what is laplace transform of $\sqrt{(1+\sin t)}$

Question

i don't know how to start with this question as it is under root sign. in text book answer is given as $$\frac{s}{s^2+(1/2)^2}+\frac{1/2}{s^2+(1/2)^2}$$ so from answer we can figure out that it is $$\cos(t/2)+\sin(t/2)$$ but i don't know how to transform the question in above form.

Answer 1

It is the same function: this is evident when you square them:

$$1+sin(t) \ \ \text{is the same as} \ \ (cos(t/2)+\sin(t/2))^2$$

Indeed, when expanded, the latter becomes: $\cos(t/2)^2+\sin(t/2)^2+2\cos(t/2)\sin(t/2)$.

and you end by applying formula $2\cos(a)\sin(a)=\sin(2a).$ We know that 1+ sint = [cos(t/2)+sin(t/2)]^2 So root of 1+ sint =cos(t/2)+sin(t/2) Its that simple and now just apply Laplace

Answer 2

Sqrt 1+sint = sqrt.(sint/2 +cost/2)^2 Which gives .(sint/2 +cost/2). As square root and square cancel out. And Laplace of (sint/2 +cost/2) will give you the answer.