Question

Laplace question help!
http://prntscr.com/n6ymqm

Answer 1

I'm not super familiar with these but perhaps @sillybilly123 may be able to help

Answer 2

Can't OTOH see a trick so this is brute force from the definiton:

\( \mathcal L \{ cos ( 2t + \pi) \cdot u ( t - \frac{\pi}{2}) \}\)

\(= \mathcal R e ~ ~ ~ \mathcal{ L} \{ e^{ j ( 2t + \pi)} \cdot u ( t - \frac{\pi}{2}) \}\)

\(= \mathcal R e ~ ~ ~ \int\limits_{\pi/2}^{\infty} ~ dt ~~~ e^{ j ( 2t + \pi)} e^{- st }\)

\(= \mathcal R e ~ ~ ~ \int\limits_{\pi/2}^{\infty} ~ dt ~~~ e^{ (2j - s)t } \cdot e^{ j \pi}\)

\(= \mathcal R e ~ ~ ~ - \left[ \dfrac{ e^{ (2j - s)t}}{2 j - s } \right]_{\pi/2}^{\infty} \qquad \because ~ e^{j \pi} = -1 \)

\(= \mathcal R e ~ ~ ~ \left[ (\dfrac{s}{s^2 + 4} + \dfrac{2 j}{s^2 + 4}) e^{- st} ( \cos 2 t + j \sin 2t) \right]_{\pi/2}^{\infty} \)

\(= \left[ \dfrac{s \cos 2t - 2 \sin 2t}{s^2 + 4} e^{- st} \right]_{\pi/2}^{\infty} \)

\(= 0 - \dfrac{s \cos \pi - 2 \sin \pi}{s^2 + 4} e^{- \frac{\pi}{2}s} \)

\(= \dfrac{s e^{- \frac{\pi}{2}s}}{s^2 + 4} \)

You can do 2 rounds of IBP instead of using Complex approach. Either way is super tedious

Answer 3

Lol
sorry 4 the late reply,
Thanks for your help @sillybilly123
I managed to solve the answer using this method:

\(cos(t+\frac{\pi}{2}-\pi)=-cos(t-\frac{\pi}{2})\)