Fouriercoefficients, piecewise continous functions, f(t)=t

How does one find the Fourier coefficients of the following functions $f:[-\pi,\pi]\rightarrow \mathbb{R}$

1. 
   

   $f(t)=t$ for $-\pi < t < \pi$ and $f(-\pi)=f(\pi)=0$

   
   


2. 
   

   $f(t)=-1$ for $-\pi < t < 0$, $f(t)=0$ for $t=-\pi,0,\pi$ and $f(t)=1$ for $0<t<\pi$ , $2\pi$ periodically extended to functions $\mathbb{R}\rightarrow \mathbb{R}$

   
   

What I have tried:

The coefficients for a function defined on $-\pi < x < \pi$ :$$a_k = \frac{1}{\pi}\int_{-\pi}^{\pi}f(x)cos(kx)dx ; $$ $$b_k=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)sin(kx)dx$$$$a_0=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x) dx $$

i think $$a_k+ib_k = \frac{1}{\pi}\int_{-\pi}^{\pi}f(x)e^{ikx}dx$$

for 1) : $\int x cos(kx) dx$= $$\frac{1}{k}sin(kx)x -\frac{1}{k} \int sin(kx)dx = \frac{1}{k}(xsin(kx)+\frac{cos(kx)}{k^2})$$ so: $$ a_k = \frac{1}{\pi}\int _{-\pi}^{\pi} xcos(kx)dx = \frac{(k\pi sin(k\pi))+ cos(k \pi) + k\pi sin(-k\pi)-cos(-k\pi)}{k^2}=0 $$

for $$\int x sin(kx)dx = \frac{-kxcos(kx)+sin(kx)}{k^2} $$ : $$\Rightarrow b_k= \frac{1}{\pi}\int_{-\pi}^{\pi} x sin(kx)dx = \frac{(-k \pi cos(k\pi)+sin(k \pi) - (k\pi cos(kx)+sin(-k\pi)}{\pi k^2}) = \frac{2 sin (\pi k)-2k\pi cos(kx)}{\pi k^2}$$

and $a_0= \int_{-\pi}^{\pi} x dx = 0$

is this correct for 1) ?

And, how does one find the Fourier coefficients of a piecewise given function? Take the coefficients for each boundary and sum them up e.g. : $$a_0 = \int_{-\pi}^{0}-1 dx + \int_{0}^{\pi}1 dx $$

and analogous for $a_k$, $b_k$ ?

via Recent Questions - Mathematics - Stack Exchange http://math.stackexchange.com/questions/290506/fouriercoefficients-piecewise-continous-functions-ft-t