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.
 A: 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
$$ \boxed{\mathcal{L}[\sqrt{1+\sin 4t}] = \frac{1}{s^{2}+4}\left(s + 2 + \frac{4\sqrt{2}\,e^{-3\pi s/8}}{1-e^{-\pi s/2}}\right)}.$$
A: With CAS help and for general $a$ and $a\neq 0$ we have:
$\mathcal{L}_t\left[\sqrt{1+\sin (a t)}\right](s)=\frac{e^{-\frac{\pi  s}{2 a}} \left(-a \left(\sqrt{2}-2 \cosh \left(\frac{\pi  s}{2
   a}\right)\right)+\frac{\left(a^2+4 s^2\right) \cosh \left(\frac{\pi  s}{2 a}\right) \, _3F_2\left(-\frac{1}{4},\frac{1}{4},1;1-\frac{i s}{2
   a},1+\frac{i s}{2 a};1\right)}{s}\right) \left(1+\tanh \left(\frac{\pi  s}{2 a}\right)\right)}{a^2+4 s^2}$
Mathematica code:
HoldForm[LaplaceTransform[Sqrt[1 + Sin[a t]], t, s] == (E^(-((\[Pi] s)/(
2 a))) (-a (Sqrt[2] - 2 Cosh[(\[Pi] s)/(2 a)]) + ((a^2 + 
4 s^2) Cosh[(\[Pi] s)/(
2 a)] HypergeometricPFQ[{-(1/4), 1/4, 1}, {1 - (I s)/(2 a), 
1 + (I s)/(2 a)}, 1])/s) (1 + Tanh[(\[Pi] s)/(2 a)]))/(
a^2 + 4 s^2)] // TeXForm

For $a=4$:
(E^(-((\[Pi] s)/8)) (-s (Sqrt[2] - 2 Cosh[(\[Pi] s)/8]) + (4 + s^2) Cosh[(\[Pi] s)/
8] HypergeometricPFQ[{-(1/4), 1/4, 1}, {1 - (I s)/8, 
1 + (I s)/8}, 1]) (1 + Tanh[(\[Pi] s)/8]))/(s (4 + s^2))


$\mathcal{L}_t\left[\sqrt{1+\sin (4 t)}\right](s)=\frac{e^{-\frac{1}{8} (\pi  s)} \left(-s \left(\sqrt{2}-2 \cosh \left(\frac{\pi 
   s}{8}\right)\right)+\left(4+s^2\right) \cosh \left(\frac{\pi  s}{8}\right) \, _3F_2\left(-\frac{1}{4},\frac{1}{4},1;1-\frac{i s}{8},1+\frac{i
   s}{8};1\right)\right) \left(1+\tanh \left(\frac{\pi  s}{8}\right)\right)}{s \left(4+s^2\right)}$

