Answer to Question #126855 in Differential Equations for ibr

Question

Find ℒ{f(t)}

by first using a trigonometric identity. (Write your answer as a function of s.)
f(t) = sin(5t + 6)

Answer 1

$\mathcal{L} (sin(5t+6) ) = \int_0^{\infty} e^{-st} sin(5t+6) dt$

Let, $I = \int e^{-st} sin(5t+6) dt$

$= [ \frac{-1}{s} e^{-st} sin(5t+6)] - \int\frac{-5}{s} e^{-st} cos(5t+6)dt$

$= [ \frac{-1}{s} e^{-st} sin(5t+6)] - [\frac{-5}{s^2} e^{-st} cos(5t+6) + \int \frac{25}{s^2} e^{-st} sin(5t+6)dt]$

$I = [ \frac{-1}{s} e^{-st} sin(5t+6)] - [\frac{-5}{s^2} e^{-st} cos(5t+6)] - \frac{25}{s^2}I$

then solving it,

we get,

$I = \frac{e^{-st}}{s^2 + 25}[ -ssin(5t+6) - 5cos(5t+6) ]$

Now, putting the limits,

$\mathcal{L} (sin(5t+6) ) = \int_0^{\infty} e^{-st} sin(5t+6) dt = \frac{1}{s^2 + 25} [ ssin(6) + 5cos(6)]$