# Find the Laplace transform of $g(t) = 1+cos^2(2t)$ by direct integration

I'm having trouble finding out how to directly integrate the function $f(t)$ because of the $\cos^2(2t)$ term. I understand that $\cos^2(2t) = \frac{1}{2} + \frac{1}{2}\cos(4t)$ but I don't understand how this simplifies the problem so that

$\int_0^\infty {e^{-st}} + \int_0^\infty {e^{-st}(\frac{1}{2} + \frac{1}{2}\cos(4t))}$

Is an easier integral to solve.

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$\int e^{ax}\cos(bx)$ can integrated by parts or by looking at it as the real part of $\int e^{(a+ib)x}$. – Javier Dec 6 '12 at 2:33

You have that $\cos(4 t) = \Re e^{4 i t}$ so \begin{align} \int_0^\infty e^{-st} \cos (4 t) dt &= \Re \int_0^\infty e^{(-s+4i)t} dt = \Re\frac{e^{(-s+4i)t}}{-s+4i}\Bigg|_{t=0}^{t=\infty} = -\Re \frac{1}{-s+4i}\\ &= \Re \frac{s + 4i}{s^2 +16} = \frac{s}{s^2 +16} \end{align}

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