I need to find the Laplace transform of sine(t), and it is proving rather difficult. I integrate by parts, then integrate by parts again so that the original integral is on both sides of the equation, then I get it wrong.
$ f\left(x\right) = \left\{ \begin{array}{lr} \sin t & : 0 \leq t \le \pi\\ 0 & : t > 1 \end{array} \right.\\$
Because the Laplace Transform of 0 is 0, I concern myself with: $$\int{e^{-st}\sin{t}\space \mathrm{dt}}$$ and proceed with integration by parts. I have $u=\sin{t}$,$du=\cos{t}$,$dv=e^{-st}$ and last but not least $dv=\frac{e^{-st}}{s}$. This yields the equation:
$$\int{e^{-st}\sin{t}\space \mathrm{dt}}=\frac{1}{s}\left(-\sin{t}e^{-st}+\int{e^{-st}\cos{t}\space \mathrm{dt}}\right)$$ and so I begin to solve the Laplace transform of cosine.
For this, I use $u=\cos{t}$,$du=-\sin{t}$,$dv=e^{-st}$,$v=\frac{e^{-st}}{s}$. This yields the equation: $$\int{e^{-st}\cos{t}\space \mathrm{dt}}=\frac{-\cos{t}e^{-st}}{s}-\frac{1}{s}\int{e^{-st}\sin{t}\space \mathrm{dt}}$$.
Now, we obviously use this solution to replace the prior integral that we just solved, $$\int{e^{-st}\sin{t}\space \mathrm{dt}}=-\frac{\sin{t}e^{-st}}{s}+\frac{1}{s}\left( \frac{-\cos{t}e^{-st}}{s}-\frac{1}{s}\int{e^{-st}\sin{t}\space \mathrm{dt}})\right)$$
Which simplifies to: $$(1+s^2)\int{e^{-st}\sin{t}\space \mathrm{dt}}=\frac{-\sin{t}e^{-st}}{s}+\frac{-\cos{t}e^{-st}}{s^2}$$.
Now let us suppose the bounds of our equation are $0$ and $\pi$. This leads our terms to be as such: $$\left[\frac{-\cos{t}e^{-st}}{s^2}\right]^{\pi}_0=\frac{e^{-\pi s}}{s^2+1}\frac{1}{s^2}$$.
The book gives the answer as merely $$\frac{e^{-\pi s}}{s^2+1}$$. How do I get rid of my extraneous $\frac{1}{s^2}$?