<s>[INST] 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.
 [/INST] Simply
$\int \sqrt{1+\sin t} \, dt=\int \dfrac{\sqrt{1+\sin t}\sqrt{1-\sin t}}{\sqrt{1-\sin t}}\,dt=\int \dfrac{\cos t}{\sqrt{1-\sin t}}\,dt$
Substitute $\sin t=u \to \cos t\, dt = du$
$\int \dfrac{du}{\sqrt{1-u}}=-2 \sqrt{1-u}+C=-2\sqrt{1-\sin t}+C$
thus $\int_0^x \sqrt{1+\sin t} \, dt=-2 \sqrt{1-\sin x}+2$
therefore $g(x)=-2 \sqrt{1-\sin x}+2-\sqrt{\sin x+1}$
$g'(x)=-\frac{1}{2} \left(\sqrt{1-\sin x}-2 \sqrt{\sin x+1}\right)$
and we have $g'(x)=0$ when
$\sqrt{1-\sin x}-2 \sqrt{\sin x+1}=0$
that is $\sqrt{1-\sin x}=2 \sqrt{\sin x+1}=0$
and squaring both sides
$1-\sin x=4(\sin x +1)\to \sin x =-\dfrac{3}{5}$
$g''(x)=\frac{1}{4} \left(2 \sqrt{1-\sin (x)}+\sqrt{1+\sin (x)}\right)$
and $g''\left(-\arcsin \dfrac{3}{5}\right)=\sqrt{\dfrac{2}{5}}+\dfrac{1}{2 \sqrt{10}}>0$ so at $\sin x =-\dfrac{3}{5}$ the function $g(x)$ has a minimum
$g\left(-\arcsin \dfrac{3}{5}\right)=2-\sqrt{10}$
and in conclusion we can say that
$g(x)\geq 2-\sqrt{10}$ for any $x>0$
as $ 2\sqrt{2} -10<2-\sqrt{10}$ we can also say that
$g(x) \geq 2\sqrt{2} -10$
even if I suspect that there is a typo somewhere
Hope this helps
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