Answer

$f(x) = [ _{(\pi-x) \ \pi/2 \le x\le \pi}^{x\ 0 \le x\le \pi/2} $

Half range cosine series

$f(x) =a_0 +\sum a_n cos \dfrac{n\pi x}{l} $

$a_0 =\dfrac{1}{\pi}\int_0^{\pi/2} f(x).dx = \dfrac{1}{\pi} [ \int_0^{\pi/2} xdx +\int_{\pi/2}^l (\pi-x)dx ] $

$\dfrac{1}{\pi}([\dfrac{x^2}{2}]_0 ^{\pi/2} + [\pi x-\dfrac{x^2}{2}]_{\pi/2}^{\pi})$

$\dfrac{1}{\pi} [\dfrac{\pi^2}{8} + \pi^2 - \dfrac{\pi^2}{2}- \dfrac{\pi^2}{2}+ \dfrac{\pi^2}{8} ]\\ = \dfrac{\pi}{4} \\ \therefore a_0= \dfrac{\pi}{4} $

$a_n = \dfrac{2}{\pi} \int_0^\pi f(x) cos \dfrac{n\pi x}{\pi} .dx\\ = \dfrac{2}{\pi}[ \int_0^{\pi/2} x cos { x} .dx + \int_{\pi/2}^\pi (\pi-x) cos { x} .dx] $

$\dfrac{2}{\pi} [ ( x.sin {n x}.\dfrac{l}{ n} -(1)(-cos nx .\dfrac{1}{ n^2})]_0^{\pi/2} + [((\pi-x) .sinn x \dfrac{1}{ n} -(-1) (-)(-cosnx \dfrac{1}{ n^2})]_{\pi/2}^{\pi} $

$\dfrac{2}{\pi} [\dfrac{\pi}{2}.sin\dfrac{n\pi}{2}.\dfrac{l}{ n} +cos\dfrac{n\pi}{2}. \dfrac{1} {n^2}- \dfrac{1}{n^2}] +[0-cos\dfrac{n\pi}{2}. \dfrac{1} {n^2}- (\dfrac{\pi}{2} (sin\dfrac{n\pi}{2}.\dfrac{\pi}{ n}) +cos\dfrac{n\pi}{2}. \dfrac{1} {n^2}] $

$\dfrac{2}{\pi}[ 2 cos\dfrac{n\pi}{2}.\dfrac{1}{n^2} -cos n\pi.\dfrac{1}{ n^2}-\dfrac{1}{ n^2} ] $

$\dfrac{2}{\pi n^2} [ 2cos\dfrac{ n\pi}{2} -cos\pi n -1]$

$a_1 =0 \because cos \dfrac{\pi}{2} =0, cos\pi =-1$

$a_2 = \dfrac{2}{\pi 2^2}.[2(-1) -(+1)-1)]\\ =\dfrac{-8}{\pi2^2}\\ similerly\ a_3 , a_4 ,a_5 =0 $

$a_6 =\dfrac{2}{\pi6^2}( 2Cos 3\pi -cos 6 \pi-1)\\ \dfrac{2}{\pi6^2 }(-2 -1-1) \\ =- \dfrac{8}{\pi6^2 }\\ a_7=0,,a_8 =0 a_9 =0, a_{10} = =- \dfrac{8}{\pi{10}^2 }\\ $

$f(x) = a_0 + \displaystyle\sum_{n=1}^{\infty} a_n cos{n\pi x}\\ a_0 +a_1 cos\pi x+a_2 cos{2\pi x}.......... +a_6 cos{6\pi x}+........a_{10} cos10\pi x.....] $

Substituing the value of  $a_0,a_1 ------ a_{10}$

$\therefore\ f(x)= \dfrac{\pi}{4} -\dfrac{8}{\pi}[\dfrac{1}{2^2}.cos{2\pi x}+\dfrac{1}{6^2}.cos{6\pi x}+\dfrac{1}{(10)^2}.cos{10\pi x}-----] $