Solve L(sin4t) ?

Apr 12, 2018

Guessing this is a Laplace transform request

Explanation:

$m a t h \boldsymbol{L} \left\{\cos \omega t \textcolor{red}{-} i \sin \omega t\right\} = m a t h \boldsymbol{L} \left\{{e}^{- i \omega t}\right\}$

$= {\int}_{0}^{\infty} {e}^{- \left(i \omega + s\right) t} \setminus \mathrm{dt}$

$= - \frac{1}{i \omega + s} {\left[{e}^{- \left(i \omega + s\right) t}\right]}_{0}^{\infty}$

$= \frac{1}{i \omega + s} = \frac{s - i \omega}{{s}^{2} + {\omega}^{2}}$

$\implies m a t h \boldsymbol{L} \left\{\cos \omega t\right\} = \setminus m a t h c a l \left(R e\right) \left(m a t h \boldsymbol{L} \left\{\cos \omega t \textcolor{red}{-} i \sin \omega t\right\}\right) = \frac{s}{{s}^{2} + {\omega}^{2}}$

AND

$\implies m a t h \boldsymbol{L} \left\{\sin \omega t\right\} = - \setminus m a t h c a l \left(I m\right) \left(m a t h \boldsymbol{L} \left\{\cos \omega t \textcolor{red}{-} i \sin \omega t\right\}\right) = \frac{\omega}{{s}^{2} + {\omega}^{2}}$

Repeat for: $\omega = 4$