$\displaystyle Let\ 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 a_0=\frac{1}{\pi{}}\int_0^{2\pi{}}\sqrt{1-\cos x}dx\\[2ex] \displaystyle a_0 = \frac{1}{\pi{}}\ \int_0^{2\pi{}} \sqrt{2 {\sin}^2{(\frac{x}{2})}}dx \\[2ex] $

$\displaystyle Since,\cos{2\theta{}=1-2\ {\sin}^2{\theta{}}\ } \\[2ex] \displaystyle =\ \frac{\sqrt{2}}{\pi{}}\int_0^{2\pi{}}\sin{(\frac{x}{2})}\ dx\\[3ex]=\frac{\sqrt{2}}{\pi{}}\ {\left[-\frac{\cos{\left(\frac{x}{2}\right)}}{\frac{1}{2}}\right]}_0^{2\pi{}}\\[3ex]=\ \frac{-2\sqrt{2}}{\pi{}}\left[\cos{\pi{}-\cos0}\right]\\[3ex]=\frac{-2\sqrt{2}}{\pi{}}[-1-1] \\[2ex] \displaystyle \therefore{}a_0=\frac{4\sqrt{2}}{\pi{}}\ \ \\[2ex] \displaystyle a_n=\frac{1}{l}\int_c^{c+2l}f\left(x\right)\cos{\frac{n\pi{}x}{l}}dx \\[2ex] \displaystyle a_n=\frac{1}{\pi{}}\int_0^{2\pi{}}\sqrt{1-\cos x}\cos{nx}dx\\[3ex] \displaystyle a_n=\frac{\sqrt{2}}{\pi{}}\int_0^{2\pi{}}\sin{(\frac{x}{2})}\cos{nx}dx\ \\[2ex] $

$\displaystyle a_n =\frac{\sqrt{2}}{\pi{}}\int_0^{\pi{}}\left[\sin{\left(2nx+\frac{x}{2}\right)-\sin{\left(2nx-\frac{x}{2}\right)}}\right]dx \\[2ex] \displaystyle a_n=\frac{\sqrt{2}}{\pi{}}\int_0^{\pi{}}\left[\sin{\frac{\left(4n+1\right)x}{2}-\sin{\frac{\left(4n-1\right)x}{2}}}\right]dx \\[2ex] \displaystyle a_n=\frac{\sqrt{2}}{\pi{}}{\left[-\frac{\cos{\frac{\left(4n+1\right)x}{2}}}{\frac{\left(4n+1\right)}{2}}{+\frac{\cos{\frac{\left(4n-1\right)x}{2}}}{\frac{\left(4n-1\right)}{2}}}\right]}_0^{\pi{}} \\[2ex] \displaystyle =\frac{\sqrt{2}}{\pi{}}\left\{\left[-\frac{2\cos{\frac{\left(4n+1\right)\pi{}}{2}}}{\frac{\left(4n+1\right)}{2}}{+\frac{2\cos{\frac{\left(4n-1\right)\pi{}}{2}}}{\frac{\left(4n-1\right)}{2}}}\right] \\[3ex] \ \ \ \ \ \ -\left[-\ \frac{2\cos0}{\left(4n+1\right)}{+\frac{2\cos0}{\left(4n-1\right)}}\right]\right\} \\[2ex] $

$\displaystyle Consider \\[2ex] \displaystyle \cos{\frac{\left(4n\pm{}1\right)\pi{}}{2}=\cos\left(2n\pi{}\pm{}\frac{\pi{}}{2}\right)}\\[2ex] \displaystyle =\cos\frac{\pi{}}{2}\\[2ex]=0 \\[2ex] $

$\displaystyle \therefore{}\ a_n=\frac{\sqrt{2}}{\pi{}}\left\{0-0+\frac{2}{4n+1}-\frac{2}{4n-1}\right\} \\[3ex] \displaystyle a_n=\frac{2\sqrt{2}}{\pi{}}\left\{\frac{4n-1-4n-1}{\left(4n+1\right)\left(4n-1\right)}\right\} \\[2ex] a_n\displaystyle =\frac{2\sqrt{2}}{\pi{}}\times{}\frac{-2}{16n^2-1}=\frac{-4\sqrt{2}}{\pi{}\left(16n^2-1\right)} \\[3ex] $

