Is there a faster way to calculate the Laplace Transform of this function, rather than using the definition? Consider the following function:
$$ 
f(t) = \begin{cases} \cos(t) &\mbox{if } 0 \leq t \le π  \\
\sin(t) & \mbox{if }  π \leq t \le 2π  \\
e^t & \mbox{if } t > 2\pi   \end{cases} 
$$ 
A way to approach the solution would be using the definition, that is:
$$\int_{0}^{π} \cos(t)e^{-st} dt + \int_{π}^{2π} \sin(t) e^{-st}dt + \int_{2π}^{+\infty} e^{-t}e^{-st} dt$$ 
then solve the integrals and find the solution.

 My question is: 
Can we use basic laplace transforms on a cased function and if not why? 
With the term basic laplace transforms I mean
paradigms such as:
$$ L[\cos(at)] = \frac{s}{s^2+a^2}$$
$$ L[\sin(at)] = \frac{a}{s^2+a^2}$$
$$ L[e^{at}]= \frac{1}{s-a}$$

In that case the Laplace transform of f(t) would be the following:

$$ L[f(t)] = 2*\frac {s^2}{s^2+1} + \frac {1}{s - 1} $$ but it just doesn't feel right. 
 A: $$\begin{align*}
f(t)&=\begin{cases}\cos t&\text{for }0\le t\le\pi\\\sin t&\text{for }\pi<t\le2\pi\\e^t&\text{for }t>2\pi\end{cases}\\[1ex]
&=\cos t(u(t)-u(t-\pi))+\sin t(u(t-\pi)-u(t-2\pi))+e^tu(t-2\pi)
\end{align*}$$
where $u(t)$ is the unit step (Heaviside step) function defined by
$$u(t-c)=\begin{cases}1&\text{for }t\ge c\\0&\text{for }t<c\end{cases}$$
It's easy to show
$$\mathcal L_s\{f(t-c)u(t-c)\}=e^{-cs}\mathcal L_s\{f(t)\}\quad\quad(*)$$
where $\mathcal L_s\{f(t)\}$ is the Laplace transform of $f(t)$. We can expand $\mathcal L_s\{f(t)\}$ as
$$\mathcal L_s\left\{\cos t\,u(t)+\cos(t-\pi)u(t-\pi)-\sin(t-\pi)u(t-\pi)+\sin(t-2\pi)u(t-2\pi)+e^{2\pi}e^{t-2\pi}u(t-2\pi)\right\}$$
since $-\cos t=\cos(t-\pi)$, $\sin t=-\sin(t-\pi)$, and $\sin(t-2\pi)=\sin t$. Then use $(*)$ to simplify this accordingly.

 $$\mathcal L_s\{f(t)\}=\frac s{1+s^2}+\frac{e^{-\pi s}s}{1+s^2}-\frac{e^{-\pi s}}{1+s^2}-\frac{e^{-2\pi s}}{1+s^2}-\frac{e^{-2\pi (s-1)}}{s-1}$$

A: Well yes, there are formulas for that, or just do it $\cdots$ 
$$\begin{aligned} \int_0^\pi \cos t\,  e^{-st} \, dt &= \int_0^\infty \cos t \, e^{-st} \, dt- \int_\pi^\infty \cos t \, e^{-st} \, dt \\&=\int_0^\infty \cos t \, e^{-st} \, dt- \int_0^\infty \cos (\pi + q) \, e^{-s(\pi+q)} \, dq \\ &=(1+e^{-s\pi})\mathcal{L}(\cos t)\\&=(1+e^{-s\pi})\frac{s}{s^2+1}
\end{aligned}$$
$$\begin{aligned} \int_\pi^{2\pi} \sin t\,  e^{-st} \, dt &= \int_\pi^\infty \sin t \, e^{-st} \, dt- \int_{2\pi}^\infty \sin t \, e^{-st} \, dt \\
&=\int_0^\infty \sin (\pi+q) \, e^{-s(\pi+q)} \, dt- \int_0^\infty \sin (2\pi + q) \, e^{-s(2\pi+q)} \, dq \\ 
&=-(e^{-s\pi}+e^{-2s\pi})\mathcal{L}(\sin t)\\&=-(e^{-s\pi}+e^{-2s\pi})\frac{1}{s^2+1}
\end{aligned}$$
$$\begin{aligned} \int_{2\pi}^\infty e^t\,  e^{-st} \, dt &= \int_{2\pi}^\infty e^{-(s-1)t} \,dt = -\int_{2\pi}^\infty \frac{de^{-(s-1)t}}{s-1} = \left.\frac{e^{-(s-1)t}}{s-1} \right|_{2\pi} = \frac{e^{-2\pi(s-1)}}{s-1}
\end{aligned}$$
Summing it up $\cdots$
$$\mathcal{L} \{f(t)\} =\frac{(1+e^{-s\pi})s -(e^{-s\pi}+e^{-2s\pi})}{s^2+1} +  \frac{e^{-2\pi(s-1)}}{s-1}
$$
