This makes no sense. The coefficient can't be a function of x. Did you forget to evaluate at \(\displaystyle \pi\) and \(\displaystyle -\pi\)?

cos(ax) is an even function- it will have **no** "sin(nx)" terms in it. The coefficient of cos(nx) will be

\(\displaystyle \frac{1}{\pi}\int_{-\pi}^{\pi} cos(ax) cos(nx) dx=\)\(\displaystyle \frac{1}{2\pi}\left(\int_{-\pi}^\pi cos((n+a)x)dx+ \int_{-\pi}^\pi cos(n- a)x dx\right)\)

for n> 0.

The n= 0 term is \(\displaystyle \frac{1}{2\pi}\int_{-pi}^{\pi} cos(ax)dx\).

As I said above, all coefficients of "sin(nx)" will be 0.

You can also use by-parts here.

\(\displaystyle C_n= \frac{1}{\pi}\int_{-\pi}^{\pi} cos(ax) cos(nx) dx \)

\(\displaystyle U = cos(ax) \) and \(\displaystyle dV = cos(nx)dx \)

\(\displaystyle du = -asin(ax) \) and \(\displaystyle V= \frac{ sin(nx) }{n} \)

\(\displaystyle \frac{1}{\pi}\int_{-\pi}^{\pi} cos(ax) cos(nx) dx = \frac{a}{\pi} [ \frac{ sin(nx)cos(ax) }{n} + \frac{a}{n} \int_{-\pi}^{\pi} sin(ax)sin(nx) dx] \)

\(\displaystyle \frac{a^2}{n \pi} [ \int_{-\pi}^{\pi} sin(ax)sin(nx) dx] \)

\(\displaystyle U = sin(ax) \) and \(\displaystyle dV = sin(nx) dx \)

\(\displaystyle du = acos(ax) \) and \(\displaystyle V= - \frac{ cos(nx) }{n} \)

\(\displaystyle \frac{a^2}{n \pi} [ \int_{-\pi}^{\pi} sin(ax)sin(nx) dx] = \frac{a^2}{n \pi} [ -\frac{ cos(nx)sin(ax) }{n} + \frac{1}{n} \int_{-\pi}^{\pi} cos(ax)cos(nx)dx ] \)

\(\displaystyle

C_n = \frac{a^2}{n \pi} [ -( \frac{ (-1)^n sin(a \pi ) }{n} - \frac{ (-1)^n sin(- a \pi ) }{n} ) + \frac{ \pi }{n} C_n] \)

\(\displaystyle

C_n = \frac{a^2}{n^2 \pi} [ -2 (-1)^n sin(a \pi ) + \pi C_n] \)

\(\displaystyle C_n (1 - \frac{a^2}{n^2} ) = -2 (-1)^n sin(a \pi ) \)

\(\displaystyle C_n = - \frac{ 2 (-1)^n sin(a \pi ) } { (1 - \frac{a^2}{n^2 } ) } \)

I'm fairly certain I've missed something or forgot something major (spidy sense is tingling...).