Range of the integral of a trigonometric function Given $$g(x) = \int_0^{x} \sqrt{1+\sin t} dt -\sqrt{1+\sin x}$$
I would like to show that $g(x) \geq 2\sqrt{2} -10,\,\forall x\in \left(0,\infty\right)$.
Any help is appreciated.

$\int_{0}^{x}\sqrt{1+\sin t}\,dt$ is clearly a non-negative, differentiable and increasing function on $\mathbb{R}^+$.
The minimum of $g(x)$ occurs at a zero of $g'(x)$, and since $g'(x)=\sqrt{1+\sin x}-\frac{\cos x}{2\sqrt{1+\sin x}}$, the minimum value of $g(x)$ is attained at some point such that $1+\sin x-\frac{1}{2}\cos x=0$, or
$$ \frac{2}{\sqrt{5}}\sin x-\frac{1}{\sqrt{5}}\cos x = \sin\left(x-\arcsin\frac{1}{\sqrt{5}}\right)=-\frac{2}{\sqrt{5}} $$
so the first interesting point is $x=\pi+\arcsin\frac{1}{\sqrt{5}}+\arcsin\frac{2}{\sqrt{5}}=\pi+\arctan\frac{1}{2}+\arctan 2=\frac{3\pi}{2}$, at which $\sqrt{1+\sin(x)}$ equals zero, $g(x)$ is positive and $g'(x)$ has a jump discontinuity. Not a minimum for sure. The second interesting point is 
$x=2\pi+\arcsin\frac{1}{\sqrt{5}}-\arcsin\frac{2}{\sqrt{5}}=\frac{3\pi}{2}+2\arctan\frac{1}{2}$ or $x=2\pi-\arctan\frac{3}{4}$, at which $\sqrt{1+\sin x}=\sqrt{\frac{2}{5}}$. We have $\int_{0}^{2\pi}\sqrt{1+\sin t}\,dt=4\sqrt{2}$ and
$$ \int_{0}^{\arctan\frac{3}{4}}\sqrt{1+\sin t}\,dt = \int_{0}^{2\arctan\frac{1}{3}}\sqrt{1+\sin t}\,dt = 2\int_{0}^{\arctan\frac{1}{3}}\sqrt{1+\sin(2z)}\,dz $$
equals:
$$ 2\int_{0}^{\frac{1}{3}}\sqrt{1+\frac{2t}{1+t^2}}\,dt=2\int_{0}^{1/3}\frac{1+t}{\sqrt{1+t^2}}\,dt =2-2\sqrt{\frac{2}{5}}$$
and the given function is positive at $x=2\pi-2\arctan\frac{1}{3}$, too. The other interesting points provide larger values, hence the minimum of the given function is attained at $x=0$ and trivially equals $-1$:
$$ \forall x\geq 0,\qquad g(x)=\int_{0}^{x}\sqrt{1+\sin t}\,dt-\sqrt{1+\sin x}\geq -1.$$
Additionally, $g(x)$ roughly behaves like $-1+\frac{2\sqrt{2}}{\pi}x$ on $\mathbb{R}^+$:

