \displaystyle +\sum_{n=1}^{\infty{}}\frac{1-\cos{2\pi{}p}}{2\pi{}}\left[\frac{1}{p+n}-\frac{1}{p-n}\right]\sin{\frac{n\pi{}x}{\pi{}}} \\[2ex] $

$\displaystyle \therefore{}\cos\ px=\frac{\sin{2\pi{}p}}{2\pi{}p}+\sin\frac{2\pi{}p}{2\pi{}}\sum_{n=1}^{\infty{}}\left[\frac{1}{p+n}+\frac{1}{p-n}\right]\cos{nx \\[4ex] \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \displaystyle +\ \frac{\left[1-\cos{2\pi{}p}\right]}{2\pi{}}}\sum_{n=1}^{\infty{}}\left[\frac{1}{p+n}-\frac{1}{p-n}\right]\sin{nx} \\[2ex] $