Answer:

$f(x)=x(2-x)=2x-x^2$

For half range cosine series, bn=0

Here, l=2

$\displaystyle a_0 = \dfrac {2}{l} \displaystyle \int^t_0 f(x) dx $

$\displaystyle = \dfrac {2}{2} \int^2_0 (2x- x^2) dx$

$= \left [ 2 \cdot \dfrac {x^2}{2} - \dfrac {x^3}{3} \right ]^2_0 $

$= \left ( 2^2 - \dfrac {2^3}{3} \right )- (0-0)$

$= \dfrac {4}{3}$

$a_n = \dfrac {2}{l} \displaystyle \int^1_0 f(x) \cos \dfrac {n \pi x}{l}dx$

$\displaystyle =\dfrac {2}{2} \int^2_0 (2x-x^2) \cos \dfrac {n \pi x} {2} dx$

$= \left [ (2x-x^2)\sin \dfrac {n\pi x}{2} \cdot \dfrac {2}{n\pi} - (2-2x)\cdot - \cos \dfrac {n \pi x}{2} \cdot \dfrac {2^2}{n^2 \pi^2} + (-2)\cdot -\sin \dfrac {n\pi x}{2} \cdot \dfrac {2^3}{n^3 \pi^3} \right ]^2_0$

$=\left [ 0+ (-2)\cdot \cos n \pi \cdot \dfrac {4}{n^2 \pi^2} + 0 \right ]- \left [ 0+(2)\cdot \dfrac {4}{n^2 \pi^2} +0 \right ]$

$= (-1)^n \dfrac {-8}{n^2\pi^2 } - \dfrac {8}{n^2 \pi^2}$

∴ Half range Fourier cosine series is, 

$\displaystyle f(x) = \dfrac {a_0}{2} + \sum^{\infty}_{n=1} a_n \cos \dfrac {n \pi x}{l}$

$\therefore 2x - x^2 = \dfrac {2}{3}+ \displaystyle \sum^{\infty}_{n=1} \dfrac {-8}{n^2 \pi^2} \Big [(-1)^n + 1 \Big ] \cos \dfrac {n \pi x}{2}$

$= \dfrac {2}{3} - \dfrac {8}{\pi^2} \left [ 0+\dfrac {2}{2^2} \cos \dfrac {2 \pi x}{2} + 0+ \dfrac {2}{4^2} \cos \dfrac {4 \pi x}{2} + \cdots \right ]$

$= \dfrac {2}{3} - \dfrac {8} {\pi^2} \times \dfrac {2}{2^2} \left [ \dfrac {\cos \pi x}{1^2}+ \dfrac {\cos 2 \pi x}{2^2} + \cdots \right ]$

$= \dfrac {2}{3} - \dfrac {4}{\pi^2} \left [ \dfrac {\cos \pi x}{1^2} + \dfrac {\cos 2 \pi x}{2^2} + \dfrac {\cos 3 \pi x }{3^2} + \cdots \right ]$