What is the Laplace transform of $\cos(4t+8)$?

Question

Could someone please explain how to transform this to the Laplace domain? I've tried to use the definition of Laplace (not sure this is the easiest way).

$$\int_{0}^{t}e^{-st}f(t)\,dt$$

$$\int_{0}^{t}e^{-st} \cos(4t+8)\,dt$$

But got stuck in a loop because, the integral of cosine is sine and the integral if sine is cosine. More or less same store for the $e^{-st}$...

I have a table with the standard laplace transformations. The closest I have is $$\mathcal{L}\{\cos(a)\} = \frac{s}{s^2+a^2}$$

Answer 1

You can also use the addition formula $\cos(x+y)=\cos x\cos y-\sin x\sin y$, so \begin{align} \mathscr{L}\{\cos(4t+8)\}&=\mathscr{L}\{\cos 8\cdot\cos(4t)-\sin 8\cdot\sin (4t)\}\\ &=(\cos 8)\mathscr{L}\{\cos(4t)\}-(\sin 8)\mathscr{L}\{\sin (4t)\}\\ &=(\cos 8)\left(\frac{s}{s^2+16}\right)-(\sin 8)\left(\frac{4}{s^2+16}\right)\\ &=\frac{s\cos 8-4\sin 8}{s^2+16} \end{align}

Answer 2

$$\int e^{-st}\cos(4t+8) dt = \frac{e^{-st}(-s\cos(4t+8))+4 \sin(4t+8))}{16+s^2}$$

This comes from integration by parts (twice).

Now evaluate from $t=0$ to $t=\infty$.

Answer 3

An alternative to the partial integration twice is the substitution using the Euler formula $$ \cos x=\frac{e^{ix}+e^{-ix}}{2} $$ where you get simple exponential integrands.

Answer 4

$$\int_{0}^{\infty}e^{-st} \cos(4t+8)\,dt = \Re\int_{0}^{\infty}e^{-st} e^{i(4t+8)}\,dt = \Re\, \dfrac{e^{8i}}{4i-s}\left.e^{(4i-s)t}\right|_0^\infty = \Re \dfrac{e^{8i}}{s-4i}=\Re\dfrac{(\cos 8 + i\sin 8)(s+4i)}{s^2+16} = \dfrac{s\cos8 - 4\sin 8}{s^2+16}$$