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

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 $$\mathcal{L}\{\cos(a)\} = \frac{s}{s^2+a^2}$$

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}

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.

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

This comes from integration by parts (twice).

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

• A good answer. If you do the second integration by parts with the wrong choice of factors, you get back to the original function and get no information. If that happens, try the opposite choice for the second integration by parts. – GEdgar Dec 28 '15 at 18:52

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