<s>[INST] Laplace transform of square root of a trigonometric function Need help with this question from my university paper.


My question : Find Laplace Transform of the following:   $\sqrt{1 + \sin(4t)}$

I do know how to solve $\sqrt{1 + \sin(t)}$ 

By taking $1 = \sin^2 \left(\frac{t}{2}\right) + \cos^2 \left(\frac{t}{2}\right)$ 

Further taking $\sin(t) = 2\sin \left(\frac{t}{2}\right) \cos \left(\frac{t}{2}\right)$

But I'm a bit confused about the above question.
Thanks in advance.
 [/INST] The function $\sqrt{1+\sin at}$ is periodic with period $T = 2\pi/a$ and the first zero at $3\pi/2a$. Following your steps, we can write this as
$$ \sqrt{1+\sin at} = \sqrt{\cos^{2}\frac{at}{2} + \sin^{2}\frac{at}{2} + 2\sin\frac{at}{2}\cos\frac{at}{2}} = \left|\cos\frac{at}{2}+\sin\frac{at}{2}\right|. $$
Using the Laplace transform of a periodic function, we split the integral into two pieces. This is necessary because of the absolute value sign. To get rid of it in the region $[3\pi/2a, 2\pi/a]$, we have to take the negative of the function.
$$\begin{aligned} I &= \int_{0}^{\infty}\sqrt{1+\sin at}\,e^{-st}\,\mathrm{d}t = \frac{1}{1-e^{-2\pi s/a}}\int_{0}^{2\pi/a}\left|\cos\frac{at}{2} + \sin\frac{at}{2}\right|e^{-st}\,\mathrm{d}t \\
 &= \frac{1}{1-e^{-2\pi s/a}}\left[\int_{0}^{3\pi/2a}\left(\cos\frac{at}{2} + \sin\frac{at}{2}\right)e^{-st}\,\mathrm{d}t - \int_{3\pi/2a}^{2\pi/a}\left(\cos\frac{at}{2} + \sin\frac{at}{2}\right)e^{-st}\,\mathrm{d}t \right]\end{aligned}$$
These integrals can be done by considering the integral
$$ \int_{0}^{3\pi/2a}e^{iat/2}e^{-st}\,\mathrm{d}t $$
and summing the real and imaginary contributions. One should then obtain
$$ \mathcal{L}[\sqrt{1+\sin at}] = \frac{1}{s^{2}+a^{2}/4}\left(s + \frac{a}{2} + \frac{\sqrt{2}\,ae^{-3\pi s/2a}}{1-e^{-2\pi s/a}}\right). $$
For $a = 4$, we have
$$ \boxed{\mathcal{L}[\sqrt{1+\sin 4t}] = \frac{1}{s^{2}+4}\left(s + 2 + \frac{4\sqrt{2}\,e^{-3\pi s/8}}{1-e^{-\pi s/2}}\right)}.$$
</s>