$\displaystyle Consider\\[2ex] \sqrt {1+\sin t}=\sqrt{{\ \left[\cos{\left(\frac{t}{2}\right)}+\ \sin\left(\frac{t}{2}\right)\right]}^2} \\[2ex] \displaystyle \sqrt{1+\sin t}=\cos{\left(\frac{t}{2}\right)}+\sin\left(\frac{t}{2}\right) \\[2ex] $

$\displaystyle \therefore{}L\left[\sqrt{1+ \sin t} \ \right ]\\[2ex] =L\left[\cos{\left(\frac{t}{2}\right)}+\sin\left(\frac{t}{2}\right)\right] \\[2ex] \displaystyle =\frac{s}{s^2+\ {\left(\frac{1}{2}\right)}^2}+\frac{\frac{1}{2}}{s^2+\ {\left(\frac{1}{2}\right)}^2} \\[2ex] \displaystyle =\frac{4s}{{4s}^2+\ 1}+\frac{2}{{4s}^2+\ 1} \\[2ex] \displaystyle =\frac{4s+2}{{4s}^2+\ 1} \\[2ex] $

$\displaystyle \therefore{}\ L\left[t\times{}\sqrt{1+sint}\right]\\[3ex] \displaystyle =\left(-1\right)\frac{d}{ds}\left(\frac{\left(4s+2\right)}{\left({4s}^2+\ 1\right)}\right) \\[2ex] \displaystyle =(-1)\times{}\frac{\left({4s}^2+\ 1\right)\times{}4-\left(4s+2\right)(4\times{}2s)}{{\left({4s}^2+\ 1\right)}^2} \\[2ex] \displaystyle =\frac{-(16s^2+4-32s^2-16s}{{\left({4s}^2+\ 1\right)}^2} \\[2ex] \displaystyle =\frac{-(-16s^2+4-16s)}{{\left({4s}^2+\ 1\right)}^2} \\[2ex] \displaystyle =\frac{(4s^2+4s-1)}{{\left({4s}^2+\ 1\right)}^2} \\[2ex]$