Now first by trigonometry property we will adjust the $\sqrt{1 + sint}$

Now $sint = 2sin\frac{t}{2}cos\frac{t}{2}$

And 1 can be written as $sin^2\frac{t}{2} + cos^2\frac{t}{2}$

$\sqrt{1 + sint}$ will become

$\sqrt{sin^2\frac{t}{2} + cos^2\frac{t}{2} +2sin\frac{t}{2}cos\frac{t}{2} }$

Which can be of the form $(a + b)^2$

$\therefore \sqrt{\bigg(sin\frac{t}{2} + cos\frac{t}{2}\bigg)}$

$\therefore$ The given question will now become $t\bigg(sin\frac{t}{2} + cos\frac{t}{2}\bigg)$

Now $L\bigg[sin\frac{t}{2} + cos\frac{t}{2}\bigg] = L\bigg[sin\frac{t}{2}\bigg] + L\bigg[cos\frac{t}{2}\bigg] = \frac{\frac{1}{2}}{s^2 + \frac{1}{4}} + \frac{s}{s^2 + \frac{1}{4}}$

$L\bigg[t\bigg(sin\frac{t}{2} + cos\frac{t}{2}\bigg)\bigg] = -1\frac{d}{ds}\bigg[\frac{\frac{1}{2} + s}{s^2 + \frac{1}{4}}\bigg]$ By multiplication by t property

$L[t\sqrt{1 + sint}] = -1\frac{d}{ds}\bigg[\frac{\frac{1}{2} + s}{s^2 + \frac{1}{4}}\bigg]$

$\therefore L[t\sqrt{1 + sint}] = -1\bigg[\frac{\bigg(s^2 + \frac{1}{4}\bigg)\frac{d}{ds}\bigg(s + \frac{1}{2}\bigg) - \bigg(s + \frac{1}{2}\bigg)\frac{d}{ds}(s^2 + \frac{1}{4})}{(s^2 + \frac{1}{4})^2}\bigg]$

$\therefore L[t\sqrt{1 + sint}] = -1\bigg[\frac{\bigg(s^2 + \frac{1}{4}\bigg) - \bigg(s + \frac{1}{2}\bigg)2s}{\bigg(s^2 + \frac{1}{4}\bigg)^2}\bigg]$

$\therefore L[t\sqrt{1 + sint}] = -1\frac{\bigg[s^2 + \frac{1}{4} - 2s^2 - s\bigg]}{\bigg(s^2 + \frac{1}{4}\bigg)^2}$

$\therefore L[t\sqrt{1 + sint}] = \frac{s^2 + s - 1/4}{\bigg(s^2 + \frac{1}{4}\bigg)^2}$

$\therefore L[t\sqrt{1 + sint}] = \frac{4[4s^2 + 4s - 1]}{(4s^2 + 1)^2}$