As $L(e^{at})=\frac1{s-a}$
So putting $a=0,L(1)=\frac1s$
and putting $a=c+id,L(e^{(c+id)t})=\frac1{s-(c+id)}$
so, $L(e^{ct}\cos dt)+iL(e^{ct}\sin dt)=\frac{s-c+id}{(s-c)^2+d^2}$
$cos^2t=\frac{1+\cos2t}2$
So, $L(cos^2t)=\frac12 L(1)+\frac12 L(\cos2t)=\frac1{2s}+\frac s{2(s^2+2^2)}$
$\cos3t\sin3t=\frac{\sin6t}2$
So, $L(\cos3t\sin3t)=\frac{L(\sin6t)}2=\frac{6}{2(s^2+6^2)}$
We know, $L(e^{at}f(t))=F(s-a)$ where $F(s)=L(f(t))$
So, $L(e^{at} t^n)=\frac{n!}{(s-a)^{n+1}}$
Hence $L(te^t)=\frac1{(s-a)^2}$ putting $a=n=1$
Putting $a=2i,n=1; L(e^{2it} t)=\frac1{(s-2i)^2}$
$L(t\cos 2t)+iL(t\sin 2t)=\frac{(s+2i)^2}{(s^2+4)^2}=\frac{s^2-4+i4s}{(s^2+4)^2}$
Now compare the real and the imaginary parts.
Note that you should make a distinction between the functions ${f_1}(t) = \sin (\omega t + \theta )U(t)$ and ${f_2}(t) = \sin (\omega t + \theta )U(t - {t_0})$, specifically, the shift rules you mention apply to the totally shifted function (${f_2}(t)$) not ${f_1}(t)$. The most straight forward approach that I recommend to you for calculating the Laplace transform of a function is "Direct Integration", as you have noted. For example $$\begin{array}{l}\mathfrak{L}\{ \sin (\omega t + \theta )\} = \int\limits_0^{ + \infty } {\sin (\omega t + \theta ){e^{ - st}}dt} = \int\limits_0^{ + \infty } {\frac{{{e^{i(\omega t + \theta )}} - {e^{ - i(\omega t + \theta )}}}}{{2i}}{e^{ - st}}dt} = \\\frac{{{e^{i\theta }}}}{{2i}}\int\limits_0^{ + \infty } {{e^{(i\omega - s)t}}dt} - \frac{{{e^{ - i\theta }}}}{{2i}}\int\limits_0^{ + \infty } {{e^{ - (i\omega + s)t}}dt} = - \frac{1}{{2i}}\left( {\frac{{{e^{i\theta }}}}{{i\omega - s}} + \frac{{{e^{ - i\theta }}}}{{i\omega + s}}} \right) = \\\frac{1}{{2i}}\left( {\frac{{{e^{i\theta }}(i\omega + s) + {e^{ - i\theta }}(i\omega - s)}}{{{\omega ^2} + {s^2}}}} \right) = \frac{\omega }{{{\omega ^2} + {s^2}}}\left( {\frac{{{e^{i\theta }} + {e^{ - i\theta }}}}{2}} \right) + \frac{s}{{{\omega ^2} + {s^2}}}\left( {\frac{{{e^{i\theta }} - {e^{ - i\theta }}}}{{2i}}} \right) = \\\frac{\omega }{{{\omega ^2} + {s^2}}}\cos \theta + \frac{s}{{{\omega ^2} + {s^2}}}\sin \theta \end{array}$$Hope it helps ;)
Best Answer
$\sin(5t) \cos(5t) = \sin(10t)/2$ You can take the transform of the above.
There is no general straight forward rule to finding the Laplace transform of a product of two functions. The best strategy is to keep the general Laplace Transforms close at hand and try to convert a given function to a linear combination of those forms.
Methods like partial fractions, writing sine, cosine as exponents .. etc, help.