<s>[INST] Area bounded by circle $x^2+y^2=4$, $y=x^2+x+1$ and the curve $ \left\lfloor \sin^2\frac{x}{4}+\cos\frac{x}{4} \right\rfloor $ and the $x$ axis is Prove that the area bounded by circle $x^2+y^2=4$,the parabola $y=x^2+x+1$ and the curve $ \left\lfloor \sin^2\frac{x}{4}+\cos\frac{x}{4} \right\rfloor $ and the $x$ axis is $\sqrt3+\frac{2\pi}{3}-\frac{1}{6}.$

In this question,$x^2+y^2=4$ is a circle centered at $(0,0)$ and radius is $2$ and parabola $y=x^2+x+1$ faces upwards whose vertex is $(\frac{-1}{2},\frac{3}{4})$.
$I=2\int_{0}^{1}\sqrt{4-y^2}dy-\int_{-1}^{0}1-(x^2+x+1)dx=\sqrt3+\frac{2\pi}{3}-\frac{1}{6}$
Thanks i now understood after help from RobertZ.
 [/INST] Hint. Inside the circle we have that $x\in [-2,2]$ which implies that $\cos(x/4)\in  [0,1]$. Therefore 
$$\sin^2\frac{x}{4}+\cos\frac{x}{4}=1+\cos\frac{x}{4}(1-\cos\frac{x}{4})\in [1,1+\frac{1}{4}]$$
Hence $\left\lfloor \sin^2\frac{x}{4}+\cos\frac{x}{4} \right\rfloor=1$ for $x\in [-2,2]$, which means that inside the circle, that curve is simply an horizontal segment.
Now make a drawing and try to find the area as an integral. 
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