$\displaystyle Consider \\[2ex] \displaystyle b_n=\frac{1}{l}\int_c^{c+2l}f\left(x\right)\sin\left(\frac{n\pi{}x}{l}\right)dx\ \ \ \\[2ex] \displaystyle =\frac{1}{\frac{\pi{}}{2}}\int_0^{\pi{}}\sqrt{2}\sin{\left(\frac{x}{2}\right)\sin\left(\frac{n\pi{}x}{\frac{\pi{}}{2}}\right)}dx \\[2ex] \displaystyle =\frac{\sqrt{2}}{\pi{}}\int_0^{\pi{}}2\sin\left(2nx\right)\sin\left(\frac{x}{2}\right)dx \\[2ex] \displaystyle =\frac{\sqrt{2}}{\pi{}}\int_0^{\pi{}}\left[\cos{\left(2nx+\frac{x}{2}\right)-\cos{\left(2nx-\frac{x}{2}\right)}}\right]dx \\[2ex] \displaystyle =\frac{\sqrt{2}}{\pi{}}\int_0^{\pi{}}\left[\cos{\frac{\left(4n+1\right)x}{2}-\cos{\frac{\left(4n-1\right)x}{2}}}\right]dx \\[2ex] \displaystyle =\frac{\sqrt{2}}{\pi{}}{\left[-\frac{\sin{\frac{\left(4n+1\right)x}{2}}}{\frac{\left(4n+1\right)}{2}}{+\frac{\sin{\frac{\left(4n-1\right)x}{2}}}{\frac{\left(4n-1\right)}{2}}}\right]}_0^{\pi{}} \\[2ex] \displaystyle =\frac{\sqrt{2}}{\pi{}}\left\{\left[-\frac{2\sin{\frac{\left(4n+1\right)\pi{}}{2}}}{\frac{\left(4n+1\right)}{2}}{+\frac{2\sin{\frac{\left(4n-1\right)\pi{}}{2}}}{\frac{\left(4n-1\right)}{2}}}\right]\\[2ex] \ \ \ \ \ \ \ \ \ \ \ -\left[-\ \frac{2\sin0}{\left(4n+1\right)}{+\frac{2\sin0}{\left(4n-1\right)}}\right]\right\} \\[2ex] $

$\displaystyle Consider \\[2ex] \displaystyle \sin{\frac{\left(4n\pm{}1\right)\pi{}}{2}=\sin\left(2n\pi{}\pm{}\frac{\pi{}}{2}\right)}=\pm{}\sin\frac{\pi{}}{2}=\pm{}1 \\[2ex] \displaystyle \therefore{}\ b_n=\frac{\sqrt{2}}{\pi{}}\left\{\left[\frac{-1}{4n+1}-\frac{1}{4n-1}\right]-\left[0+0\right]\right\} \\[2ex] \displaystyle =\frac{\sqrt{2}}{\pi{}}\left\{\frac{-4n-1-4n+1}{\left(4n+1\right)\left(4n-1\right)}\right\} \\[2ex] \displaystyle =\frac{\sqrt{2}}{\pi{}}\times{}\frac{-8n}{16n^2-1}=\frac{-8n\sqrt{2}}{\pi{}\left(16n^2-1\right)} \\[2ex] $

$\displaystyle In\ Fourier\ series, \\[4ex] \displaystyle f(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty{}}a_n\cos\left(\frac{n\pi{}x}{l}\right)+\sum_{n=1}^{\infty{}}b_n\sin\left(\frac{n\pi{}x}{l}\right) \\[2ex] \displaystyle \sqrt{1-\cos x}=\frac{2\sqrt{2}}{\pi{}}+\sum_{n=1}^{\infty{}}\frac{-4\sqrt{2}}{\pi{}\left(16n^2-1\right)}\cos\left(\frac{n\pi{}x}{\frac{\pi{}}{2}}\right)+\sum_{n=1}^{\infty{}}\frac{-8n\sqrt{2}}{\pi{}\left(16n^2-1\right)}\sin\left(\frac{n\pi{}x}{\frac{\pi{}}{2}}\right) \\[3ex] \displaystyle =\frac{2\sqrt{2}}{\pi{}}-\frac{4\sqrt{2}}{\pi{}}\sum_{n=1}^{\infty{}}\frac{\cos2nx}{\left(16n^2-1\right)}-\frac{8\sqrt{2}}{\pi{}}\sum_{n=1}^{\infty{}}\frac{n\sin 2nx}{\left(16n^2-1\right)}\ \ \ \ \ \ \rightarrow{}(i) \\[2ex] \displaystyle =\frac{2\sqrt{2}}{\pi{}}+\frac{4\sqrt{2}}{\pi{}}\left[\frac{\cos2x}{\left(16\bullet{}1^2-1\right)}+\frac{\cos4x}{\left(16{\bullet{}2}^2-1\right)}+…\right]-\frac{8\sqrt{2}}{\pi{}}\left[\frac{\sin2x}{\left(16{\bullet{}1}^2-1\right)}+\frac{2\sin 2nx}{\left(16\bullet{}2^2-1\right)}+…\right] \\[5ex] \displaystyle Deduction:\ Put\ x=0\ in\ (i) \\[2ex] \displaystyle \therefore{}\sqrt{1-\cos0}=\frac{2\sqrt{2}}{\pi{}}-\frac{4\sqrt{2}}{\pi{}}\sum_{n=1}^{\infty{}}\frac{\cos0}{\left(16n^2-1\right)} \\[2ex] \displaystyle \therefore{}0-\frac{2\sqrt{2}}{\pi{}}=-\frac{4\sqrt{2}}{\pi{}}\sum_{n=1}^{\infty{}}\frac{1}{\left(16n^2-1\right)} \\[2ex] \displaystyle \therefore{}\frac{1}{2}=\sum_{n=1}^{\infty{}}\frac{1}{{\left(4n\right)}^2-1} \\[2ex]$